SECTION D - QUANTITATIVE APTITUDE - CHAPTER 14

(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:8)(cid:9)(cid:8)(cid:10)(cid:11) (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9) (cid:10)(cid:6)(cid:11)(cid:1)(cid:5)(cid:6)(cid:12)(cid:13)(cid:1)(cid:6)(cid:4)(cid:14)(cid:11) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) LLLLLEEEEEAAAAARRRRRNNNNNIIIIINNNNNGGGGG OOOOOBBBBBJJJJJEEEEECCCCCTTTTTIIIIIVVVVVEEEEESSSSS The Students will be introduced in this chapter to the techniques of developing discrete and continuous probability distributions and its applications. 1111144444.....11111 IIIIINNNNNTTTTTRRRRROOOOODDDDDUUUUUCCCCCTTTTTIIIIIOOOOONNNNN In chapter ten, it may be recalled, we discussed frequency distribution. In a similar manner, we may think of a probability distribution where just like distributing the total frequency to different class intervals, the total probability (i.e. one) is distributed to different mass points in case of a discrete random variable or to different class intervals in case of a continuous random variable. Such a probability distribution is known as Theoretical Probability Distribution, since such a distribution exists only in theory. We need study theoretical probability distribution for the following important factors: (a) An observed frequency distribution, in many a case, may be regarded as a sample i.e. a representative part of a large, unknown, boundless universe or population and we may be interested to know the form of such a distribution. By fitting a theoretical probability distribution to an observed frequency distribution of, say, the lamps produced by a manufacturer, it may be possible for the manufacturer to specify the length of life of the lamps produced by him up to a reasonable degree of accuracy. By studying the effect of a particular type of missiles, it may be possible for our scientist to suggest the number of such missiles necessary to destroy an army position. By knowing the distribution of smokers, a social activist may warn the people of a locality about the nuisance of active and passive smoking and so on. (b) Theoretical probability distribution may be profitably employed to make short term projections for the future. (c) Statistical analysis is possible only on the basis of theoretical probability distribution. Setting confidence limits or testing statistical hypothesis about population parameter(s) is based on the probability distribution of the population under consideration. A probability distribution also possesses all the characteristics of an observed distribution. We (μ ) define population mean (µ), population median , population mode (µ ), population standard 0 0 deviation (σ) etc. exactly same way we have done earlier. These characteristics are known as population parameters. Again a probability distribution may be either a discrete probability distribution or a Continuous probability distribution depending on the random variable under study. TTTTTwwwwwooooo iiiiimmmmmpppppooooorrrrrtttttaaaaannnnnttttt dddddiiiiissssscccccrrrrreeeeettttteeeee ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaarrrrreeeee (((((aaaaa))))) BBBBBiiiiinnnnnooooommmmmiiiiiaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaannnnnddddd (((((bbbbb))))) PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn..... SSSSSooooommmmmeeeee iiiiimmmmmpppppooooorrrrrtttttaaaaannnnnttttt cccccooooonnnnntttttiiiiinnnnnuuuuuooooouuuuusssss ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnnsssss aaaaarrrrreeeee (((((aaaaa))))) NNNNNooooorrrrrmmmmmaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn (((((bbbbb))))) CCCCChhhhhiiiii-----sssssqqqqquuuuuaaaaarrrrreeeee DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn (((((ccccc))))) SSSSStttttuuuuudddddeeeeennnnntttttsssss-----DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn (((((ddddd))))) FFFFF-----DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn (cid:10)(cid:11)(cid:19)(cid:20) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 1111144444.....22222 BBBBBIIIIINNNNNOOOOOMMMMMIIIIIAAAAALLLLL DDDDDIIIIISSSSSTTTTTRRRRRIIIIIBBBBBUUUUUTTTTTIIIIIOOOOONNNNN One of the most important and frequently used discrete probability distribution is Binomial Distribution. It is derived from a particular type of random experiment known as Bernoulli process after the famous mathematician Bernoulli. Noting that a 'trial' is an attempt to produce a particular outcome which is neither certain nor impossible, the characteristics of Bernoulli trials are stated below: (i) Each trial is associated with two mutually exclusive and exhaustive outcomes, the occurrence of one of which is known as a 'success' and as such its non occurrence as a 'failure'. As an example, when a coin is tossed, usually occurrence of a head is known as a success and its non–occurrence i.e. occurrence of a tail is known as a failure. (ii) The trials are independent. (iii) The probability of a success, usually denoted by p, and hence that of a failure, usually denoted by q = 1–p, remain unchanged throughout the process. (iv) The number of trials is a finite, positive integer. A discrete random variable x is defined to follow binomial distribution with parameters n and p, to be denoted by x ~ B (n, p), if the probability mass function of x is given by f (x) = p (X = x) = nc pxqn-x for x = 0, 1, 2, …., n x = 0, otherwise ……… (14.1) WWWWWeeeee mmmmmaaaaayyyyy nnnnnooooottttteeeee ttttthhhhheeeee fffffooooollllllllllooooowwwwwiiiiinnnnnggggg iiiiimmmmmpppppooooorrrrrtttttaaaaannnnnttttt pppppoooooiiiiinnnnntttttsssss iiiiinnnnn cccccooooonnnnnnnnnneeeeeccccctttttiiiiiooooonnnnn wwwwwiiiiittttthhhhh bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn::::: (((((aaaaa))))) AAAAAsssss nnnnn >>>>>00000,,,,, ppppp,,,,, qqqqq ≥≥≥≥≥ 00000,,,,, iiiiittttt fffffooooollllllllllooooowwwwwsssss ttttthhhhhaaaaattttt fffff(((((xxxxx))))) ≥≥≥≥≥ 00000 fffffooooorrrrr eeeeevvvvveeeeerrrrryyyyy xxxxx ∑ AAAAAlllllsssssooooo fffff(((((xxxxx))))) ===== fffff(((((00000))))) +++++ fffff(((((11111))))) +++++ fffff(((((22222))))) +++++ ……………..........+++++ fffff(((((nnnnn))))) ===== 11111………………………………………(((((1111144444.....22222))))) x (((((bbbbb))))) BBBBBiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss kkkkknnnnnooooowwwwwnnnnn aaaaasssss bbbbbiiiiipppppaaaaarrrrraaaaammmmmeeeeetttttrrrrriiiiiccccc dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaasssss iiiiittttt iiiiisssss ccccchhhhhaaaaarrrrraaaaacccccttttteeeeerrrrriiiiissssseeeeeddddd bbbbbyyyyy tttttwwwwwooooo pppppaaaaarrrrraaaaammmmmeeeeettttteeeeerrrrrsssss nnnnn aaaaannnnnddddd ppppp..... TTTTThhhhhiiiiisssss mmmmmeeeeeaaaaannnnnsssss ttttthhhhhaaaaattttt iiiiifffff ttttthhhhheeeee vvvvvaaaaallllluuuuueeeeesssss ooooofffff nnnnn aaaaannnnnddddd ppppp aaaaarrrrreeeee kkkkknnnnnooooowwwwwnnnnn,,,,, ttttthhhhheeeeennnnn ttttthhhhheeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss kkkkknnnnnooooowwwwwnnnnn cccccooooommmmmpppppllllleeeeettttteeeeelllllyyyyy..... (((((ccccc))))) TTTTThhhhheeeee mmmmmeeeeeaaaaannnnn ooooofffff ttttthhhhheeeee bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss gggggiiiiivvvvveeeeennnnn bbbbbyyyyy µ ===== nnnnnppppp ……………..... (((((1111144444.....33333))))) (((((ddddd))))) DDDDDeeeeepppppeeeeennnnndddddiiiiinnnnnggggg ooooonnnnn ttttthhhhheeeee vvvvvaaaaallllluuuuueeeeesssss ooooofffff ttttthhhhheeeee tttttwwwwwooooo pppppaaaaarrrrraaaaammmmmeeeeettttteeeeerrrrrsssss,,,,, bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn mmmmmaaaaayyyyy bbbbbeeeee uuuuunnnnniiiiimmmmmooooodddddaaaaalllll ooooorrrrr bbbbbiiiii----- mmmmmooooodddddaaaaalllll..... µ ,,,,, ttttthhhhheeeee mmmmmooooodddddeeeee ooooofffff bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn,,,,, iiiiisssss gggggiiiiivvvvveeeeennnnn bbbbbyyyyy µ ===== ttttthhhhheeeee lllllaaaaarrrrrgggggeeeeesssssttttt 0 0 iiiiinnnnnttttteeeeegggggeeeeerrrrr cccccooooonnnnntttttaaaaaiiiiinnnnneeeeeddddd iiiiinnnnn (((((nnnnn+++++11111)))))ppppp iiiiifffff (((((nnnnn+++++11111)))))ppppp iiiiisssss aaaaa nnnnnooooonnnnn-----iiiiinnnnnttttteeeeegggggeeeeerrrrr ===== (((((nnnnn+++++11111)))))ppppp aaaaannnnnddddd (((((nnnnn+++++11111)))))ppppp ----- 11111 iiiiifffff (((((nnnnn+++++11111)))))ppppp iiiiisssss aaaaannnnn iiiiinnnnnttttteeeeegggggeeeeerrrrr …………….....(((((1111144444.....44444))))) (((((eeeee))))) TTTTThhhhheeeee vvvvvaaaaarrrrriiiiiaaaaannnnnccccceeeee ooooofffff ttttthhhhheeeee bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss gggggiiiiivvvvveeeeennnnn bbbbbyyyyy σ2 ===== nnnnnpppppqqqqq ………………………………………..... (((((1111144444.....55555))))) SSSSSiiiiinnnnnccccceeeee ppppp aaaaannnnnddddd qqqqq aaaaarrrrreeeee nnnnnuuuuummmmmeeeeerrrrriiiiicccccaaaaallllllllllyyyyy llllleeeeessssssssss ttttthhhhhaaaaannnnn ooooorrrrr eeeeeqqqqquuuuuaaaaalllll tttttooooo 11111,,,,, nnnnnpppppqqqqq <<<<< nnnnnppppp ⇒vvvvvaaaaarrrrriiiiiaaaaannnnnccccceeeee ooooofffff aaaaa bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll vvvvvaaaaarrrrriiiiiaaaaabbbbbllllleeeee iiiiisssss aaaaalllllwwwwwaaaaayyyyysssss llllleeeeessssssssss ttttthhhhhaaaaannnnn iiiiitttttsssss mmmmmeeeeeaaaaannnnn..... AAAAAlllllsssssooooo vvvvvaaaaarrrrriiiiiaaaaannnnnccccceeeee ooooofffff XXXXX aaaaattttttttttaaaaaiiiiinnnnnsssss iiiiitttttsssss mmmmmaaaaaxxxxxiiiiimmmmmuuuuummmmm vvvvvaaaaallllluuuuueeeee aaaaattttt ppppp ===== qqqqq ===== 00000.....55555 aaaaannnnnddddd ttttthhhhhiiiiisssss mmmmmaaaaaxxxxxiiiiimmmmmuuuuummmmm vvvvvaaaaallllluuuuueeeee iiiiisssss nnnnn/////44444..... (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:21) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) (((((fffff))))) AAAAAddddddddddiiiiitttttiiiiivvvvveeeee ppppprrrrrooooopppppeeeeerrrrrtttttyyyyy ooooofffff bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn..... IIIIIfffff XXXXX aaaaannnnnddddd yyyyy aaaaarrrrreeeee tttttwwwwwooooo iiiiinnnnndddddeeeeepppppeeeeennnnndddddeeeeennnnnttttt vvvvvaaaaarrrrriiiiiaaaaabbbbbllllleeeeesssss sssssuuuuuccccchhhhh ttttthhhhhaaaaattttt XXXXX ∼β (((((nnnnn,,,,, PPPPP))))) 11111 aaaaannnnnddddd yyyyy ∼β (((((nnnnn,,,,, PPPPP))))) 22222 TTTTThhhhheeeeennnnn (((((XXXXX+++++yyyyy))))) ∼β (((((nnnnn +++++ nnnnn +++++,,,,, PPPPP))))) ………………………………………………………………………………………………………………………………………………….......... (((((1111144444.....66666))))) 11111 22222 AAAAAppppppppppllllliiiiicccccaaaaatttttiiiiiooooonnnnnsssss ooooofffff BBBBBiiiiinnnnnooooommmmmiiiiiaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn Binomial distribution is applicable when the trials are independent and each trial has just two outcomes success and failure. It is applied in coin tossing experiments, sampling inspection plan, genetic experiments and so on. