Full Text Transcript (Pages 1–50 of 83)
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8) (cid:9)(cid:8) (cid:10)(cid:10)
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)
(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)
(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)
(cid:3)(cid:11)(cid:14)
(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
LLLLLEEEEEAAAAARRRRRNNNNNIIIIINNNNNGGGGG OOOOOBBBBBJJJJJEEEEECCCCCTTTTTIIIIIVVVVVEEEEESSSSS
After reading this Chapter , a student will be able to understand different measures of
central tendency, i.e. Arithmetic Mean, Median, Mode, Geometric Mean and Harmonic
Mean, and computational techniques of these measures.
They will also learn comparative advantages and disadvantages of these measures and
therefore which measures to use in which circumstance.
However, to understand a set of observation, it is equally important to have knowledge of
dispersion which indicates the volatility. In advanced stage of chartered accountancy course,
volatility measures will be useful in understanding risk involved in financial decision making.
This chapter will also guide the students to know details about various measures of dispersion.
1111111111.....11111 DDDDDEEEEEFFFFFIIIIINNNNNIIIIITTTTTIIIIIOOOOONNNNN OOOOOFFFFF CCCCCEEEEENNNNNTTTTTRRRRRAAAAALLLLL TTTTTEEEEENNNNNDDDDDEEEEENNNNNCCCCCYYYYY
In many a case, like the distributions of height, weight, marks, profit, wage and so on, it has
been noted that starting with rather low frequency, the class frequency gradually increases till
it reaches its maximum somewhere near the central part of the distribution and after which
the class frequency steadily falls to its minimum value towards the end. Thus, central tendency
may be defined as the tendency of a given set of observations to cluster around a single central
or middle value and the single value that represents the given set of observations is described
as a measure of central tendency or, location or average. Hence, it is possible to condense a
vast mass of data by a single representative value. The computation of a measure of central
tendency plays a very important part in many a sphere. A company is recognized by its high
average profit, an educational institution is judged on the basis of average marks obtained by
its students and so on. Furthermore, the central tendency also facilitates us in providing a basis
for comparison between different distribution. Following are the different measures of central
tendency:
(i) Arithmetic Mean (AM)
(ii) Median (Me)
(iii) Mode (Mo)
(iv) Geometric Mean (GM)
(v) Harmonic Mean (HM)
1111111111.....22222 CCCCCRRRRRIIIIITTTTTEEEEERRRRRIIIIIAAAAA FFFFFOOOOORRRRR AAAAANNNNN IIIIIDDDDDEEEEEAAAAALLLLL MMMMMEEEEEAAAAASSSSSUUUUURRRRREEEEE OOOOOFFFFF CCCCCEEEEENNNNNTTTTTRRRRRAAAAALLLLL TTTTTEEEEENNNNNDDDDDEEEEENNNNNCCCCCYYYYY
Following are the criteria for an ideal measure of central tendency:
(i) It should be properly and unambiguously defined.
(ii) It should be easy to comprehend.
(iii) It should be simple to compute.
(iv) It should be based on all the observations.
(cid:10)(cid:10)(cid:11)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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(v) It should have certain desirable mathematical properties.
(vi) It should be least affected by the presence of extreme observations.
1111111111.....33333 AAAAARRRRRIIIIITTTTTHHHHHMMMMMEEEEETTTTTIIIIICCCCC MMMMMEEEEEAAAAANNNNN
For a given set of observations, the AM may be defined as the sum of all the observations to be
divided by the number of observations. Thus, if a variable x assumes n values x , x , x ,………..x ,
1 2 3 n
then the AM of x, to be denoted by X, is given by,
(cid:2) (cid:2) (cid:2) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:2)
+ + + +
(cid:4) (cid:7) (cid:6) (cid:5) (cid:1)
= (cid:1)
(cid:1)
∑(cid:2)
(cid:8)
= (cid:8)=(cid:7)
(cid:1)
∑(cid:2)
(cid:4) (cid:8)
= ……………………..(11.1)
(cid:1)
In case of a simple frequency distribution relating to an attribute, we have
(cid:9)(cid:2) (cid:9) (cid:2) (cid:9) (cid:2) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:9) (cid:2)
+ + + +
(cid:2) (cid:7) (cid:7) (cid:6) (cid:6) (cid:5) (cid:5) (cid:1) (cid:1)
=
(cid:9) (cid:9) (cid:9) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:9)
+ + + +
(cid:7) (cid:6) (cid:5) (cid:1)
∑(cid:9)(cid:2)
(cid:8) (cid:8)
= ∑(cid:9)
(cid:8)
∑(cid:9)(cid:2)
(cid:4) (cid:8) (cid:8)
= ……………………..(11.2)
(cid:10)
Assuming the observation x occurs f times, i=1,2,3,……..n and N=≤f
i i i
In case of grouped frequency distribution also we may use formula (11.2) with x as the mid
i
value of the i-th class interval, on the assumption that all the values belonging to the i-th
class interval are equal to x.
i
However, in most cases, if the classification is uniform, we consider the following formula
for the computation of AM from grouped frequency distribution:
∑fd
…………………………..(11.3)
x =A + i i ×C
N
(cid:2) (cid:12)
−
Where, (cid:13) (cid:8) = (cid:8) (cid:11)
A = Assumed Mean
C = Class Length
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:20)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnnsssss
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....11111::::: Following are the daily wages in rupees of a sample of 9 workers: 58, 62, 48, 53,
70, 52, 60, 84, 75. Compute the mean wage.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let x denote the daily wage in rupees.
Then as given, x =58, x =62, x =48, x =53, x =70 x =52, x =60, x =84 and x =75.
1 2 3 4 5 , 6 7 8 9
Applying (11.1) the mean wage is given by,
9
∑ x
i
x= i=1
9
(cid:22)(cid:17)(cid:18) (cid:20)(cid:6) (cid:19)(cid:18) (cid:17)(cid:5) (cid:16)(cid:21) (cid:17)(cid:6) (cid:20)(cid:21) (cid:18)(cid:19) (cid:16)(cid:17)(cid:15)
+ + + + + + + +
= Rs. (cid:14)
(cid:17)(cid:20)(cid:6)
= Rs. (cid:14)
= Rs. 62.44.
EEEEExxxxxaaaaammmmmpppppllllleeeee..... 1111111111.....22222::::: Compute the mean weight of a group of BBA students of St. Xavier’s College
from the following data :
Weight in kgs. 44 – 48 49 – 53 54 – 58 59 – 63 64 – 68 69 – 73
No. of Students 3 4 5 7 9 8
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Computation of mean weight of 36 BBA students
No. of
Weight in kgs. Student (f1) Mid-Value (x) fx
i i i
(1) (2) (3) (4) = (2) x (3)
44 – 48 3 46 138
49 – 53 4 51 204
54 – 58 5 56 280
59 – 63 7 61 427
64 – 68 9 66 594
69 – 73 8 71 568
Total 36 – 2211
Applying (11.2), we get the average weight as
∑(cid:9)(cid:2)
(cid:2) (cid:8) (cid:8)
=
(cid:10)
(cid:6)(cid:6)(cid:7)(cid:7)
= (cid:5)(cid:20) kgs.
= 61.42 kgs.
(cid:10)(cid:10)(cid:11)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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EEEEExxxxxaaaaammmmmpppppllllleeeee..... 1111111111.....33333::::: Find the AM for the following distribution:
Class Interval 350 – 369 370 – 389 390 – 409 410 – 429 430 – 449 450 – 469 470 – 489
Frequency 23 38 58 82 65 31 11
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We apply formula (11.3) since the amount of computation involved in finding the
AM is much more compared to EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....22222. Any mid value can be taken as A. However,
usually A is taken as the middle most mid-value for an odd number of class intervals and any
one of the two middle most mid-values for an even number of class intervals. The class length
is taken as C.
TTTTTaaaaabbbbbllllleeeee 1111111111.....22222 CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff AAAAAMMMMM
(cid:2) (cid:12)
−
(cid:13) (cid:8)
Class Interval Frequency(f) Mid-Value(x) (cid:8) = (cid:23) fd
i i i i
(cid:2) (cid:19)(cid:7)(cid:14)(cid:3)(cid:17)(cid:21)
−
(cid:8)
= (cid:6)(cid:21)
(1) (2) (3) (4) (5) = (2)X(4)
350 – 369 23 359.50 – 3 – 69
370 – 389 38 379.50 – 2 – 76
390 – 409 58 399.50 – 1 – 58
410 – 429 82 419.50 (A) 0 0
430 – 449 65 439.50 1 65
450 – 469 31 459.50 2 62
470 – 489 11 479.50 3 33
Total 308 – – – 43
The required AM is given by
∑(cid:9)(cid:13)
(cid:2)
=
(cid:12)
+
(cid:8) (cid:8)
×
(cid:11)
(cid:10)
((cid:24)(cid:19)(cid:5))
= 419.50 + (cid:5)(cid:21)(cid:18) ×20
= 419.50 – 2.79
= 416.71
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:22)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
EEEEExxxxxaaaaammmmmpppppllllleeeee..... 1111111111.....44444::::: Given that the mean height of a group of students is 67.45 inches. Find the
missing frequencies for the following incomplete distribution of height of 100 students.
Height in inches 60 – 62 63 – 65 66 – 68 69 – 71 72 – 74
No. of Students 5 18 – – 8
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Let x denote the height and f and f as the two missing frequencies.
3 4
TTTTTaaaaabbbbbllllleeeee 1111111111.....33333
EEEEEssssstttttiiiiimmmmmaaaaatttttiiiiiooooonnnnn ooooofffff mmmmmiiiiissssssssssiiiiinnnnnggggg fffffrrrrreeeeeqqqqquuuuueeeeennnnnccccciiiiieeeeesssss.....