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....11111::::: A coin is tossed 8 times. Assuming the coin to be unbiased, what is the probability of getting? (i) 4 heads (ii) at least 4 heads (iii) at most 3 heads SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We apply binomial distribution as the tossing are independent of each other. With every tossing, there are just two outcomes either a head, which we call a success or a tail, which we call a failure and the probability of a success (or failure) remains constant throughout. Let X denotes the no. of heads. Then X follows binomial distribution with parameter n = 8 and p = 1/2 (since the coin is unbiased). Hence q = 1 – p = 1/2 The probability mass function of X is given by f(x) = nc px qn-x x = 10c . (1/2)x . (1/2)10-x x 10 c x = 10 2 = 10c / 1024 for x = 0, 1, 2, ……….10 x (i) probability of getting 4 heads = f (4) = 10c / 1024 4 = 210 / 1024 = 105 / 512 (cid:10)(cid:11)(cid:19)(cid:11) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India (ii) probability of getting at least 4 heads = P (X ≥ 4) = P (X = 4) + P (X = 5) + P (X = 6) + P(X = 7) +P (X = 8) = 10c / 1024 + 10c / 1024 + 10c / 1024 + 10c / 1024 + 10c /1024 4 5 6 7 8 210+252+210+120+45 = 1024 = 837 / 1024 (iii ) probability of getting at most 3 heads = P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = f (0) + f (1) + f (2) + f (3) = 10c / 1024 + 10c / 1024 + 10c / 1024 +10c / 1024 0 1 2 3 1+10+45+120 = 1024 = 176 / 1024 = 11/64 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....22222 ::::: If 15 dates are selected at random, what is the probability of getting two Sundays? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: If X denotes the number at Sundays, then it is obvious that X follows binomial distribution with parameter n = 15 and p = probability of a Sunday in a week = 1/7 and q = 1 – p = 6 / 7. Then f(x) = 15c (1/7)x. (6/7)15–x. x for x = 0, 1, 2,……….. 15. Hence the probability of getting two Sundays = f(2) = 15c (1/7)2 . (6/7)15–2 2 105×613 = 715 ≅ 0.29 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....33333 ::::: The incidence of occupational disease in an industry is such that the workmen have a 10% chance of suffering from it. What is the probability that out of 5 workmen, 3 or more will contract the disease? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let X denote the number of workmen in the sample. X follows binomial with (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) parameters n = 5 and p = probability that a workman suffers from the occupational disease = 0.1 Hence q = 1 – 0.1 = 0.9. Thus f (x) = 5c (0.1)x. (0.9)5-x x . For x = 0, 1, 2,…….,5. The probability that 3 or more workmen will contract the disease = P (x ≥3) = f (3) + f (4) + f (5) = 5c (0.1)3 (0.9)5-3 + 5c (0.1)4. (0.9) 5-4 + 5c (0.1)5 3 4 5 = 10 x 0.001 x 0.81 + 5 x 0.0001 x 0.9 + 1 x 0.00001 = 0.0081 + 0.00045 + 0.00001 ≅ 0.0086. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....44444 ::::: Find the probability of a success for the binomial distribution satisfying the following relation 4 P (x = 4) = P (x = 2) and having the other parameter as six. SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: We are given that n = 6. The probability mass function of x is given by f (x) = nc px q n–x x = 6c px q n–x x for x = 0, 1, …… ,6. Thus P (x = 4) = f (4): = 6c p4 q 6–4 4 = 15 p4 q2 and P (x = 2) = f (2) = 6c p2 q 6-2 2 = 15p2 q4 Hence 4 P (x = 4) = P (x = 2) ⇒ 60 p4 q2 = 15 p2 q4 ⇒ 15 p2 q2 (4p2 – q2) = 0 ⇒ 4p2 – q2 = 0 (as p ≠ 0, q ≠ 0 ) ⇒ 4p2 – (1 – p)2 = 0 (as q = 1 – p) ⇒ (2p + 1 – p) = 0 or (2p – 1 + p) = 0 ⇒ p = –1 or p = 1/3 Thus p = 1/3 (as p ≠ –1) (cid:10)(cid:11)(cid:19)(cid:23) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....55555 ::::: Find the binomial distribution for which mean and standard deviation are 6 and 2 respectively. SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Let x ~ B (n, p) Given that mean of x = np = 6 … ( 1 ) and SD of x = 2 ⇒ variance of x = npq = 4 ….. ( 2 ) 2 Dividing ( 2 ) by ( 1 ), we get q = 3 1 Hence p = 1 – q = 3 1 1 Replacing p by in equation ( 1 ), we get n × = 6 3 3 ⇒ n = 18 Thus the probability mass function of x is given by f( x ) = nc px q n–x x = 18c ( 1/3 )x . ( 2/3 )18–x x for x = 0, 1, 2,…… ,18 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....66666 ::::: Fit a binomial distribution to the following data: x: 0 1 2 3 4 5 f: 3 6 10 8 3 2 SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: In order to fit a theoretical probability distribution to an observed frequency distribution it is necessary to estimate the parameters of the probability distribution. There are several methods of estimating population parameters. One rather, convenient method is ‘Method of Moments’. This comprises equating p moments of a probability distribution to p moments of the observed frequency distribution, where p is the number of parameters to be estimated. Since n = 5 is given, we need estimate only one parameter p. We equate the first moment about origin i.e. AM of the probability distribution to the AM of the given distribution and estimate p. i.e. npˆ = x x ⇒ pˆ = (pˆ is read as p hat ) n The fitted binomial distribution is then given by f( x ) = nc pˆ x ( 1 – pˆ )n-x x For x = 0, 1, 2, …… n On the basis of the given data, we have (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) fx x= ∑ i i N × × × × × × 3 0+6 1+10 2+8 3+3 4+2 5 = =2.25 3+6+10+8+3+2 2.25 Thus pˆ = x/n = =0.45 n and qˆ = 1 – pˆ = 0.55 The fitted binomial distribution is f (x) = 5c (0.45)x (0.55)5-x x For x = 0, 1, 2, 3, 4, 5. TTTTTaaaaabbbbbllllleeeee 1111144444.....11111 FFFFFiiiiittttttttttiiiiinnnnnggggg BBBBBiiiiinnnnnooooommmmmiiiiiaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn tttttooooo aaaaannnnn OOOOObbbbbssssseeeeerrrrrvvvvveeeeeddddd DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn XXXXX fffff ((((( xxxxx ))))) EEEEExxxxxpppppeeeeecccccttttteeeeeddddd fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy OOOOObbbbbssssseeeeerrrrrvvvvveeeeeddddd fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy = 5c ( 0.4 )x ( 0.6 )5–x Nf ( x ) = 32 f ( x ) x 0 0.07776 2.49 ≅ 3 3 1 0.25920 8.29 ≅ 8 6 2 0.34560 11.06 ≅ 11 10 3 0.23040 7.37 ≅ 7 8 4 0.07680 2.46 ≅ 3 3 5 0.01024 0.33 ≅ 0 2 Total 1.000 00 32 32 A look at table 14.1 suggests that the fitting of binomial distribution to the given frequency distribution is satisfactory. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....77777 ::::: 6 coin are tossed 512 times. Find the expected frequencies of heads. Also, compute the mean and SD of the number of heads. SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: If x denotes the number of heads, then x follows binomial distribution with parameters n = 6 and p = prob. of a head = ½, assuming the coins to be unbiased. The probability mass function of x is given by f ( x ) = 6c (1/2)x. (1/2)6–x x = 6c /26 x for x = 0, 1, …..6. The expected frequencies are given by Nf ( x ). (cid:10)(cid:11)(cid:19)(cid:25) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India TTTTTAAAAABBBBBLLLLLEEEEE 1111144444.....22222 FFFFFiiiiinnnnndddddiiiiinnnnnggggg EEEEExxxxxpppppeeeeecccccttttteeeeeddddd FFFFFrrrrreeeeeqqqqquuuuueeeeennnnnccccciiiiieeeeesssss wwwwwhhhhheeeeennnnn 66666 cccccoooooiiiiinnnnnsssss aaaaarrrrreeeee tttttooooosssssssssseeeeeddddd 555551111122222 tttttiiiiimmmmmeeeeesssss xxxxx fffff (((((xxxxx))))) NNNNNfffff (((((xxxxx))))) xxxxx fffff (((((xxxxx))))) xxxxx22222fffff (((((xxxxx))))) EEEEExxxxxpppppeeeeecccccttttteeeeeddddd fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy 0 1/64 8 0 0 1 6/64 48 6/64 6/64 2 15/64 120 30/64 60/64 3 20/64 160 60/64 180/64 4 15/64 120 60/64 240/64 5 6/64 48 30/64 150/64 6 1/64 8 6/64 36/64 Total 1 512 3 10.50 ∑ Thus mean = µ = xf (x) = 3 x ∑ E (x2) = x2 f (x) = 10.50 x ∑ Thus σ2 = x2 f (x) – µ2 x = 10.50 – 32 = 1.50 ∴ SD = σ = 1.50 ≅ 1.22 Applying formula for mean and SD, we get µ = np = 6 × 1/2 = 3 and σ = npq = 6× 1 × 1 = 1.50 ≅ 1.22 2 2 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....88888 ::::: An experiment succeeds thrice as after it fails. If the experiment is repeated 5 times, what is the probability of having no success at all ? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Denoting the probability of a success and failure by p and q respectively, we have, p = 3q ⇒p = 3 ( 1 – p ) ⇒p = 3/4 ∴ q = 1 – p = 1/4 when n = 5 and p = 3/4,,,,, we have (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:26) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) f (x) = 5c (3/4)x (1/4)5–x x for n = 0, 1, .......... , 5. So probability of having no success = f ( 0 ) = 5c (3/4)0 (1/4 )5–0 0 = 1/1024 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....99999 ::::: What is the mode of the distribution for which mean and SD are 10 and 5 respectively. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: As given np = 10 .......... (1) and npq = 5 ⇒ npq= 5 ...................... (2) on solving (1) and (2), we get n = 20 and p = 1/2 Hence mode = Largest integer contained in (n+1)p = Largest integer contained in (20+1) × 1/2 = Largest integer contained in 10.50 = 10. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111100000 ::::: If x and y are 2 independent binomial variables with parameters 6 and 1/2 and 4 and 1/2 respectively, what is P ( x + y ≥ 1 )? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let z = x + y. It follows that z also follows binomial distribution with parameters ( 6 + 4 ) and 1/2 i.e. 10 and 1/2 Hence P ( z ≥ 1 ) = 1 – P ( z < 1 ) = 1 – P ( z = 0 ) = 1 – 10c (1/2 )0. (1/2 )10–0 0 = 1 – 1 / 210 = 1023 / 1024 1111144444.....33333 PPPPPOOOOOIIIIISSSSSSSSSSOOOOONNNNN DDDDDIIIIISSSSSTTTTTRRRRRIIIIIBBBBBUUUUUTTTTTIIIIIOOOOONNNNN Poisson distribution is a theoretical discrete probability distribution which can describe many processes. Simon Denis Poisson of France introduced this distribution way back in the year 1837. (cid:10)(cid:11)(cid:19)(cid:10)(cid:27) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India PPPPPoooooiiiiissssssssssooooonnnnn MMMMMooooodddddeeeeelllll LLLLLeeeeettttt uuuuusssss ttttthhhhhiiiiinnnnnkkkkk ooooofffff aaaaa rrrrraaaaannnnndddddooooommmmm eeeeexxxxxpppppeeeeerrrrriiiiimmmmmeeeeennnnnttttt uuuuunnnnndddddeeeeerrrrr ttttthhhhheeeee fffffooooollllllllllooooowwwwwiiiiinnnnnggggg cccccooooonnnnndddddiiiiitttttiiiiiooooonnnnnsssss::::: IIIII..... TTTTThhhhheeeee ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy ooooofffff fffffiiiiinnnnndddddiiiiinnnnnggggg sssssuuuuucccccccccceeeeessssssssss iiiiinnnnn aaaaa vvvvveeeeerrrrryyyyy sssssmmmmmaaaaallllllllll tttttiiiiimmmmmeeeee iiiiinnnnnttttteeeeerrrrrvvvvvaaaaalllll ((((( ttttt,,,,, ttttt +++++ dddddttttt ))))) iiiiisssss kkkkkttttt,,,,, wwwwwhhhhheeeeerrrrreeeee kkkkk (((((>>>>>00000))))) iiiiisssss aaaaa cccccooooonnnnnssssstttttaaaaannnnnttttt..... IIIIIIIIII..... TTTTThhhhheeeee ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy ooooofffff hhhhhaaaaavvvvviiiiinnnnnggggg mmmmmooooorrrrreeeee ttttthhhhhaaaaannnnn ooooonnnnneeeee sssssuuuuucccccccccceeeeessssssssss iiiiinnnnn ttttthhhhhiiiiisssss tttttiiiiimmmmmeeeee iiiiinnnnnttttteeeeerrrrrvvvvvaaaaalllll iiiiisssss vvvvveeeeerrrrryyyyy lllllooooowwwww..... IIIIIIIIIIIIIII..... TTTTThhhhheeeee ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy ooooofffff hhhhhaaaaavvvvviiiiinnnnnggggg sssssuuuuucccccccccceeeeessssssssss iiiiinnnnn ttttthhhhhiiiiisssss tttttiiiiimmmmmeeeee iiiiinnnnnttttteeeeerrrrrvvvvvaaaaalllll iiiiisssss iiiiinnnnndddddeeeeepppppeeeeennnnndddddeeeeennnnnttttt ooooofffff ttttt aaaaasssss wwwwweeeeellllllllll aaaaasssss eeeeeaaaaarrrrrllllliiiiieeeeerrrrr sssssuuuuucccccccccceeeeesssssssssseeeeesssss..... The above model is known as Poisson Model. The probability of getting x successes in a relatively long time interval T containing m small time intervals t i.e. T = mt. is given by e–kt.