(cid:2) (cid:12)
−
(cid:13) (cid:8)
CI Frequency Mid - Value (x) (cid:8) = (cid:23) fd
i i i
(cid:2) (cid:20)(cid:16)
−
(cid:8)
(f) (cid:5)
i
(1) (2) (3) (4) (5) = (2) x (4)
60-62 5 61 -2 -10
63 – 65 18 64 – 1 – 18
66 – 68 f 67 (A) 0 0
3
69 – 71 f 70 1 f
4 4
72 – 74 8 73 2 16
Total 31+ f + f – – – 12+f
3 4 4
As given, we have
(cid:5)(cid:7) (cid:9) (cid:9) (cid:7)(cid:21)(cid:21)
+ + =
(cid:5) (cid:19)
⇒ (cid:9) + (cid:9) = (cid:20)(cid:14)(cid:25) ………………………………..(1)
(cid:5) (cid:19)
and x=67.45
∑(cid:9)(cid:13)
⇒ (cid:12) + (cid:10) (cid:8) (cid:8) × (cid:11) = (cid:20)(cid:16)(cid:3)(cid:19)(cid:17)
(cid:22) (cid:7)(cid:6) (cid:9) (cid:15)
− +
⇒ (cid:20)(cid:16) + (cid:7)(cid:21)(cid:21) (cid:19) × (cid:5) = (cid:20)(cid:16)(cid:3)(cid:19)(cid:17)
⇒ ((cid:24)(cid:7)(cid:6)(cid:25) + (cid:25)(cid:9) )× (cid:5) =((cid:20)(cid:16)(cid:3)(cid:19)(cid:17)(cid:24)(cid:20)(cid:16))× (cid:7)(cid:21)(cid:21)
(cid:19)
⇒ (cid:24)(cid:7)(cid:6) + (cid:9) = (cid:7)(cid:17)
(cid:19)
⇒ f =27
4
On substituting 27 for f in (1), we get
4
(cid:9)
+
(cid:6)(cid:16)
=
(cid:20)(cid:14)
⇒
(cid:9) (cid:25)
=
(cid:25)(cid:19)(cid:6)
(cid:5) (cid:5)
Thus, the missing frequencies would be 42 and 27.
(cid:10)(cid:10)(cid:11)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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PPPPPrrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff AAAAAMMMMM
(i) IIIIIfffff aaaaallllllllll ttttthhhhheeeee ooooobbbbbssssseeeeerrrrrvvvvvaaaaatttttiiiiiooooonnnnnsssss aaaaassssssssssuuuuummmmmeeeeeddddd bbbbbyyyyy aaaaa vvvvvaaaaarrrrriiiiiaaaaabbbbbllllleeeee aaaaarrrrreeeee cccccooooonnnnnssssstttttaaaaannnnntttttsssss,,,,, sssssaaaaayyyyy kkkkk,,,,, ttttthhhhheeeeennnnn ttttthhhhheeeee AAAAAMMMMM iiiiisssss
aaaaalllllsssssooooo kkkkk..... For example, if the height of every student in a group of 10 students is 170 cm,
then the mean height is, of course, 170 cm.
(ii) the algebraic sum of deviations of a set of observations from their AM is zero
i.e. for unclassified data ,
∑(cid:22)(cid:2)
−
(cid:2)(cid:15)
=
(cid:21) }
(cid:8)
.........(11.4)
and for grouped frequency distribution,
∑(cid:9)(cid:22)(cid:2)
−
(cid:2)(cid:15)
=
(cid:21)
(cid:8) (cid:8)
For example, if a variable x assumes five observations, say 58,63,37,45,29, then x=46.4.
Hence, the deviations of the observations from the AM i.e. (x −x) are 11.60, 16.60, –9.40,
i
–1.40 and –17.40, then ∑(x −x)=11.60 + 16.60 + (–9.40) + (–1.40) + (–17.40) = 0 .
i
(iii) AM is affected due to a change of origin and/or scale which implies that if the original
variable x is changed to another variable y by effecting a change of origin, say a, and
scale say b, of x i.e. y=a+bx, then the AM of y is given by y =a+bx.
For example, if it is known that two variables x and y are related by 2x+3y+7=0 and
(cid:16) (cid:6)(cid:2)
(cid:26) − −
x=15, then the AM of y is given by = (cid:5)
(cid:16) (cid:6) (cid:7)(cid:17) (cid:5)(cid:16)
− − × − (cid:7)(cid:6)(cid:3)(cid:5)(cid:5)
= (cid:5) = (cid:5) =− .
(iv) If there are two groups containing n and n observations and x and x as the respective
1 2 1 2
arithmetic means, then the combined AM is given by
(cid:1) (cid:2) (cid:1) (cid:2)
+
(cid:2) (cid:7) (cid:7) (cid:6) (cid:6)
= (cid:1) (cid:1) ………………………………(11.5)
+
(cid:7) (cid:6)
This property could be extended to k(72) groups and we may write
∑(cid:1) (cid:2)
(cid:2) (cid:8) (cid:8)
=
∑(cid:1) ……………………………….(11.6)
(cid:8)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....55555 ::::: The mean salary for a group of 40 female workers is Rs.5200 per month and
that for a group of 60 male workers is Rs.6800 per month. What is the combined salary?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: As given n = 40, n = 60, (cid:2)(cid:25) = Rs.5200 and (cid:2)(cid:25) = Rs.6800 hence, the combined
(cid:7) (cid:6)
1 2
mean salary per month is
(cid:1) (cid:2) (cid:1) (cid:2)
+
(cid:2) (cid:7) (cid:7) (cid:6) (cid:6)
=
(cid:1) (cid:1)
+
(cid:7) (cid:6)
(cid:19)(cid:21) (cid:27)(cid:28)(cid:3)(cid:25)(cid:25)(cid:17)(cid:6)(cid:21)(cid:21) (cid:20)(cid:21) (cid:27)(cid:28)(cid:3)(cid:25)(cid:25)(cid:20)(cid:18)(cid:21)(cid:21)
× + ×
= (cid:19)(cid:21) (cid:20)(cid:21) = Rs.6160.
+
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:24)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
1111111111.....44444 MMMMMEEEEEDDDDDIIIIIAAAAANNNNN ––––– PPPPPAAAAARRRRRTTTTTIIIIITTTTTIIIIIOOOOONNNNN VVVVVAAAAALLLLLUUUUUEEEEESSSSS
As compared to AM, median is a positional average which means that the value of the median
is dependent upon the position of the given set of observations for which the median is wanted.
Median, for a given set of observations, may be defined as the middle-most value when the
observations are arranged either in an ascending order or a descending order of magnitude.
As for example, if the marks of the 7 students are 72, 85,56,80,65,52 and 68, then in order to
find the median mark, we arrange these observations in the following ascending order of
magnitude: 52, 56, 65, 68, 72, 80, 85.
Since the 4th term i.e. 68 in this new arrangement is the middle most value, the median mark is
68 i.e. Me= 68.
As a second example, if the wages of 8 workers, expressed in rupees are
56, 82, 96, 120, 110, 82, 106, 100 then arranging the wages as before, in an ascending order of
magnitude, we get Rs.56, Rs.82, Rs.82, Rs.96, Rs.100, Rs.106, Rs.110, Rs.120. Since there are
two middle-most values, namely, Rs.96, and Rs.100 any value between Rs.96 and Rs.100 may
be, theoretically, regarded as median wage. However, to bring uniqueness, we take the arithmetic
mean of the two middle-most values, whenever the number of the observations is an even
number. Thus, the median wage in this example, would be
Rs. 96 + Rs. 100
M = =Rs. 98
2
In case of a grouped frequency distribution, we find median from the cumulative frequency
distribution of the variable under consideration. We may consider the following formula, which
can be derived from the basic definition of median.
N
−N
M =l + 2 l ×C……………………………………………(11.7)
1 N −N
u l
Where,
l = lower class boundary of the median class i.e. the class containing median.
1
N = total frequency.
N = less than cumulative frequency corresponding to l.
l 1
N = less than cumulative frequency corresponding to l.
u 2
l being the upper class boundary of the median class.
2
C = l – l = length of the median class.
2 1
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....66666 ::::: Compute the median for the distribution as given in EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....33333.....
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: First, we find the cumulative frequency distribution which is exhibited in
TTTTTaaaaabbbbbllllleeeee 1111111111.....44444.....
(cid:10)(cid:10)(cid:11)(cid:25) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
TTTTTaaaaabbbbbllllleeeee 1111111111.....44444
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff MMMMMeeeeedddddiiiiiaaaaannnnn
Less than
Class boundary cumulative
frequency
349.50 0
369.50 23
389.50 61
409.50 (l) 119 (N)
1 l
429.50 (l) 201(N )
2 u
449.50 266
469.50 297
489.50 308
(cid:10) (cid:5)(cid:21)(cid:18)
We find, from the TTTTTaaaaabbbbbllllleeeee 1111111111.....44444, (cid:6) = (cid:6) = 154 lies between the two cumulative frequencies
119 and 201 i.e. 119 < 154 < 201 . Thus, we have N = 119, N = 201 l = 409.50 and l =
l u 1 2
429.50. Hence C = 429.50 – 409.50 =20.
Substituting these values in (11.7), we get,
(cid:7)(cid:17)(cid:19)(cid:24)(cid:7)(cid:7)(cid:14)
(cid:19)(cid:21)(cid:14)(cid:3)(cid:17)(cid:21) (cid:6)(cid:21)
M = + (cid:6)(cid:21)(cid:7)(cid:24)(cid:7)(cid:7)(cid:14) ×
= 409.50+8.54
= 418.04.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....77777::::: Find the missing frequency from the following data, given that the median mark
is 23.
Mark : 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
No. of students: 5 8 ? 6 3
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: Let us denote the missing frequency by f . Table 11.5 shows the relevant computation.
3
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:26)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
TTTTTaaaaabbbbbllllleeeee 1111111111.....55555
(((((EEEEEssssstttttiiiiimmmmmaaaaatttttiiiiiooooonnnnn ooooofffff mmmmmiiiiissssssssssiiiiinnnnnggggg fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy)))))
Less than
Mark cumulative frequency
0 0
10 5
20(l) 13(N)
1 l
30(l) 13+f (N )
2 3 u
40 19+f
3
50 22+f
3
Going through the mark column, we find that 20<23<30. Hence l=20, l =30 and accordingly
1 2
N=13, N =13+f . Also the total frequency i.e. N is 22+f . Thus,
l u 3 3
N
−N
M =l + 2 l ×C
1 N −N
u l
22+f
3 –13
⇒ 23 = 20+ 2 ×10
(13+f )–13
3
(cid:6)(cid:6) (cid:9) (cid:24)(cid:6)(cid:20)
+
(cid:5) (cid:5) (cid:17)
⇒ = (cid:9) ×
(cid:5)
⇒ (cid:5)(cid:9) = (cid:17)(cid:9) (cid:24)(cid:6)(cid:21)
(cid:5) (cid:5)
⇒ (cid:6)(cid:9) = (cid:6)(cid:21)
(cid:5)
⇒ (cid:9) = (cid:7)(cid:21)
(cid:5)
So, the missing frequency is 10.
PPPPPrrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff mmmmmeeeeedddddiiiiiaaaaannnnn
We cannot treat median mathematically, the way we can do with arithmetic mean. We consider
below two important features of median.
(i) If x and y are two variables, to be related by y=a+bx for any two constants a and b,
then the median of y is given by
y = a + bx
me me
For example, if the relationship between x and y is given by 2x – 5y = 10 and if x
me
i.e. the median of x is known to be 16.