(kt)x x! for x = 0, 1, 2, ......…∞…… ( 14.7 ) Taking kT = m, the above form is reduced to e–m.mx x! for x = 0, 1, 2, ...... ∞…..... (14.8) DDDDDeeeeefffffiiiiinnnnniiiiitttttiiiiiooooonnnnn ooooofffff PPPPPoooooiiiiissssssssssooooonnnnn DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn A random variable X is defined to follow Poisson distribution with parameter λ, to be denoted by X ~ P (λ) if the probability mass function of x is given by e–m.mx f (x) = P (X = x) = for x = 0, 1, 2, ... ∞ x! = 0 otherwise ..... (14.9) Here e is a transcendental quantity with an approximate value as 2.71828. It is wiser to remember the following important points in connection with Poisson distribution: (((((iiiii))))) SSSSSiiiiinnnnnccccceeeee eeeee–––––mmmmm ===== 11111/////eeeeemmmmm >>>>>00000,,,,, wwwwwhhhhhaaaaattttteeeeevvvvveeeeerrrrr mmmmmaaaaayyyyy bbbbbeeeee ttttthhhhheeeee vvvvvaaaaallllluuuuueeeee ooooofffff mmmmm,,,,, mmmmm >>>>> 00000,,,,, iiiiittttt fffffooooollllllllllooooowwwwwsssss ttttthhhhhaaaaattttt fffff (((((xxxxx))))) ≥ 00000 fffffooooorrrrr eeeeevvvvveeeeerrrrryyyyy xxxxx..... AAAAAlllllsssssooooo iiiiittttt cccccaaaaannnnn bbbbbeeeee eeeeessssstttttaaaaabbbbbllllliiiiissssshhhhheeeeeddddd ttttthhhhhaaaaattttt ∑ fffff(((((xxxxx))))) ===== 11111 iiiii.....eeeee..... fffff(((((00000))))) +++++ fffff(((((11111))))) +++++ fffff(((((22222))))) +++++................................... ===== 11111.................... (((((1111144444.....1111100000))))) x (((((iiiiiiiiii))))) PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss kkkkknnnnnooooowwwwwnnnnn aaaaasssss aaaaa uuuuunnnnniiiiipppppaaaaarrrrraaaaammmmmeeeeetttttrrrrriiiiiccccc dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaasssss iiiiittttt iiiiisssss ccccchhhhhaaaaarrrrraaaaacccccttttteeeeerrrrriiiiisssss e eeeeddddd b bbbbyyyyy ooooonnnnnlllllyyyyy ooooonnnnneeeee pppppaaaaarrrrraaaaammmmmeeeeettttteeeeerrrrr mmmmm..... (((((iiiiiiiiiiiiiii))))) TTTTThhhhheeeee mmmmmeeeeeaaaaannnnn ooooofffff PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss gggggiiiiivvvvveeeeennnnn bbbbbyyyyy mmmmm iiiii,,,,,eeeee µµµµµ ===== mmmmm..... (((((1111144444.....1111111111))))) (((((iiiiivvvvv))))) TTTTThhhhheeeee vvvvvaaaaarrrrriiiiiaaaaannnnnccccceeeee ooooofffff PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss gggggiiiiivvvvveeeeennnnn bbbbbyyyyy σσσσσ22222 ===== mmmmm (((((1111144444.....1111122222))))) (((((vvvvv))))) LLLLLiiiiikkkkkeeeee bbbbbiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn,,,,, PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn cccccooooouuuuulllllddddd bbbbbeeeee aaaaalllllsssssooooo uuuuunnnnniiiiimmmmmooooodddddaaaaalllll ooooorrrrr bbbbbiiiiimmmmmooooodddddaaaaalllll dddddeeeeepppppeeeeennnnndddddiiiiinnnnnggggg uuuuupppppooooonnnnn ttttthhhhheeeee vvvvvaaaaallllluuuuueeeee ooooofffff ttttthhhhheeeee pppppaaaaarrrrraaaaammmmmeeeeettttteeeeerrrrr mmmmm..... (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:10)(cid:10) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) We have µ = The largest integer contained in m if m is a non-integer 0 = m and m–1 if m is an integer ............ (14.13) (vi) PPPPPoooooiiiiissssssssssooooonnnnn aaaaapppppppppprrrrroooooxxxxxiiiiimmmmmaaaaatttttiiiiiooooonnnnn tttttooooo BBBBBiiiiinnnnnooooommmmmiiiiiaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn If n, the number of independent trials of a binomial distribution, tends to infinity and p, the probability of a success, tends to zero, so that m = np remains finite, then a binomial distribution with parameters n and p can be approximated by Poisson distribution with parameter m (= np). In other words when n is rather large and p is rather small so that m = np is moderate then β (n, p) ≅ P (m). ...................... (14.14) (vii) AAAAAddddddddddiiiiitttttiiiiivvvvveeeee ppppprrrrrooooopppppeeeeerrrrrtttttyyyyy ooooofffff PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn If X and y are two independent variables following Poisson distribution with parameters m and m respectively, then z = X + y also follows Poisson distribution with parameter 1 2 (m + m ). 1 2 i.e. if x ~ p (m ) 1 and y ~ p (m ) 2 and X and y are independent, then z = X + y ~ p (m + m ) ....... (14.15) 1 2 AAAAAppppppppppllllliiiiicccccaaaaatttttiiiiiooooonnnnn ooooofffff PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn Poisson distribution is applied when the total number of events is pretty large but the probability of occurrence is very small. Thus we can apply Poisson distribution, rather profitably, for the following cases: a) The distribution of the no. of printing mistakes per page of a large book. b) The distribution of the no. of road accidents on a busy road per minute. c) The distribution of the no. of radio-active elements per minute in a fusion process. d) The distribution of the no. of demands per minute for health centre and so on. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111111111 ::::: Find the mean and standard deviation of x where x is a Poisson variate satisfying the condition P (x = 2) = P ( x = 3). SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x be a Poisson variate with parameter m. The probability max function of x is then given by e-m.mx f (x) = for x = 0, 1, 2, ........ ∞ x! now, P (x = 2) = P (x = 3) ⇒f(2) = f(3) (cid:10)(cid:11)(cid:19)(cid:10)(cid:20) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India e–m.m2 e–m.m3 ⇒ = 2! 3! e–m.m2 ⇒ (1-m/3) = 0 2 ⇒ 1 – m / 3 = 0 ( as e–m > 0, m > 0 ) ⇒ m = 3 Thus the mean of this distribution is m = 3 and standard deviation = 3 ≅ 1.73. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111122222 ::::: The probability that a random variable x following Poisson distribution would assume a positive value is (1 – e–2.7). What is the mode of the distribution? SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: If x ~ P (m), then its probability mass function is given by e–m.m2 f(x) = for x = 0, 1, 2, .......... ∞ x! The probability that x assumes a positive value = P (x > 0) = 1– P (x ≤ 0) = 1 – P (x = 0) = 1 – f(0) = 1 – e–m As given, 1 – e–m = 1 – e–2.7 ⇒ e–m = e–2.7 ⇒ m = 2.7 Thus µ = largest integer contained in 2.7 0 = 2 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111133333 ::::: The standard deviation of a Poisson variate is 1.732. What is the probability that the variate lies between –2.3 to 3.68? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x be a Poisson variate with parameter m. Then SD of x is m . As given m = 1.732 ⇒m = (1.732)2 ≅ 3. The probability that x lies between –2.3 and 3.68 (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:10)(cid:21) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) = P(– 2.3 < x < 3.68) = f(0) + f(1) + f(2) + f(3) (As x can assume 0, 1, 2, 3, 4 .....) e–3.30 e–3.31 e–3.32 e–3.33 = + + + 0! 1! 2! 3! = e–3 (1 + 3 + 9/2 + 27/6) = 13e–3 13 = e3 13 = (as e = 2.71828) (2.71828)3 ≅ 0.65 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111144444 ::::: X is a Poisson variate satisfying the following relation: P (X = 2) = 9P (X = 4) + 90P (X = 6). What is the standard deviation of X? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let X be a Poisson variate with parameter m. Then the probability mass function of X is e–m.mx P (X = x) = f(x) = for x = 0, 1, 2, ..... ∞ x! Thus P (X = 2) = 9P (X = 4) + 90P (X = 6) ⇒ f(2) = 9 f(4) + 90 f(6) e–mm2 9e–m.m4 90. e–mm6 ⇒ = + 2! 4! 6! e–m m2 90m4 9m2  ⇒  + −1=0 2  360 12  e–mm2 ⇒ (m4+3m2−4)=0 8 ⇒ e–m .m2 (m2 + 4) (m2 – 1) = 0 ⇒ m2 – 1 = 0 (as e–m > 0 m > 0 and m2 + 4 ≠0) ⇒ m =1 (as m > 0, m ≠ –1) Thus the standard deviation of X is 1 = 1 (cid:10)(cid:11)(cid:19)(cid:10)(cid:11) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111155555 ::::: Between 9 and 10 AM, the average number of phone calls per minute coming into the switchboard of a company is 4. Find the probability that during one particular minute, there will be, 1. no phone calls 2. at most 3 phone calls (given e–4 = 0.018316) SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let X be the number of phone calls per minute coming into the switchboard of the company. We assume that X follows Poisson distribution with parameters m = average number of phone calls per minute = 4. 1. The probability that there will be no phone call during a particular minute = P (X = 0) e–4.40 = 0! = e– 4 = 0.018316 2. The probability that there will be at most 3 phone calls = P ( X ≤ 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3) e–4.40 e–4.41 e–4.42 e–4.43 = + + + 0! 1! 2! 3! = e– 4 ( 1 + 4 + 16/2 + 64/6) = e– 4 × 71/3 = 0.018316 × 71/3 ≅ 0.43 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111166666 ::::: If 2 per cent of electric bulbs manufactured by a company are known to be defectives, what is the probability that a sample of 150 electric bulbs taken from the production process of that company would contain 1. exactly one defective bulb? 2. more than 2 defective bulbs? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x be the number of bulbs produced by the company. Since the bulbs could be either defective or non-defective and the probability of bulb being defective remains the same, it follows that x is a binomial variate with parameters n = 150 and p = probability of a bulb being defective = 0.02. However since n is large and p is very small, we can approximate this binomial distribution with Poisson distribution with parameter m = np = 150 x 0.02 = 3. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:10)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 1. The probability that exactly one bulb would be defective = P ( X = 1 ) e–3.31 = 1! = e–3 × 3 3 = e3 = 3/(2.71828)3 ≅ 0.15 (ii) The probability that there would be more than 2 defective bulbs = P ( X > 2 ) = 1 – P ( X ≤ 2 ) = 1 – [ f ( 0 ) + f ( 1 ) + f ( 2 )] e–3× 30 e–3× 31 e–3× 32  = 1 –  + +   0! 1! 2!  = 1 – 8.5 × e–3 = 1 – 0.4232 = 0.5768 ≅ 0.58 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111177777 ::::: The manufacturer of a certain electronic component is certain that two per cent of his product is defective. He sells the components in boxes of 120 and guarantees that not more than two per cent in any box will be defective. Find the probability that a box, selected at random, would fail to meet the guarantee? Given that e–2.40 = 0.0907. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x denote the number of electric components. Then x follows binomial distribution with n = 120 and p = probability of a component being defective = 0.02. As before since n is quite large and p is rather small, we approximate the binomial distribution with parameters n and p by a Poisson distribution with parameter m = n.p = 120 × 0.02 = 2.40. Probability that a box, selected at random, would fail to meet the specification = probability that a sample of 120 items would contain more than 2.40 defective items. = P (X > 2.40) = 1 – P ( X ≤ 2.40) = 1 – [ P ( X = 0 ) + P (X = 1 ) + P (X = 2) ] 2 2.40 = 1 – [ e–2.40 + e–2.40 × 2.4 + e–2.40 ×   ]  2  (cid:10)(cid:11)(cid:19)(cid:10)(cid:23) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 2 (2.40) = 1 – e–2.40 (1 + 2.40 + ) 2 = 1 – 0.0907 × 6.28 ≅ 0.43 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111188888 ::::: A discrete random variable x follows Poisson distribution. Find the values of (i) P (X = at least 1 ) (ii) P ( X ≤ 2/ X ≥ 1 ) You are given E ( x ) = 2.20 and e–2.20 = 0.1108. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since X follows Poisson distribution, its probability mass function is given by e–m.mx f ( x ) = for x = 0, 1, 2, …… ∞ x! (i) P ( X = at least 1 ) = P (X ≥ 1 ) = 1 – P ( X < 1 ) = 1 – P ( X = 0 ) = 1 – e–m = 1 – e–2.20 (as E ( x ) = m = 2.20, given) = 1 – 0.1108 (as e–2.20 = 0.1108 as given) ≅ 0.89. (ii) P ( x ≤ 2 / x ≥ 1 ) [(X≤2)∩(X ≥1)] (A∩B) = P (as P (A/B)=P P (B) P(X ≥1) P(X=1)+P(X=2) = 1–P(X<1) f(1)+f(2) = 1–f(0) e–m.m+e–m.m2/2 = 1−e–m (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:10)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) e–2.20× 2.2 + e–2.20×(2.20)2/2 = ( ∵ m=2.2) 1–e–2.20 0.5119 = 0.8892 ≅ 0.58 FFFFFiiiiittttttttttiiiiinnnnnggggg aaaaa PPPPPoooooiiiiissssssssssooooonnnnn dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn As explained earlier, we can apply the method of moments to fit a Poisson distribution to an observed frequency distribution. Since Poisson distribution is uniparametric, we equate m, the parameter of Poisson distribution, to the arithmetic mean of the observed distribution and get the estimate of m. i.e. mˆ =x The fitted Poisson distribution is then given by ˆ e–mˆ .(mˆ )x f(x) = forx = 0, 1, 2..................∞ x! EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....1111199999::::: Fit a Poisson distribution to the following data : Number of death: 0 1 2 3 4 Frequency: 122 46 23 8 1 ( Given that e–0.6 = 0.5488 ) SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: The mean of the observed frequency distribution is ∑fx x = i i N 122×0+46×1+23×2+8×3+1×4 = − 122+46+23+8+1 120 = 200 = 0.6 Thus mˆ = 0.6 Hence f ˆ ( 0 ) = e–mˆ = e–0.6 = 0.5488 e–mˆ ×m f ˆ ( 1 ) = = 0.6 × e–0.6 = 0.3293 1! (cid:10)(cid:11)(cid:19)(cid:10)(cid:25) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India (0.6)2 ×0.5488=0.0988 2! (0.6)3 ×0.5488=0.0198 3! Lastly P ( X ≥ 4 ) = 1 – P ( X < 4 ). TTTTTaaaaabbbbbllllleeeee 1111144444.....33333 FFFFFiiiiittttttttttiiiiinnnnnggggg PPPPPoooooiiiiissssssssssooooonnnnn DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn tttttooooo aaaaannnnn OOOOObbbbbssssseeeeerrrrrvvvvveeeeeddddd FFFFFrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn ooooofffff DDDDDeeeeeaaaaattttthhhhhsssss XXXXX fffff (((((xxxxx))))) EEEEExxxxxpppppeeeeecccccttttteeeeeddddd OOOOObbbbbssssseeeeerrrrrvvvvveeeeeddddd fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy NNNNN × fffff ((((( xxxxx ))))) 0 0.5488 109.76 = 110 122 1 0.6 x 0.5488 = 0.3293 65.86 = 65 46 2 (0.6)2/2 x 0.5488 = 0.0.0988 19.76 = 20 23 3 (0.6)3/3 x 0.5488 = 0.0.0198 3.96 = 4 8 4 or more 0.0033 (By subtraction) 0.66 = 1 1 Total 1 200 200 1111144444.....44444 NNNNNOOOOORRRRRMMMMMAAAAALLLLL OOOOORRRRR GGGGGAAAAAUUUUUSSSSSSSSSSIIIIIAAAAANNNNN DDDDDIIIIISSSSSTTTTTRRRRRIIIIIBBBBBUUUUUTTTTTIIIIIOOOOONNNNN The two distributions discussed so far, namely binomial and Poisson, are applicable when the random variable is discrete. In case of a continuous random variable like height or weight, it is impossible to distribute the total probability among different mass points because between any two unequal values, there remains an infinite number of values. Thus a continuous random variable is defined in term of its probability density function f (x), provided, of course, such a function really exists f (x) satisfies the following condition: f(x) ≥ 0 for x ∈ (α,β) β and ∫f(x)= 1 (α, β), β>α, being the domain of the continuous variable x. α The most important and universally accepted continuous probability distribution is known as normal distribution. Though many mathematicians like De-Moivre, Laplace etc. contributed towards the development of normal distribution, Karl Gauss was instrumental for deriving normal distribution and as such normal distribution is also referred to as Gaussian Distribution. A continuous random variable x is defined to follow normal distribution with parameters µ and σ 2, to be denoted by (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:10)(cid:26) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) X ~ N (µ, σ 2 ) ………………. (14.16) If the probability density function of the random variable x is given by 1 .e−(x−u)2/2σ2 f(x) = σ 2π for −∞<x<∞……… (14.17) Some important points relating to normal distribution are listed below: (a) The name Normal Distribution has its origin some two hundred years back as the then mathematician were in search for a normal model that can describe the probability distribution of most of the continuous random variables. (b) If we plot the probability function y = f (x), then the curve, known as probability curve, takes the following shape: −∞ µ ∞ Figure 14.1 Showing Normal Probability Curve A quick look at figure 14.1 reveals that the normal curve is bell shaped and has one peak, which implies that the normal distribution has one unique mode. The line drawn through x = µ has divided the normal curve into two parts which are equal in all respect. Such a curve is known as symmetrical curve and the corresponding distribution is known as Symmetrical distribution. Thus, we find that the normal distribution is symmetrical about x = µ. It may also be noted that the binomial distribution is also symmetrical about p = 0.5. We next note that the two tails of the normal curve extend indefinitely on both sides of the curve and both the left and right tails never touch the horizontal axis. The total area of the normal curve or for that any probability curve is taken to be unity i.e. one. Since the vertical line drawn through x = µ divides the curve into two equal halves, it automatically follows that, (cid:10)(cid:11)(cid:19)(cid:20)(cid:27) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India The area between – ∞ to µ = the area between µ to ∞ = 0.5 When the mean is zero, we have The area between – ∞ to 0 = the area between 0 to ∞ = 0.5 (c) If we take µ = 0 and σ = 1 in (14.17), we have 1 e−z2/ 2 f(x) = for –∞< x <∞ …….. (14.18) 2π The random variable x is known as standard normal variate (or variable) or standard normal deviate. The probability that a standard normal variate X would take a value less than or equal to a particular value say X = x is given by φ (x) = p ( X ≤ x ) ……. (14.19) φ (x) is known as the cumulative distribution function. We also have φ (0) = P ( X ≤ 0 ) = Area of the standard normal curve between –∞ and 0 = 0.5 …….. (14.20) (d) TTTTThhhhheeeee nnnnnooooorrrrrmmmmmaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss kkkkknnnnnooooowwwwwnnnnn aaaaasssss bbbbbiiiiipppppaaaaarrrrraaaaammmmmeeeeetttttrrrrriiiiiccccc dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaasssss iiiiittttt iiiiisssss ccccchhhhhaaaaarrrrraaaaacccccttttteeeeerrrrriiiiissssseeeeeddddd bbbbbyyyyy tttttwwwwwooooo pppppaaaaarrrrraaaaammmmmeeeeettttteeeeerrrrrsssss µµµµµ aaaaannnnnddddd σσσσσ 22222..... OOOOOnnnnnccccceeeee ttttthhhhheeeee tttttwwwwwooooo pppppaaaaarrrrraaaaammmmmeeeeettttteeeeerrrrrsssss aaaaarrrrreeeee kkkkknnnnnooooowwwwwnnnnn,,,,, ttttthhhhheeeee nnnnnooooorrrrrmmmmmaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss cccccooooommmmmpppppllllleeeeettttteeeeelllllyyyyy ssssspppppeeeeeccccciiiiifffffiiiiieeeeeddddd..... PPPPPrrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff NNNNNooooorrrrrmmmmmaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn 1. Since π = 22/7 , e–θ = 1 / eθ > 0, whatever θ may be, it follows that f (x) ≥ 0 for every x. It can be shown that ∞ ∫ f(x)dx =1 −∞ 2. The mean of the normal distribution is given by µ. Further, since the distribution is symmetrical about x = µ, it follows that the mean, median and mode of a normal distribution coincide, all being equal to µ. 3. The standard deviation of the normal distribution is given by σ. Mean deviation of normal distribution is σ 2 (cid:1) ≅ (cid:2)(cid:3)(cid:4) σ ………… (14.21) The first and third quartiles are given by q = µ – 0.675 σ …………. (14.22) 1 and q = µ + 0.675 σ …………. (14.23) 3 so that, quartile deviation = 0.675 σ ………. (14.24) (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:20)(cid:10) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 4. The normal distribution is symmetrical about x = µ. As such, its skewness is zero i.e. the normal curve is neither inclined move towards the right (negatively skewed) nor towards the left (positively skewed). 5. The normal curve y = f (x) has two points of inflexion to be given by x = µ – σ and x = µ + σ i.e. at these two points, the normal curve changes its curvature from concave to convex and from convex to concave. 6. If x ~ N (µ, σ2) then z = x – µ/σ ~ N (0, 1), z is known as standardised normal variate or normal deviate. We also have P (z ≤ k ) = φ (k) ……………. (14.25) The values of φ(k) for different k are given in a table known as “Biometrika.” Because of symmetry, we have φ (– k) = 1 – φ (k) …………………. (14.26) We can evaluate the different probabilities in the following manner: P (x < a ) = P ( x – µ/σ < a – µ/σ) = P (z < k ), ( k = a – µ/σ) = φ ( k) ………………….. (14.27) Also P ( x ≤ a ) = P ( x < a ) as x is continuous. P ( x > b ) = 1 – P ( x ≤ b ) = 1 – φ ( b – µ/σ ) …………. (14.28) and P ( a < x < b ) = φ ( b – µ/σ ) – φ ( a – µ/σ ) …. (14.29) ordinate at x = a is given by (1/σ) φ (a – µ/ σ) …………….. (14.30) Also, φ (– k) = φ (k) ……….. (14.31) The values of φ (k) for different k are also provided in the Biometrika Table. 7. Area under the normal curve is shown in the following figure : µ – 3σ µ – 2σ µ – σ x = µ µ + σ µ + 2σ µ + 3σ (z = –3) (z = –2) (z = –1) (z = 0) (z = 1) (z = 2) (z = 3) (cid:10)(cid:11)(cid:19)(cid:20)(cid:20) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India µ−3σ µ−2σ µ−σ x=µ µ+σ µ+2σ µ+3σ (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) ( = –3) ( = –2) ( = –1) (( = –0) (( = 1) ( = 2) ( = 3) FFFFFiiiiiggggguuuuurrrrreeeee 1111144444.....22222 AAAAArrrrreeeeeaaaaa UUUUUnnnnndddddeeeeerrrrr NNNNNooooorrrrrmmmmmaaaaalllll CCCCCuuuuurrrrrvvvvveeeee From this figure, we find that P ( µ – σ < x < µ ) = P (µ < x < µ + σ ) = 0.34135 or alternatively, P (–1 < z < 0 ) = P ( 0 < z < 1 ) = 0.34135 P (µ – 2 σ < x < µ ) = P ( µ < x < µ + 2 σ ) = 0.47725 i.e. P (– 2 < z < 1 ) = P (1 < z < 2 ) = 0.47725 P ( µ – 3 σ < x < µ ) = P (µ < x < µ + 3σ ) = 0.49865 i.e. P(–3 < z < 0 ) = P ( 0 < z < 3 ) = 0.49865 ……. (14.32) combining these results, we have P (µ – σ < x < µ + σ ) = 0.6828 => P (–1 < z < 1 ) = 0.6828 P ( µ – 2 σ < x < µ + 2σ ) = 0.9546 => P (– 2 < z < 2 ) = 0.9546 and P ( µ – 3 σ < x < µ + 3 σ ) = 0.9973 => P (– 3 < z < 3 ) = 0.9973. ………… (14.33) We note that 99.73 per cent of the values of a normal variable lies between (µ – 3 σ) and (µ + 3 σ). Thus the probability that a value of x lies outside that limit is as low as 0.0027. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:20)(cid:21) %531.0 %41.2 %95.31 %531.43 %531.43 %95.31 %41.2 %531.0 − α −X X α Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 8. If x and y are independent normal variables with means and standard deviations as µ 1 and µ and σ , and σ respectively, then z = x + y also follows normal distribution with 2 1 2 mean (µ + µ ) and SD = σ2+σ2 respectively. 1 2 1 2 i.e. If x ~ N (µ , σ 2) 1 1 and y ~ N ( µ , σ 2) and x and y are independent, 2 2 then z = x + y ~ N ( µ + µ , σ 2 + σ 2 ) 1 2 1 2 …… ( 14.34 ) AAAAAppppppppppllllliiiiicccccaaaaatttttiiiiiooooonnnnnsssss ooooofffff NNNNNooooorrrrrmmmmmaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn The applications of normal distributions is not restricted to statistics only. Many science subjects, social science subjects, management, commerce etc. find many applications of normal distributions. Most of the continuous variables like height, weight, wage, profit etc. follow normal distribution. If the variable under study does not follow normal distribution, a simple transformation of the variable, in many a case, would lead to the normal distribution of the changed variable. When n, the number of trials of a binomial distribution, is large and p, the probability of a success, is moderate i.e. neither too large nor too small then the binomial distribution, also, tends to normal distribution. Poisson distribution, also for large value of m approaches normal distribution. Such transformations become necessary as it is easier to compute probabilities under the assumption of a normal distribution. Not only the distribution of discrete random variable, the probability distributions of t, chi-square and F also tend to normal distribution under certain specific conditions. In order to infer about the unknown universe, we take recourse to sampling and inferences regarding the universe is made possible only on the basis of normality assumption. Also the distributions of many a sample statistic approach normal distribution for large sample size. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222200000::::: For a random variable x, the probability density function is given by e−(x−4)2 f ( x ) = (cid:1) for – ∞ < x < ∞ . Identify the distribution and find its mean and variance. SSSSSooooollllluuuuutttttiiiiiooooonnnnn: The given probability density function may be written as 1 e−(x−4)2/2×1/2 f ( x ) = (cid:1) for – ∞ < x < ∞ 1/ 2× 2 1 −(x−µ)2 e = for – ∞ < x < ∞ σ× 2 (cid:1) 2σ2 with µ = 4 and σ2 = ½ (cid:10)(cid:11)(cid:19)(cid:20)(cid:11) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India Thus the given probability density function is that of a normal distribution with µ = 4 and variance = ½. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222211111::::: If the two quartiles of a normal distribution are 47.30 and 52.70 respectively, what is the mode of the distribution? Also find the mean deviation about median of this distribution. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: The 1st and 3rd quartiles of N (µ , σ2) are given by (µ – 0.675 σ) and (µ + 0.675 σ) respectively. As given, µ – 0.675 σ = 47.30 …. (1) µ + 0.675 σ = 52.70 …. (2) Adding these two equations, we get 2 µ = 100 or µ = 50 Thus Mode = Median = Mean = 50. Also σ = 4. Also Mean deviation about median = mean deviation about mode = mean deviation about mean ≅ 0.80 σ = 3.20 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222222222::::: Find the points of inflexion of the normal curve 1 f(x) = .e-(x-10)2/32 (cid:1) 4 2 for – ∞ < x < ∞ SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Comparing f (x) to the probability densities function of a normal variable x , we find that µ = 10 and σ = 4. The points of inflexion are given by µ – σ and µ + σ i.e. 10 – 4 and 10 + 4 i.e. 6 and 14. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222233333 ::::: If x is a standard normal variable such that P (0 ≤ x ≤ b) = a, what is the value of P (|x|≥ b)? SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: P ((x) ≥ b) = 1 – P (|x|≤ b) = 1 – P (– b ≤ x ≤ b) = 1 – [ P ( 0 ≤ x ≤ b ) – P (– b ≤ x ≤ 0)] (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:20)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) = 1 – [ P ( 0 ≤ x ≤ b ) + P ( 0 ≤ x ≤ b ) ] = 1 – 2a EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222244444::::: X follows normal distribution with mean as 50 and variance as 100. What is P (x ≥ 60)? Given φ ( 1 ) = 0.8413 SSSSSooooollllluuuuutttttiiiiiooooonnnnn: We are given that x ~ N ( µ, σ2 ) where µ = 50 and σ2 = 100 = > σ = 10 Thus P ( x ≥ 60 ) = 1 – P ( x ≤ 60 ) x–50 60–50 = 1 – P  ≤ = 1 – P (z ≤ 1 )  10 10  = 1 – φ ( 1 ) (From 14.27 ) = 1 – 0.8413 ≅ 0.16 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222255555::::: If a random variable x follows normal distribution with mean as 120 and standard deviation as 40, what is the probability that P ( x ≤ 150 / x > 120 )? Given that the area of the normal curve between z = 0 to z = 0.75 is 0.3734. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: P ( x ≤ 150 / x > 120 ) P(120<x≤150) = P(x>120) P(120<x≤150) = 1−P(x≤120) 120−120 x−120 150−120 P ≤ ≤   40 40 40  = x−120 120−120 1−P ≤   40 40  P(0< z ≤0.75) = 1−P(z ≤0) φ(0.75)–φ(0) = (From 14.29) 1−φ(0) (cid:10)(cid:11)(cid:19)(cid:20)(cid:23) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 0.8734−0.50 = 1−0.50 ≅ 0.75 (φ ( 0.75) = Area of the normal curve between z = – ∞ to z = 0.75 = area between – ∞ to 0 + Area between 0 to 0.75 = 0.50 + 0.3734 = 0.8734 ) EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222266666::::: X is a normal variable with mean = 5 and SD 10. Find the value of b such that the probability of the interval [ 2 5, b ] is 0.4772 given φ ( 2 ) = 0.9772. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We are given that x ~ N ( µ, σ2 ) where µ = 25 and σ = 10 and P [ 25 < x < b ] = 0.4772 25−25 x−25 b−25 ⇒ < < =0.4772    10 10 10  b−25 ⇒P[0<z< ]=0.4772 10 b−25 ⇒φ −φ(0)=0.4772  10   b−25 ⇒φ  −0.50=0.4772  10   b−25 ⇒φ   =0.9772  10  b−25 ⇒φ =φ(2) ( as given) 10 b−25 ⇒ =2 10 ⇒ b = 25 + 2 × 10 = 45. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222277777::::: In a sample of 500 workers of a factory, the mean wage and SD of wages are found to be Rs. 500 and Rs. 48 respectively. Find the number of workers having wages: (i) more than Rs. 600 (ii) less than Rs. 450 (iii) between Rs. 548 and Rs. 600. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let X denote the wage of the workers in the factory. We assume that X is normally distributed with mean wage as Rs. 500 and standard deviation of wages as Rs. 48 respectively. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:20)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) (i) Probability that a worker selected at random would have wage more than Rs. 600 = P ( X > 600 ) = 1 – P ( X ≤ 600 ) X–500 600–500 = 1 – P  ≤   48 48  = 1 – P (z ≤ 2.08 ) = 1 – φ ( 2.08 ) = 1 – 0.9812 (From Biometrika Table) = 0.0188 Thus the number of workers having wages less than Rs. 600 = 500 × 0.0188 = 9.4 ≅ 9 (ii) Probability of a worker having wage less than Rs. 450 = P ( X < 450 ) X-500 450-500 = P  <   48 48  = P(z < – 1.04 ) = φ ( – 1.04 ) = 1 – φ ( 1.04 ) (from 14.26) = 1 – 0.8508 (from Biometrika Table) = 0.1492 Hence the number of workers having wages less than Rs. 450 = 500 × 0.1492 ≅ 75 (iii) Probability of a worker having wage between Rs. 548 and Rs. 600. = P ( 548 < x < 600 ) 548–500 x–500 600–500 = P  < <   48 48 48  (cid:10)(cid:11)(cid:19)(cid:20)(cid:25) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India = P ( 1 < z < 2.08 ) = φ ( 2.08 ) – φ ( 1 ) = 0.9812 – 0.8413 (consulting Biometrika) = 0.1399 So the number of workers with wages between Rs. 548 and Rs. 600 = 500 × 0.1399 ≅ 70. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222288888::::: The distribution of wages of a group of workers is known to be normal with mean Rs. 500 and SD Rs. 100. If the wages of 100 workers in the group are less than Rs. 430, what is the total number of workers in the group? SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Let X denote the wage. It is given that X is normally distributed with mean as Rs. 500 and SD as Rs. 100 and P ( X < 430 ) = 100/N, N being the total no. of workers in the group X−500 430−500 100 ⇒P < =  100 100  N 100 ⇒P(z <–0.70)= N 100 ⇒φ(− 0.70)= N 100 ⇒1−φ(0.70)= N 100 ⇒1− 0.758= N 100 ⇒0.242= N ⇒N≅ 413. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....2222299999::::: The mean height of 2000 students at a certain college is 165 cms and SD 9 cms. What is the probability that in a group of 5 students of that college, 3 or more students would have height more than 174 cm? SSSSSooooollllluuuuutttttiiiiiooooonnnnn: Let X denote the height of the students of the college. We assume that X is normally distributed with mean (µ) 165 cms and SD (σ) as 9 cms. If p denotes the probability that a student selected at random would have height more than 174 cms., then (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:20)(cid:26) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) p = P ( X > 174 ) = 1 – P ( X ≤ 174 ) X−165 174−165 =1 – P  ≤   9 9  = 1 – P (z ≤ 1 ) = 1 – φ ( 1 ) = 1 – 0.8413 = 0.1587 If y denotes the number of students having height more than 174 cm. in a group of 5 students then y ~ β (n, p) where n = 5 and p = 0.1587. Thus the probability that 3 or more students would be more than 174 cm. = p ( y ≥ 3 ) = p ( y = 3 ) + p ( y = 4 ) + p ( y = 5 ) = 5 (0.1587 )3. ( 0.8413 )2 + 5 ( 0.1587 )4 x ( 0.8413 ) + 5 ( 0.1587 )5 C3 C4 C5 = 0.02829 + 0.002668 + 0.000100 = 0.03106. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....3333300000::::: The mean of a normal distribution is 500 and 16 per cent of the values are greater than 600. What is the standard deviation of the distribution? (Given that the area between z = 0 to z = 1 is 0.34) SSSSSooooollllluuuuutttttiiiiiooooonnnnn : Let σ denote the standard deviation of the distribution. We are given that P ( X > 600 ) = 0.16 ⇒ 1 – P ( X ≤ 600 ) = 0.16 ⇒ P ( X ≤ 600 ) = 0.84 X−500 600−500 ⇒P  ≤ =0.84  σ σ   100 ⇒P  z ≤ =0.84  σ  100 ⇒ φ  =φ(1)  σ  (cid:10)(cid:11)(cid:19)(cid:21)(cid:27) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India (100) ⇒ =1 (cid:1) ⇒ σ = 100. EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....3333311111::::: In a business, it is assumed that the average daily sales expressed in rupees follows normal distribution. Find the coefficient of variation of sales given that the probability that the average daily sales is less than Rs. 124 is 0.0287 and the probability that the average daily sales is more than Rs. 270 is 0.4599. SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let us denote the average daily sales by x and the mean and SD of x by µ and σ respectively. As given, P ( x < 124 ) = 0.0287 ……..(1) P ( x > 270 ) = 0.4599 ……..(2) From (1), we have X−µ 124−µ P  < =0.0287  σ σ  124−µ ⇒P (z < )=0.0287 σ 124−µ ⇒φ  =0.0287  σ   µ−124 ⇒1−φ  =0.0287  σ   µ−124 ⇒φ  =0.9713  σ   µ−124 ⇒φ   = φ (2.085) (From Biometrika)  σ   µ−124 ⇒  = 2.085 …….