Then 2x – 5y = 10
(cid:10)(cid:10)(cid:11)(cid:10)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
⇒ y = –2 + 0.40x
⇒ y = –2 + 0.40 x
me me
⇒ y = –2 + 0.40×16
me
⇒ y = 4.40.
me
(ii) For a set of observations, the sum of absolute deviations is minimum when the deviations
are taken from the median. This property states that ∑|x–A| is minimum if we choose
i
A as the median.
PPPPPAAAAARRRRRTTTTTIIIIITTTTTIIIIIOOOOONNNNN VVVVVAAAAALLLLLUUUUUEEEEESSSSS OOOOORRRRR QQQQQUUUUUAAAAARRRRRTTTTTIIIIILLLLLEEEEESSSSS OOOOORRRRR FFFFFRRRRRAAAAACCCCCTTTTTIIIIILLLLLEEEEESSSSS
These may be defined as values dividing a given set of observations into a number of equal
parts. When we want to divide the given set of observations into two equal parts, we consider
median. Similarly, quartiles are values dividing a given set of observations into four equal
parts. So there are three quartiles – first quartile or lower quartile to be denoted by Q second
1,
quartile or median to be denoted by Q or Me and third quartile or upper quartile to be denoted
2
by Q . First quartile is the value for which one fourth of the observations are less than or equal
3
to Q and the remaining three – fourths observations are more than or equal to Q . In a similar
1 1
manner, we may define Q and Q .
2 3
Deciles are the values dividing a given set of observation into ten equal parts. Thus, there are
nine deciles to be denoted by D , D , D ,…..D . D is the value for which one-tenth of the given
1 2 3 9 1
observations are less than or equal to D and the remaining nine-tenth observations are greater
1
than or equal to D when the observations are arranged in an ascending order of magnitude.
1
Lastly, we talk about the percentiles or centiles that divide a given set of observations into 100
equal parts. The points of sub-divisions being P , P ,………..P . P is the value for which one
1 2 99 1
hundredth of the observations are less than or equal to P and the remaining ninety-nine
1
hundredths observations are greater than or equal to P once the observations are arranged in
1
an ascending order of magnitude.
For unclassified data, the pth quartile is given by the (n+1)pth value, where n denotes the total
number of observations. p = 1/4, 2/4, 3/4 for Q , Q and Q respectively. p=1/10, 2/
1 2 3
10,………….9/10. For D , D ,……,D respectively and lastly p=1/100, 2/100,….,99/100 for
1 2 9
P , P , P ….P respectively.
1 2 3 99
In case of a grouped frequency distribution, we consider the following formula for the
computation of quartiles.
Np −N
Q =l + l ×C …………………………………………… (11.8)
1 N −N
u l
The symbols, except p, have their usual interpretation which we have already discussed while
computing median and just like the unclassified data, we assign different values to p depending
on the quartile.
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:10)(cid:10)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Another way to find quartiles for a grouped frequency distribution is to draw the ogive (less
than type) for the given distribution. In order to find a particular quartile, we draw a line
parallel to the horizontal axis through the point Np. We draw perpendicular from the point of
intersection of this parallel line and the ogive. The x-value of this perpendicular line gives us
the value of the quartile under discussion.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....88888::::: Following are the wages of the labourers: Rs.82, Rs.56, Rs.90, Rs.50, Rs.120,
Rs.75, Rs.75, Rs.80, Rs.130, Rs.65. Find Q , D and P .
1 6 82
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Arranging the wages in an ascending order, we get Rs.50, Rs.56, Rs.65, Rs.75,
Rs.75, Rs.80, Rs.82, Rs.90, Rs.120, Rs.130.
Hence, we have
(cid:22)(cid:1) (cid:7)(cid:15)
+
$ "#(cid:25)(cid:25)(cid:29)(cid:30)(cid:31) !
=
(cid:7) (cid:19)
((cid:7)(cid:21)
+
(cid:7))
"#(cid:25)(cid:29)(cid:30)(cid:31) !
= (cid:19)
= 2.75th value
= 2nd value + 0.75 × difference between the third and the 2nd values.
= Rs. [56 + 0.75 × (65 – 56)]
= Rs. 62.75
(cid:20)
D = (10 + 1) × (cid:7)(cid:21) th value
6
= 6.60th value
= 6th value + 0.60 × difference between the 7th and the 6th values.
= Rs. (80 + 0.60 × 2)
= Rs. 81.20
(cid:18)(cid:6)
% (cid:22)(cid:7)(cid:21) (cid:7)(cid:15)
(cid:18)(cid:6) = + × (cid:7)(cid:21)(cid:21) th value
= 9.02th value
= 9th value + 0.02 × difference between the 10th and the 9th values
= Rs. (120 + 0.02 ×10)
= Rs.120.20
Next, let us consider one problem relating to the grouped frequency distribution.
(cid:10)(cid:10)(cid:11)(cid:10)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....99999::::: Following distribution relates to the distribution of monthly wages of 100 workers.
Wages in Rs. : less than more than
500 500–699 700–899 900–1099 1100–1499 1500
No. of workers : 5 23 29 27 10 6
Compute Q , D and P .
3 7 23
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: This is a typical example of an open end unequal classification as we find the lower
class limit of the first class interval and the upper class limit of the last class interval are not
stated, and theoretically, they can assume any value between 0 and 500 and 1500 to any
number respectively. The ideal measure of the central tendency in such a situation in median
as the median or second quartile is based on the fifty percent central values. Denoting the first
LCB and the last UCB by the L and U respectively, we construct the following cumulative
frequency distribution:
TTTTTaaaaabbbbbllllleeeee 1111111111.....77777
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff qqqqquuuuuaaaaarrrrrtttttiiiiillllleeeeesssss
Wages in rupees No. of workers
(CB) ( less than cumulative
frequency)
L 0
499.50 5
699.50 28
899.50 57
1099.50 84
1499.50 94
U 100
(cid:5)(cid:10) (cid:5) (cid:7)(cid:21)(cid:21)
× (cid:16)(cid:17)
For Q , (cid:19) = (cid:19) =
3
since, 57<75 <84, we take N = 57, N =84, l=899.50, l=1099.50, c = l–l = 200
l u 1 2 2 1
in the formula (11.8) for computing Q .
3
[ (cid:16)(cid:17) (cid:17)(cid:16) ]
(cid:18)(cid:14)(cid:14)(cid:3)(cid:17)(cid:21) − (cid:6)(cid:21)(cid:21)
Therefore, Q = Rs. + (cid:18)(cid:19) (cid:17)(cid:16) × =Rs.1032.83
3 −
(cid:16)(cid:10) (cid:16) (cid:7)(cid:21)(cid:21)
×
Similarly, for D , (cid:7)(cid:21) = (cid:7)(cid:21) = 70 which also lies between 57 and 84.
7
[ (cid:16)(cid:21) (cid:17)(cid:16) ]
& (cid:27)(cid:28)(cid:3) (cid:18)(cid:14)(cid:14)(cid:3)(cid:17)(cid:21) − (cid:6)(cid:21)(cid:21)
Thus, (cid:16) = + (cid:18)(cid:19) − (cid:17)(cid:16) × = Rs.995.80
(cid:6)(cid:5)(cid:10) (cid:6)(cid:5)
Lastly for P , (cid:7)(cid:21)(cid:21) = (cid:7)(cid:21)(cid:21) × 100 = 23 and as 5 < 23 < 28, we have
23
(cid:6)(cid:5)(cid:24)(cid:17)
P = Rs. [499.50 + (cid:6)(cid:18)(cid:24)(cid:17) × 200 ]
23
= Rs. 656.02
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:10)(cid:20)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
1111111111.....55555 MMMMMOOOOODDDDDEEEEE
For a given set of observations, mode may be defined as the value that occurs the maximum
number of times. Thus, mode is that value which has the maximum concentration of the
observations around it. This can also be described as the most common value with which,
even, a layman may be familiar with.
Thus, if the observations are 5, 3, 8, 9, 5 and 6, then Mo=5 as it occurs twice and all the other
observations occur just once. The definition for mode also leaves scope for more than one
mode. Thus sometimes we may come across a distribution having more than one mode. Such
a distribution is known as a multi-modal distribution. Bi-modal distribution is one having two
mode.
Furthermore, it also appears from the definition that mode is not always defined. As an example,
if the marks of 5 students are 50, 60, 35, 40, 56, there is no modal mark as all the observations
occur once i.e. the same number of times.
We may consider the following formula for computing mode from a grouped frequency
distribution:
f −f
0 −1 ×c
Mode =l + ……………………….(11.9)
1 2f 0 −f −1 −f 1
where,
l = LCB of the modal class.
1
i.e. the class containing mode.
f = frequency of the modal class
0
f = frequency of the pre – modal class
–1
f = frequency of the post modal class
1
C = class length of the modal class
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111100000::::: Compute mode for the distribution as described in Example. 11.3
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: The frequency distribution is shown below
TTTTTaaaaabbbbbllllleeeee 1111111111.....88888
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff mmmmmooooodddddeeeee
Class Interval Frequency
350 - 369 23
370 - 389 38
390 - 409 58 (f )
–1
410 - 429 82 (f)
0
430 - 449 65 (f)
1
450 - 469 31
470 - 489 11
Going through the frequency column, we note that the highest frequency i.e. f is 82. Hence, f
0 –1
(cid:10)(cid:10)(cid:11)(cid:10)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
= 58 and f = 65. Also the modal class i.e. the class against the highest frequency is 410 – 429.
1
Thus l = LCB=409.50 and c=429.50 – 409.50 = 20
1
Hence, applying formulas (11.9), we get
(cid:18)(cid:6) (cid:17)(cid:18)
’( (cid:19)(cid:21)(cid:14)(cid:3)(cid:17) − (cid:6)(cid:21)
= + ×
(cid:6) (cid:18)(cid:6) (cid:17)(cid:18) (cid:20)(cid:17)
× − −
= 421.21 which belongs to the modal class. (410 – 429)
When it is difficult to compute mode from a grouped frequency distribution, we may consider
the following empirical relationship between mean, median and mode:
Mean – Mode = 3(Mean – Median) …………………….(11.9A)
(11.9A) holds for a moderately skewed distribution. We also note that if y = a+bx, then
y =a+bx …………………………………….(11.10)
mo mo
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111111111::::: For a moderately skewed distribution of marks in statistics for a group of 200
students, the mean mark and median mark were found to be 55.60 and 52.40. What is the
modal mark?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since in this case, mean = 55.60 and median = 52.40, applying (11.9A), we get the
modal mark as
Mode = 3 × Median – 2 × Mean
= 3 × 52.40 – 2 × 55.60
= 46.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111122222::::: If y = 2 + 1.50x and mode of x is 15, what is the mode of y?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
By virtue of (11.10), we have
y = 2 + 1.50 × 15
mo
= 24.50.