(3)  σ  From (2) we have, 1 – P ( x ≤ 270 ) = 0.4599 (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:21)(cid:10) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12)  X−µ 270−µ  ⇒ P ≤  = 0.5401  σ σ   270−µ ⇒ φ   = 0.5401  σ   270−µ ⇒ φ   = φ (0.1)  σ   270−µ ⇒   = 0.1 …..(4)  σ  Dividing (3) by (4), we get µ−124 = 20.85 270−µ ⇒µ –124 = 5629.50 – 20.85 µ ⇒µ = 5753.50/21.85 = 263.32 Substituting this value of µ in (3), we get 263.32 -124 = 2.085 σ ⇒ σ = 66.82 Thus the coefficient of variation of sales = σ/µ × 100 66.82 = ×100 263.32 = 25.38 EEEEExxxxxaaaaammmmmpppppllllleeeee 1111144444.....3333322222::::: x and y are independent normal variables with mean 100 and 80 respectively and standard deviation as 4 and 3 respectively. What is the distribution of (x + y)? SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We know that if x ~ N (µ , σ 2 ) and y~ N (µ , σ 2 ) and they are independent, then 1 1 2 2 z = x + y follows normal with mean (µ + µ ) and 1 2 SD = σ 2+σ 2 respectively. 1 1 (cid:10)(cid:11)(cid:19)(cid:21)(cid:20) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India Thus the distribution of (x + y) is normal with mean (100 + 80) or 180 and SD 42+32 = 5 1111144444.....55555 CCCCCHHHHHIIIII-----SSSSSQQQQQUUUUUAAAAARRRRREEEEE DDDDDIIIIISSSSSTTTTTRRRRRIIIIIBBBBBUUUUUTTTTTIIIIIOOOOONNNNN,,,,, TTTTT-----DDDDDIIIIISSSSSTTTTTRRRRRIIIIIBBBBBUUUUUTTTTTIIIIIOOOOONNNNN AAAAANNNNNDDDDD FFFFF ––––– DDDDDIIIIISSSSSTTTTTRRRRRIIIIIBBBBBUUUUUTTTTTIIIIIOOOOONNNNN We are going to study statistical inference in the concluding chapter. For statistical inference, we need some basic ideas about three more continuous theoretical probability distributions, namely, chi-square distribution, t – distribution and F – distribution. Before discussing this distribution, let us review standard normal distribution. SSSSStttttaaaaannnnndddddaaaaarrrrrddddd NNNNNooooorrrrrmmmmmaaaaalllll DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn If a continuous random variable z follows standard normal distribution, to be denoted by z ~ N(0, 1), then the probability density function of z is given by 1 e−z2/2 f(z) = for - ∞ < z < ∞….. (14.35) 2π SSSSSooooommmmmeeeee iiiiimmmmmpppppooooorrrrrtttttaaaaannnnnttttt ppppprrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff zzzzz aaaaarrrrreeeee llllliiiiisssssttttteeeeeddddd bbbbbeeeeelllllooooowwwww ::::: (((((iiiii))))) zzzzz hhhhhaaaaasssss mmmmmeeeeeaaaaannnnn,,,,, mmmmmeeeeedddddiiiiiaaaaannnnn aaaaannnnnddddd mmmmmooooodddddeeeee aaaaallllllllll eeeeeqqqqquuuuuaaaaalllll tttttooooo zzzzzeeeeerrrrrooooo..... (((((iiiiiiiiii))))) TTTTThhhhheeeee ssssstttttaaaaannnnndddddaaaaarrrrrddddd dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn ooooofffff zzzzz iiiiisssss 11111..... AAAAAlllllsssssooooo ttttthhhhheeeee aaaaapppppppppprrrrroooooxxxxxiiiiimmmmmaaaaattttteeeee vvvvvaaaaallllluuuuueeeeesssss ooooofffff mmmmmeeeeeaaaaannnnn dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn aaaaannnnnddddd qqqqquuuuuaaaaarrrrrtttttiiiiillllleeeee dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn aaaaarrrrreeeee 00000.....88888 aaaaannnnnddddd 00000.....666667777755555 rrrrreeeeessssspppppeeeeeccccctttttiiiiivvvvveeeeelllllyyyyy..... (((((iiiiiiiiiiiiiii))))) TTTTThhhhheeeee ssssstttttaaaaannnnndddddaaaaarrrrrddddd nnnnnooooorrrrrmmmmmaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss sssssyyyyymmmmmmmmmmeeeeetttttrrrrriiiiicccccaaaaalllll aaaaabbbbbooooouuuuuttttt zzzzz ===== 00000..... (((((iiiiivvvvv))))) TTTTThhhhheeeee tttttwwwwwooooo pppppoooooiiiiinnnnntttttsssss ooooofffff iiiiinnnnnfffffllllleeeeexxxxxiiiiiooooonnnnn ooooofffff ttttthhhhheeeee ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy cccccuuuuurrrrrvvvvveeeee ooooofffff ttttthhhhheeeee ssssstttttaaaaannnnndddddaaaaarrrrrddddd nnnnnooooorrrrrmmmmmaaaaalllll dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaarrrrreeeee –––––11111 aaaaannnnnddddd 11111..... (((((vvvvv))))) TTTTThhhhheeeee tttttwwwwwooooo tttttaaaaaiiiiilllllsssss ooooofffff ttttthhhhheeeee ssssstttttaaaaannnnndddddaaaaarrrrrddddd nnnnnooooorrrrrmmmmmaaaaalllll cccccuuuuurrrrrvvvvveeeee nnnnneeeeevvvvveeeeerrrrr tttttooooouuuuuccccchhhhh ttttthhhhheeeee hhhhhooooorrrrriiiiizzzzzooooonnnnntttttaaaaalllll aaaaaxxxxxiiiiisssss..... (((((vvvvviiiii))))) TTTTThhhhheeeee uuuuuppppppppppeeeeerrrrr aaaaannnnnddddd lllllooooowwwwweeeeerrrrr ppppp pppppeeeeerrrrr ccccceeeeennnnnttttt pppppoooooiiiiinnnnntttttsssss ooooofffff ttttthhhhheeeee ssssstttttaaaaannnnndddddaaaaarrrrrddddd nnnnnooooorrrrrmmmmmaaaaalllll vvvvvaaaaarrrrriiiiiaaaaabbbbbllllleeeee zzzzz aaaaarrrrreeeee gggggiiiiivvvvveeeeennnnn bbbbbyyyyy P ( Z > z ) = p …….. (14.36) p And P ( Z < z ) = p 1–p i.e. P ( Z < – z ) = p respectively … (14.37) p ( since for a standard normal distribution z = – z ) 1–p p Selecting P = 0.005, 0.025, 0.01 and 0.05 respectively, We have zzzzz ===== 22222.....5555588888 00000.....000000000055555 zzzzz ===== 11111.....9999966666 00000.....000002222255555 zzzzz ===== 22222.....3333333333 00000.....0000011111 zzzzz ===== 11111.....666664444455555 ………… ( 14.38) 00000.....0000055555 These are shown in fig 14.3. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:21)(cid:21) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) (((((vvvvviiiiiiiiii)))))IIIIIfffff x dddddeeeeennnnnooooottttteeeeesssss ttttthhhhheeeee aaaaarrrrriiiiittttthhhhhmmmmmeeeeetttttiiiiiccccc mmmmmeeeeeaaaaannnnn ooooofffff aaaaa rrrrraaaaannnnndddddooooommmmm sssssaaaaammmmmpppppllllleeeee ooooofffff sssssiiiiizzzzzeeeee nnnnn dddddrrrrraaaaawwwwwnnnnn fffffrrrrrooooommmmm aaaaa nnnnnooooorrrrrmmmmmaaaaalllll pppppooooopppppuuuuulllllaaaaatttttiiiiiooooonnnnn ttttthhhhheeeeennnnn,,,,, n(x–µ) Z= ~ N ( 0, 1 ) ………………… 14.39 σ p p −∞ ∞ – z Z = 0 z p p FFFFFiiiiiggggg ..... 1111144444.....33333 Showing upper and lower p % points of the standard normal variable. CCCCChhhhhiiiii–––––sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn::::: (((((χ2 ––––– dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn))))) If a continuous random variable x follows Chi–square distribution with n degrees of freedom (df) i.e. n independent condition without any restriction or constraints, to be denoted by x~X2 n then the probability density function of x is given by f(x) = k . e –x/2 x n/2 – 1 (Where k is a constant) for 0 < x < ∞ ……….. (14.40) TTTTThhhhheeeee iiiiimmmmmpppppooooorrrrrtttttaaaaannnnnttttt ppppprrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff χ2(((((ccccchhhhhiiiii-----sssssqqqqquuuuuaaaaarrrrreeeee))))) dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn aaaaarrrrreeeee mmmmmeeeeennnnntttttiiiiiooooonnnnneeeeeddddd bbbbbeeeeelllllooooowwwww::::: (((((iiiii))))) MMMMMeeeeeaaaaannnnn ooooofffff ttttthhhhheeeee ccccchhhhhiiiii-----sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn ===== nnnnn (((((iiiiiiiiii))))) SSSSStttttaaaaannnnndddddaaaaarrrrrddddd dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn ooooofffff ccccchhhhhiiiii–––––sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn ===== 2n (((((iiiiiiiiiiiiiii))))) AAAAAddddddddddiiiiitttttiiiiivvvvveeeee ppppprrrrrooooopppppeeeeerrrrrtttttyyyyy ooooofffff ccccchhhhhiiiii–––––sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn..... IIIIIfffff xxxxx aaaaannnnnddddd yyyyy aaaaarrrrreeeee tttttwwwwwooooo iiiiinnnnndddddeeeeepppppeeeeennnnndddddeeeeennnnnttttt ccccchhhhhiiiii-----sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn wwwwwiiiiittttthhhhh mmmmm aaaaannnnnddddd nnnnn dddddeeeeegggggrrrrreeeeeeeeeesssss ooooofffff fffffrrrrreeeeeeeeeedddddooooommmmm,,,,, ttttthhhhheeeeennnnn (((((xxxxx +++++ yyyyy))))) aaaaalllllsssssooooo fffffooooollllllllllooooowwwwwsssss ccccchhhhhiiiii-----sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn wwwwwiiiiittttthhhhh (((((mmmmm +++++ nnnnn))))) dddddfffff..... i.e., if x ~ χ2 m and y ~ χ2 m and x and y are independent, then µ = x + y ~ χ2 ……….. (14.41) m+n (cid:10)(cid:11)(cid:19)(cid:21)(cid:11) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India (iv) For large n, 2x2 – 2n–1follows as approximate standard normal distribution. (v) The upper and lower p per cent points of chi-square distribution with n df are given by P ( χ2 > χ2 ) = p p,n and P (χ2 < χ2 , n ) = p ………… (14.42) 1-p (cid:1) (cid:1) (cid:1) (cid:1) (vi) If are n independent standard normal variables, then 1, 2, 3 ………… n n µ = ∑ zi2~χ n 2 Similarly, if x x x x are n independent normal variables, with 1, 2, 3 …………… n 1 a common mean µ and common variables σ2, then µ = ∑( xi - µ/σ ) 2 ~χ2 …. (14.43) n Lastly if a random sample of size n is taken from a normal population with mean µ and variance σ2, then ∑(x −x)2 µ= i ~χ 2 …………. (14.44) σ2 n−1 (((((vvvvviiiiiiiiii)))))CCCCChhhhhiiiii-----sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss pppppooooosssssiiiiitttttiiiiivvvvveeeeelllllyyyyy ssssskkkkkeeeeewwwwweeeeeddddd iiiii.....eeeee..... ttttthhhhheeeee ppppprrrrrooooobbbbbaaaaabbbbbiiiiillllliiiiitttttyyyyy cccccuuuuurrrrrvvvvveeeee ooooofffff ttttthhhhheeeee ccccchhhhhiiiii–––––sssssqqqqquuuuuaaaaarrrrreeeee dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn iiiiisssss iiiiinnnnncccccllllliiiiinnnnneeeeeddddd mmmmmooooovvvvveeeee ooooonnnnn ttttthhhhheeeee rrrrriiiiiggggghhhhhttttt..... p p χ 1 2 −p, n χ2 n p, FFFFFiiiiiggggguuuuurrrrreeeee 1111144444.....44444 Showing the upper and lower p per cent point of chi-square distribution with n df. ttttt ––––– dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn: If a continuous random variable t follows t – distribution with n df, then its probability density function is given by f (t)=k1+t2/n -(n+1)/2   (where k is a constant) for – ∞< t <∞ ……… 14.45 (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:21)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) This is denoted by t ~ t . n The important properties of t-distribution are mentioned below: (i) Mean of t-distribution is zero. (ii) Standard deviation of t-distribution n/(n−2) , n > 2 (iii) t-distribution is symmetrical about t = 0. (iv) For large n (> 30), t-distribution tends to the standard normal distribution. (v) The upper and lower p per cent points of t-distribution are given by P ( t > t , n ) = p p And P ( t < t , n ) = p …………. (14.46) p (vi) If y and z are two independent random variables such that y ~ χ2and Z ~ N (0 , 1) , then n n t= z ~t n …………(14.47) y Similarly, if a random sample of size n is taken from a normal distribution with mean m and SD σ, then n-1(x– µ) t= : t (14.48) S n–1 ………….. Here x and S denote the sample mean and sample SD respectively. p p –∞ –t t = 0 t ∞ p,n p,n FFFFFiiiiiggggguuuuurrrrreeeee 1111144444.....55555 Showing the upper and lower p per cent point pf t – distribution with n df. (cid:10)(cid:11)(cid:19)(cid:21)(cid:23) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India FFFFF ––––– DDDDDiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn If a continuous random variable F follows F – distribution with n and n degrees of freedom, to 1 2 be denoted by F ~ F then its probability density function is given by n , n , 1 2 f ( F ) = k . F n 1 /2 – 1 .( 1 + n F / n ) –( n1 + n2)/ 2 1 (where k is a constant) for 0 < F < ∞ …………….