1111111111.....66666 GGGGGEEEEEOOOOOMMMMMEEEEETTTTTRRRRRIIIIICCCCC MMMMMEEEEEAAAAANNNNN AAAAANNNNNDDDDD HHHHHAAAAARRRRRMMMMMOOOOONNNNNIIIIICCCCC MMMMMEEEEEAAAAANNNNN
For a given set of n positive observations, the geometric mean is defined as the n-th root of the
product of the observations. Thus if a variable x assumes n values x , x , x ,……….., x , all the
1 2 3 n
values being positive, then the GM of x is given by
G= (x × x × x ……….. × x )1/n .................................................(11.11)
1 2 3 n
For a grouped frequency distribution, the GM is given by
G= (x f 1 × x f 2 × x f 3 …………….. × x f n )1/N .................................................(11.12)
1 2 3 n
Where N = ∑f
i
In connection with GM, we may note the following properties :
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:10)(cid:22)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
(i) Logarithm of G for a set of observations is the Am of the logarithm of the observations; i.e.
logG=1/rΣlogx ……………………………………..(11.13)
i
(ii) If all the observations assumed by a variable are constants, say K(70), then the GM of the
observations is also K.
(cid:1)
(iii) GM of the product of two variables is the product of their GM‘s i.e. if = xy, then
(cid:1)
GM of = (GM of x) × (GM of y) ……………………………………..(11.14)
(cid:1)
(iv) GM of the ratio of two variables is the ratio of the GM’s of the two variables i.e. if = x/y
then
)’(cid:25)(cid:25)((cid:9)(cid:25)(cid:25)(cid:2)
GM of
(cid:1)=
)’(cid:25)(cid:25)((cid:9)(cid:25)(cid:25)(cid:26) ……………………………………..(11.15)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111133333::::: Find the GM of 3, 6 and 12.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: As given x =3, x =6, x =12 and n=3.
1 2 3
Applying (11.11), we have G= (3×6×12) 1/3 = (63)1/3=6.
EEEEExxxxxaaaaammmmmpppppllllleeeee..... 1111111111.....1111144444::::: Find the GM for the following distribution:
x : 2 4 8 16
f : 2 3 3 2
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: According to (11.12) , the GM is given by
G = (xf 1×xf2×xf3×xf4)1/N
1 2 3 4
= (22 × 43 × 83 × 162 ) 1/10
= (2)2.50
= 4 2
= 5.66
HHHHHaaaaarrrrrmmmmmooooonnnnniiiiiccccc MMMMMeeeeeaaaaannnnn
For a given set of non-zero observations, harmonic mean is defined as the reciprocal of the AM
of the reciprocals of the observation. So, if a variable x assumes n non-zero values x , x ,
1 2
x ,……………,x then the HM of x is given by
3 n,
n
H=
∑(1/x )
i
(cid:10)(cid:10)(cid:11)(cid:10)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
For a grouped frequency distribution, we have
N
H=
f
∑ i
x
i
PPPPPrrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff HHHHHMMMMM
(i) If all the observations taken by a variable are constants, say x, then the HM of the
observations is also x.
(ii) If there are two groups with n and n observations and H and H as respective HM’s
1 2 1 2
than the combined HM is given by
(cid:1) (cid:1)
+
(cid:7) (cid:6)
(cid:1) (cid:1) ………………(11.18)
(cid:7) (cid:6)
+
* *
(cid:7) (cid:6)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111155555::::: Find the HM for 4, 6 and 10.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Applying (11.16), we have
(cid:5)
*
=
(cid:7) (cid:7) (cid:7)
+ +
(cid:19) (cid:20) (cid:7)(cid:21)
(cid:5)
=
(cid:21)(cid:3)(cid:6)(cid:17) (cid:21)(cid:3)(cid:7)(cid:16) (cid:21)(cid:3)(cid:7)(cid:21)
+ +
(cid:25)(cid:17)(cid:3)(cid:16)(cid:16)
=
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111166666::::: Find the HM for the following data:
X: 2 4 8 16
f: 2 3 3 2
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Using (11.17), we get
(cid:7)(cid:21)
*
=
(cid:6) (cid:5) (cid:5) (cid:6)
+ + +
(cid:6) (cid:19) (cid:18) (cid:7)(cid:20)
= 4.44
RRRRReeeeelllllaaaaatttttiiiiiooooonnnnn bbbbbeeeeetttttwwwwweeeeeeeeeennnnn AAAAAMMMMM,,,,, GGGGGMMMMM,,,,, aaaaannnnnddddd HHHHHMMMMM
For any set of positive observations, we have the following inequality:
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:10)(cid:24)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
AM ≥ GM ≥ HM …………………….. (11.19)
The equality sign occurs, as we have already seen, when all the observations are equal.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111177777::::: compute AM, GM, and HM for the numbers 6, 8, 12, 36.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: In accordance with the definition, we have
(cid:20) (cid:18) (cid:7)(cid:6) (cid:5)(cid:20)
(cid:12)’
=
+ + +
=
(cid:7)(cid:17)(cid:3)(cid:17)0
(cid:19)
GM = (6 × 8 × 12 × 36)11111/////44444
= (28 × 34)1/4 =12
(cid:19)
*’ (cid:14)(cid:3)(cid:14)(cid:5)
= =
(cid:7) (cid:7) (cid:7) (cid:7)
+ + +
(cid:20) (cid:18) (cid:7)(cid:6) (cid:5)(cid:20)
The computed values of AM, GM, and HM establish (11.19).