(14.49) IIIIImmmmmpppppooooorrrrrtttttaaaaannnnnttttt ppppprrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff FFFFF ––––– dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn n 2 1. Mean of the F – distribution = , n > 2 n −2 2 2 2. Standard deviation of the F – distribution n 2(n +n −2) = 2 1 2 , n >4 n −2 n (n −4) 2 2 1 2 2(n +n ) 1 2 and for large n and n , SD = 1 2 n n 1 2 3. F – distribution has a positive skewness. 4. The upper and lower p per cent points of F – distribution are given by P = (F > F , (n , n ) ) = p p 1 2 1 and P ( F < ) = p ………. (14.50) F (n , n ) p 2 1 5. If U and V are two independent random variables such that U ~ χ2 n1 and V ~ χ 2 then n 2 U/n F = 1 ~ F (14.51) V/n 2 n1, n2 ……………… 6. For large values of n and n , F – distribution tends to normal distribution with mean, and 1 2 2(n +n ) 1 2 SD = n n 1 2 (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:21)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) p p 1 F , (n , n ) F , (n , n ) p 2 1 p 2 1 FFFFFiiiiiggggguuuuurrrrreeeee 1111144444.....66666 Showing the upper and lower p per cent points of F–distribution with n and n degree of 1 2 freedom. EEEEEXXXXXEEEEERRRRRCCCCCIIIIISSSSSEEEEE SSSSSeeeeettttt ::::: AAAAA Write down the correct answers. Each question carries 1 mark. 1. A theoretical probability distribution. (a) does not exist. (b) exists only in theory. (c) exists in real life. (d) both (b) and (c). 2. Probability distribution may be (a) discrete. (b) continuous. (c) infinite. (d) both (a) and (b). 3. An important discrete probability distribution is (a) Poisson distribution. (b) Normal distribution. (c) Cauchy distribution. (d) Log normal distribution. 4. An important continuous probability distribution (a) Binomial distribution. (b) Poisson distribution. (c) Geometric distribution. (d) Chi-square distribution. 5. Parameter is a characteristic of (a) population. (b) sample. (c) probability distribution. (d) both (a) and (b). 6. An example of a parameter is (a) sample mean. (b) population mean. (c) binomial distribution. (d) sample size. (cid:10)(cid:11)(cid:19)(cid:21)(cid:25) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 7. A trial is an attempt to (a) make something possible. (b) make something impossible. (c) prosecute an offender in a court of law. (d) produce an outcome which is neither certain nor impossible. 8. The important characteristic(s) of Bernoulli trials (a) each trial is associated with just two possible outcomes. (b) trials are independent. (c) trials are infinite. (d) both (a) and (b). 9. The probability mass function of binomial distribution is given by (a) f(x) = px q n–x. (b) f(x) = nc px q n–x. x (c) f(x) = nc qx p n–x. (d) f(x) = nc pn–x q x. x x 10. If x is a binomial variable with parameters n and p, then x can assume (a) any value between 0 and n. (b) any value between 0 and n, both inclusive. (c) any whole number between 0 and n, both inclusive. (d) any number between 0 and infinity. 11. A binomial distribution is (a) never symmetrical. (b) never positively skewed. (c) never negatively skewed. (d) symmetrical when p = 0.5. 12. The mean of a binomial distribution with parameter n and p is (a) n (1– p). (b) np (1 – p). (c) np. (d) np(1–p). 13. The variance of a binomial distribution with parameters n and p is (a) np2 (1 – p). (b) np(1−p). (c) nq (1 – q). (d) n2p2 (1– p)2. 14. An example of a bi-parametric discrete probability distribution is (a) binomial distribution. (b) poisson distribution. (c) normal distribution. (d) both (a) and (b). 15. For a binomial distribution, mean and mode (a) are never equal. (b) are always equal. (c) are equal when q = 0.50. (d) do not always exist. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:21)(cid:26) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 16. The mean of binomial distribution is (a) always more than its variance. (b) always equal to its variance. (c) always less than its variance. (d) always equal to its standard deviation. 17. For a binomial distribution, there may be (a) one mode. (b) two mode. (c) (a). (d) (a) or (b). 18. The maximum value of the variance of a binomial distribution with parameters n and p is (a) n/2. (b) n/4. (c) np (1 – p). (d) 2n. 19. The method usually applied for fitting a binomial distribution is known as (a) method of least square. (b) method of moments. (c) method of probability distribution. (d) method of deviations. 20. Which one is not a condition of Poisson model? (a) the probability of having success in a small time interval is constant. (b) the probability of having success more than one in a small time interval is very small. (c) the probability of having success in a small interval is independent of time and also of earlier success. (d) the probability of having success in a small time interval (t, t + dt) is kt for a positive constant k. 21. Which one is uniparametric distribution? (a) Binomial. (b) Poisson. (c) Normal. (d) Hyper geometric. 22. For a Poisson distribution, (a) mean and standard deviation are equal. (b) mean and variance are equal. (c) standard deviation and variance are equal. (d) both (a) and (b). 23. Poisson distribution may be (a) unimodal. (b) bimodal. (c) Multi-modal. (d) (a) or (b). 24. Poisson distribution is (a) always symmetric. (b) always positively skewed. (c) always negatively skewed. (d) symmetric only when m = 2. 25. A binomial distribution with parameters m and p can be approximated by a Poisson distribution with parameter m = np is (a) m → ∝. (b) p → 0. (c) m → ∝ and p → 0. (d) m →∝ and p → 0 so that mp remains finite.. (cid:10)(cid:11)(cid:19)(cid:11)(cid:27) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 26. For Poisson fitting to an observed frequency distribution, (a) we equate the Poisson parameter to the mean of the frequency distribution. (b) we equate the Poisson parameter to the median of the distribution. (c) we equate the Poisson parameter to the mode of the distribution. (d) none of these. 27. The most important continuous probability distribution is known as (a) Binomial distribution. (b) Normal distribution. (c) Chi-square distribution. (d) sampling distribution. 28. The probability density function of a normal variable x is given by 1 − 1 ( x−µ )2 (a) f(x) = .e 2 σ for – ∝ < x < ∝. σ 2π 1 − −(x−µ)2 (b) f(x) = f(x) = .e 2σ2 for 0 < x < ∝. σ 2π 1 − (x−µ)2 (c) f(x) = .e 2σ2 for – ∝ < x < ∝. 2πσ (d) none of these. 29. The total area of the normal curve is (a) one. (b) 50 per cent. (c) 0.50. (d) any value between 0 and 1. 30. The normal curve is (a) Bell-shaped. (b) U- shaped. (c) J- shaped. (d) Inverted J – shaped. 31. The normal curve is (a) positively skewed. (b) negatively skewed. (c) Symmetrical. (d) all these. 32. Area of the normal curve is (a) between – ∝ to µ is 0.50. (b) between µ to ∝ is 0.50. (c) between – ∝ to ∝ is 0.50. (d) both (a) and (b). (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:11)(cid:10) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 33. The cumulative distribution function of a random variable X is given by (a) F(x) = P ( X ≤ x). (b) F(X) = P ( X ≤ x). (c) F(x) = P ( X ≥ x). (d) F(x) = P ( X = x). 34. The mean and mode of a normal distribution (a) may be equal. (b) may be different. (c) are always equal. (d) (a) or (b). 35. The mean deviation about median of a standard normal variate is (a) 0.675 σ. (b) 0.675 . (c) 0.80 σ. (d) 0.80. 36. The quartile deviation of a normal distribution with mean 10 and SD 4 is (a) 0.675. (b) 67.50. (c) 2.70. (d) 3.20. 37. For a standard normal distribution, the points of inflexion are given by (a) µ – σ and µ + σ. (b) – σ and σ. (c) –1 and 1. (d) 0 and 1. 38. The symbol φ (a) indicates the area of the standard normal curve between (a) 0 to a. (b) a to ∞. (c) – ∝ to a. (d) – ∝ to ∝. 39. The interval (µ - 3σ, µ + 3σ) covers (a) 95% area of a normal distribution. (b) 96% area of a normal distribution. (c) 99% area of a normal distribution. (d) all but 0.27% area of a normal distribution. 40. Number of misprints per page of a thick book follows (a) Normal distribution. (b) Poisson distribution. (c) Binomial distribution. (d) Standard normal distribution. 41. The result of ODI matches between India and Pakistan follows (a) Binomial distribution. (b) Poisson distribution. (c) Normal distribution. (d) (b) or (c). 42. The wage of workers of a factory follow (a) Binomial distribution. (b) Poisson distribution. (c) Normal distribution . (d) Chi-square distribution. 43. If X and Y are two independent random variables such that X ~ χ 2m and Y~ χ 2 n , then the distribution of (X +Y) is (a) normal. (b) standard normal. (c) T. (d) chi-square. (cid:10)(cid:11)(cid:19)(cid:11)(cid:20) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India SSSSSeeeeettttt BBBBB ::::: Write down the correct answers. Each question carries 2 marks. 1. What is the standard deviation of the number of recoveries among 48 patients when the probability of recovering is 0.75? (a) 36. (b) 81. (c) 9. (d) 3. 2. X is a binomial variable with n = 20. What is the mean of X if it is known that x is symmetric? (a) 5. (b) 10. (c) 2. (d) 8. 3. If X ~ B (n, p), what would be the least value of the variance of x when n = 16? (a) 2. (b) 4. (c) 8. (d) 5 . 4. If x is a binomial variate with parameter 15 and 1/3, what is the value of mode of the distribution (a) 5 and 6. (b) 5. (c) 5.50. (d) 6. 5. What is the no. of trials of a binomial distribution having mean and SD as 3 and 1.5 respectively? (a) 2. (b) 4. (c) 8. (d) 12. 6. What is the probability of getting 3 heads if 6 unbiased coins are tossed simultaneously? (a) 0.50. (b) 0.25. (c) 0.3125. (d) 0.6875. 7. If the overall percentage of success in an exam is 60, what is the probability that out of a group of 4 students, at least one has passed? (a) 0.6525. (b) 0.9744. (c) 0.8704. (d) 0.0256. 8. What is the probability of making 3 correct guesses in 5 True – False answer type questions? (a) 0.3125. (b) 0.5676. (c) 0.6875. (d) 0.4325 9. If the standard deviation of a Poisson variate X is 2, what is P (1.5 < X < 2.9)? (a) 0.231. (b) 0.158. (c) 0.15. (d) 0.144. 10. If the mean of a Poisson variable X is 1, what is P (X = at least one)? (a) 0.456. (b) 0.821. (c) 0.632. (d) 0.254. 11. If X ~ P (m) and its coefficient of variation is 50, what is the probability that X would assume only non-zero values? (a) 0.018. (b) 0.982. (c) 0.989. (d) 0.976. 12. If 1.5 per cent of items produced by a manufacturing units are known to be defective, what is the probability that a sample of 200 items would contain no defective item? (a) 0.05. (b) 0.15. (c) 0.20. (d) 0.22. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:11)(cid:21) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 13. For a Poisson variate X, P (X = 1) = P (X = 2). What is the mean of X? (a) 1.00. (b) 1.50. (c) 2.00. (d) 2.50. 14. If 1 per cent of an airline‘s flights suffer a minor equipment failure in an aircraft, what is the probability that there will be exactly two such failures in the next 100 such flights? (a) 0.50. (b) 0.184. (c) 0.265. (d) 0.256. 15. If for a Poisson variable X, f(2) = 3 f(4), what is the variance of X? (a) 2. (b) 4. (c) 2 . (d) 3. 16. What is the coefficient of variation of x, characterised by the following probability density 1 (x−10)2 e− function: f(x) = for – ∝ < x < ∝ 4 2π 32 (a) 50. (b) 60. (c) 40. (d) 30. 17. What is the first quartile of X having the following probability density function? 1 − (x−10)2 e− f(x) = for – ∝ < x < ∝ 72π 72 (a) 4. (b) 5. (c) 5.95. (d) 6.75. 18. If the two quartiles of N (µ , σ2) are 14.6 and 25.4 respectively, what is the standard deviation of the distribution? (a) 9. (b) 6. (c) 10. (d) 8. 19. If the mean deviation of a normal variable is 16, what is its quartile deviation? (a) 10.00. (b) 13.50. (c) 15.00. (d) 12.05. 20. If the points of inflexion of a normal curve are 40 and 60 respectively, then its mean deviation is (a) 40. (b) 45. (c ) 50. (d) 60. 21. If the quartile deviation of a normal curve is 4.05, then its mean deviation is (a) 5.26. (b) 6.24. (c ) 4.24. (d) 4.80. 22. If the Ist quartile and mean deviation about median of a normal distribution are 13.25 and 8 respectively, then the mode of the distribution is (a) 20. (b) 10. (c) 15. (d) 12. 23. If the area of standard normal curve between z = 0 to z = 1 is 0.3413, then the value of φ (1) is (a) 0.5000. (b) 0.8413. (c) –0.5000. (d) 1. (cid:10)(cid:11)(cid:19)(cid:11)(cid:11) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 24. If X and Y are 2 independent normal variables with mean as 10 and 12 and SD as 3 and 4, then (X+Y) is normally distributed with (a) mean = 22 and SD = 7. (b) mean = 22 and SD = 25. (c) mean = 22 and SD = 5. (d) mean = 22 and SD = 49. SSSSSeeeeettttt ::::: CCCCC Answer the following questions. Each question carries 5 marks. 1. If it is known that the probability of a missile hitting a target is 1/8, what is the probability that out of 10 missiles fired, at least 2 will hit the target? (a) 0.4258. (b) 0.3968. (c) 0.5238. (d) 0.3611. 2. X is a binomial variable such that 2 P(X = 2) = P(X = 3) and mean of X is known to be 10/3. What would be the probability that X assumes at most the value 2? (a) 16/81. (b) 17/81. (c) 47/243. (d) 46/243. 3. Assuming that one-third of the population are tea drinkers and each of 1000 enumerators takes a sample of 8 individuals to find out whether they are tea drinkers or not, how many enumerators are expected to report that five or more people are tea drinkers? (a) 100. (b) 95. (c) 88. (d) 90. 4. If a random variable X follows binomial distribution with mean as 5 and satisfying the condition 10 P (X = 0) = P (X = 1), what is the value of P (X ≥ / x > 0)? (a) 0.67. (b) 0.56. (c) 0.99. (d) 0.82. 5. Out of 128 families with 4 children each, how many are expected to have at least one boy and one girl? (a) 100. (b) 105. (c) 108. (d) 112. 6. In 10 independent rollings of a biased die, the probability that an even number will appear 5 times is twice the probability that an even number will appear 4 times. What is the probability that an even number will appear twice when the die is rolled 8 times? (a) 0.0304. (b) 0.1243. (c) 0.2315. (d) 0.1926. 7. If a binomial distribution is fitted to the following data: x: 0 1 2 3 4 f: 16 25 32 17 10 then the sum of the expected frequencies for x = 2, 3 and 4 would be (a) 58. (b) 59. (c) 60. (d) 61. (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:11)(cid:22) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 8. If X follows normal distribution with µ = 50 and σ = 10, what is the value of P (x ≤ 60 / x > 50)? (a) 0.8413. (b) 0.6828. (c) 0.1587. (d) 0.7256. 9. X is a Poisson variate satisfying the following condition 9 P (X = 4) + 90 P (X = 6) = P (X = 2). What is the value of P (X £ 1)? (a) 0.5655 (b) 0.6559 (c) 0.7358 (d) 0.8201 10. A random variable x follows Poisson distribution and its coefficient of variation is 50. What is the value of P (x > 1 / x > 0)? (a) 0.1876 (b) 0.2341 (c) 0.9254 (d) 0.8756 11. A renowned hospital usually admits 200 patients every day. One per cent patients, on an average, require special room facilities. On one particular morning, it was found that only one special room is available. What is the probability that more than 3 patients would require special room facilities? (a) 0.1428 (b) 0.1732 (c) 0.2235 (d) 0.3450 12. A car hire firm has 2 cars which is hired out everyday. The number of demands per day for a car follows Poisson distribution with mean 1.20. What is the proportion of days on which some demand is refused? (Given e 1.20 = 3.32). (a) 0.25 (b) 0.3012 (c) 0.12 (d) 0.03 13. If a Poisson distribution is fitted to the following data: Mistake per page 0 1 2 3 4 5 No. of pages 76 74 29 17 3 1 Then the sum of the expected frequencies for x = 0, 1 and 2 is (a) 150. (b) 184. (c) 165. (d) 148. 14. The number of accidents in a year attributed to taxi drivers in a locality follows Poisson distribution with an average 2. Out of 500 taxi drivers of that area, what is the number of drivers with at least 3 accidents in a year? (a) 162 (b) 180 (c) 201 (d) 190 15. In a sample of 800 students, the mean weight and standard deviation of weight are found to be 50 Kg and 20 Kg respectively. On the assumption of normality, what is the number of students weighing between 46 Kg and 62 Kg? Given area of the standard normal curve between z = 0 to z = 0.20 = 0.0793 and area between z = 0 to z = 0.60 = 0.2257. (a) 250 (b) 244 (c) 240 (d) 260 16. The salary of workers of a factory is known to follow normal distribution with an average salary of Rs. 10,000 and standard deviation of salary as Rs. 2,000. If 50 workers receive salary more than Rs. 14,000, then the total no. of workers in the factory is (a) 2,193 (b) 2,000 (c) 2,200 (d) 2,500 (cid:10)(cid:11)(cid:19)(cid:11)(cid:23) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India 17. For a normal distribution with mean as 500 and SD as 120, what is the value of k so that the interval [500, k] covers 40.32 per cent area of the normal curve? Given φ (1.30) = 0.9032. (a) 740 (b) 750 (c) 760 (d) 800 18. The average weekly food expenditure of a group of families has a normal distribution with mean Rs. 1,800 and standard deviation Rs. 300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of Rs. 1,800? Given φ (1) = 0.84. (a) 0.418 (b) 0.582 (c) 0.386 (d) 0.614 19. If the weekly wages of 5000 workers in a factory follows normal distribution with mean and SD as Rs. 700 and Rs. 50 respectively, what is the expected number of workers with wages between Rs. 660 and Rs. 720? (a) 2,050 (b) 2,200 (c) 2,218 (d) 2,300 20. 50 per cent of a certain product have weight 60 Kg or more whereas 10 per cent have weight 55 Kg or less. On the assumption of normality, what is the variance of weight? Given φ (1.28) = 0.90. (a) 15.21 (b) 9.00 (c) 16.00 (d) 22.68 (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:11)(cid:24) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) AAAAANNNNNSSSSSWWWWWEEEEERRRRRSSSSS SSSSSeeeeettttt ::::: AAAAA 1. (a) 2. (d) 3. (a) 4. (d) 5. (a) 6. (b) 7. (d) 8. (d) 9. (a) 10. (c) 11. (d) 12. (c) 13. (c) 14. (a) 15. (c) 16. (a) 17. (c) 18. (b) 19. (b) 20. (a) 21. (b) 22. (b) 23. (d) 24. (b) 25. (d) 26. (a) 27. (b) 28. (a) 29. (a) 30. (a) 31. (c) 32 (d) 33. (a) 34. (c) 35. (d) 36. (c) 37. (c) 38. (c) 39. (d) 40. (b) 41. (a) 42. (c) 43. (d) SSSSSeeeeettttt ::::: BBBBB 1. (d) 2. (b) 3. (a) 4. (b) 5. (d) 6. (c) 7. (b) 8. (a) 9. (d) 10. (c) 11. (b) 12. (a) 13. (c) 14. (b) 15. (a) 16. (c) 17. (c) 18. (d) 19. (b) 20. (a) 21. (d) 22. (a) 23. (b) 24. (c) SSSSSeeeeettttt ::::: CCCCC 1. (d) 2. (b) 3. (c) 4. (c) 5. (d) 6. (a) 7. (d) 8. (b) 9. (c) 10. (c) 11. (a) 12. (d) 13. (b) 14. (a) 15. (b) 16. (a) 17. (c) 18. (b) 19. (c) 20. (a) (cid:10)(cid:11)(cid:19)(cid:11)(cid:25) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India AAAAADDDDDDDDDDIIIIITTTTTIIIIIOOOOONNNNNAAAAALLLLL QQQQQUUUUUEEEEESSSSSTTTTTIIIIIOOOOONNNNN BBBBBAAAAANNNNNKKKKK 1. When a coin is tossed 10 times then (a) Normal Distribution (b) Poisson Distribution (c) Binomial Distribution (d) None is used 2. In Binomial Distribution ‘n‘ means (a) No. of trials of the experiment (b) the probability of getting success (c) no. of success (d) none 3. Binomial Distribution is a (a) Continuous (b) discrete (c) both (d) none probability distribution. 4. When there are a fixed number of repeated trial of any experiments under identical conditions for which only one of two mutually exclusive outcomes, success or failure can result in each trial then (a) Normal Distribution (b) Binomial Distribution (c) Poisson Distribution (d) None is used 5. In Binomial Distribution ‘p’ denotes Probability of (a) Success (b) Failure (c) Both (d) None 6. When ‘p’ = 0. 5, the binomial distribution is (a) asymmetrical (b) symmetrical (c) Both (d) None 7. When ‘p’ is larger than 0. 5, the binomial distribution is (a) asymmetrical (b) symmetrical (c) Both (d) None 8. Mean of Binomial distribution is (a) npq (b) np (c) both (d) none 9. Variance of Binomial distribution is (a) npq (b) np (c) both (d) none 10. When p = 0.1 the binomial distribution is skewed to the (a) left (b) right (c) both (d) none 11. If in Binomial distribution np = 9 and npq = 2. 25 then q is equal to (a) 0.25 (b) 0.75 (c) 1 (d) none 12. In Binomial Distribution (a) mean is greater than variance (b) mean is less than variance (c) mean is equal to variance (d) none (cid:18)(cid:5)(cid:3)(cid:5)(cid:16)(cid:18)(cid:5)(cid:16)(cid:1)(cid:18) (cid:10)(cid:11)(cid:19)(cid:11)(cid:26) Copyright -The Institute of Chartered Accountants of India (cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:3)(cid:1)(cid:6)(cid:7)(cid:8)(cid:9)(cid:10)(cid:11)(cid:6)(cid:12)(cid:1)(cid:5)(cid:6)(cid:13)(cid:14)(cid:1)(cid:6)(cid:4)(cid:15)(cid:12) 13. Standard deviation of binomial distribution is (a) square of npq (b) square root of npq (c) square of np (d) square root of np 14. _________ distribution is a limiting case of Binomial distribution (a) Normal (b) Poisson (c) Both (d) none 15. When the no. of trials is large then (a) Normal (b) Poisson (c) Binomial (d) none distribution is used 16. In Poisson Distribution, probability of success is very close to (a) 1 (b) – 1 (c) 0 (d) none 17. In Poisson Distribution np is (a) finite (b) infinite (c) 0 (d) none 18. In ________________ distribution, mean = variance (a) Normal (b) Binomial (c) Poisson (d) none 19. In Poisson distribution mean is equal to (a) npq (b) np (c) square root mp (d) square root mpq 20. In Poisson distribution standard deviation is equal to (a) square root of np (b) square of np (c) square root of npq (d) square mpq 21. For continuous events _________________ distribution is used. (a) Normal (b) Poisson (c) Binomial (d) none 22. Probability density function is associated with (a) discrete cases (b) continuous cases (c) both (d) none 23. Probability density function is always (a) greater than 0 (b) greater than equal to 0 (c) less than 0 (d) less than equal to 0 24. In continuous cases probability of the entire space is (a) 0 (b) –1 (c) 1 (d) none 25. In discrete case the probability of the entire space is (a) 0 (b) 1 (c) –1 (d) none 26. Binomial distribution is symmetrical if (a) p > q (b) p < q (c) p = q (d) none (cid:10)(cid:11)(cid:19)(cid:22)(cid:27) (cid:1)(cid:12)(cid:13)(cid:13)(cid:12)(cid:14)(cid:8)(cid:4)(cid:7)(cid:12)(cid:15)(cid:16)(cid:1)(cid:16)(cid:6)(cid:14)(cid:1)(cid:17)(cid:8)(cid:5)(cid:6)(cid:18)(cid:5) Copyright -The Institute of Chartered Accountants of India