WWWWWeeeeeiiiiiggggghhhhhttttteeeeeddddd aaaaavvvvveeeeerrrrraaaaagggggeeeee
When the observations under consideration have a hierarchical order of importance, we take
recourse to computing weighted average, which could be either weighted AM or weighted
GM or weighted HM.
∑wx
i i
Weighted AM = ………………………..………(11.20)
∑w
i
∑w logx
i i
Weighted GM = Ante log ………………………..………(11.21)
∑w
i
∑w
i
w
Weighted HM = ∑ i ………………………..………(11.22)
x
i
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111188888::::: Find the weighted AM and weighted HM of first n natural numbers, the
weights being equal to the squares of the Corresponding numbers.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: As given,
x 1 2 3 …. n
w 12 22 32 …. n2
∑wx
i i
Weighted AM =
∑w
i
(cid:10)(cid:10)(cid:11)(cid:10)(cid:25) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
(cid:7) (cid:7)(cid:6) (cid:6) (cid:6)(cid:6) (cid:5) (cid:5)(cid:6) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:1) (cid:1)(cid:6)
× + × + × + ×
= (cid:7)(cid:6) (cid:6)(cid:6) (cid:5)(cid:6) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)(cid:6)
+ + + +
(cid:7)(cid:5) (cid:6)(cid:5) (cid:5)(cid:5) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)(cid:5)
+ + + +
= (cid:7)(cid:6) (cid:6)(cid:6) (cid:5)(cid:6) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)(cid:6)
+ + + +
2
n(n+1)
2
=
n(n+1)(2n+1)
6
(cid:5)(cid:1)(cid:22)(cid:1) (cid:7)(cid:15)
+
= (cid:6)(cid:22)(cid:6)(cid:1) (cid:7)(cid:15)
+
∑w
i
w
Weighted HM =∑ i
x
i
(cid:7)(cid:6) (cid:6)(cid:6) (cid:5)(cid:6) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:1)(cid:6)
+ + +
(cid:7)(cid:6) (cid:6)(cid:6) (cid:5)(cid:6) (cid:1)(cid:6)
= (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)
+ + +
(cid:7) (cid:6) (cid:5) (cid:1)
(cid:7)(cid:6) (cid:6)(cid:6) (cid:5)(cid:6) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)(cid:6)
+ + + +
=
(cid:7) (cid:6) (cid:5) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)
+ + + +
(cid:1)((cid:1)
+
(cid:7))((cid:6)(cid:1)
+
(cid:7))
(cid:20)
= (cid:1)(cid:22)(cid:1) + (cid:7)(cid:15)
(cid:6)
(cid:6)(cid:1) (cid:7)
+
= (cid:5)
AAAAA GGGGGeeeeennnnneeeeerrrrraaaaalllll rrrrreeeeevvvvviiiiieeeeewwwww ooooofffff ttttthhhhheeeee dddddiiiiiffffffffffeeeeerrrrreeeeennnnnttttt mmmmmeeeeeaaaaasssssuuuuurrrrreeeeesssss ooooofffff ccccceeeeennnnntttttrrrrraaaaalllll ttttteeeeennnnndddddeeeeennnnncccccyyyyy
After discussing the different measures of central tendency, now we are in a position to have
a review of these measures of central tendency so far as the relative merits and demerits are
concerned on the basis of the requisites of an ideal measure of central tendency which we have
already mentioned in section 11.2. The best measure of central tendency, usually, is the AM. It
is rigidly defined, based on all the observations, easy to comprehend, simple to calculate and
amenable to mathematical properties. However, AM has one drawback in the sense that it is
very much affected by sampling fluctuations. In case of frequency distribution, mean cannot
be advocated for open-end classification.
Like AM, median is also rigidly defined and easy to comprehend and compute. But median is
not based on all the observation and does not allow itself to mathematical treatment. However,
median is not much affected by sampling fluctuation and it is the most appropriate measure of
central tendency for an open-end classification.
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:10)(cid:26)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Although mode is the most popular measure of central tendency, there are cases when mode
remains undefined. Unlike mean, it has no mathematical property. Mode is also affected by
sampling fluctuations.
GM and HM, like AM, possess some mathematical properties. They are rigidly defined and
based on all the observations. But they are difficult to comprehend and compute and, as such,
have limited applications for the computation of average rates and ratios and such like things.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....1111199999 ::::: Given two positive numbers a and b, prove that AAAAAHHHHH=====GGGGG22222. Does the result hold
for any set of observations?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: For two positive numbers a and b, we have,
(cid:30) +
(cid:12) +
=
(cid:6)
G= ab
(cid:6)
*
And =
(cid:7) (cid:7)
+
(cid:30) +
(cid:6)(cid:30)+
=
(cid:30) +
+
(cid:30) + (cid:6)(cid:30)+
(cid:12)* +
Thus = (cid:6) × (cid:30) +
+
= ab = G2
No, this result holds for only two positive observations or if the observations are in arithmetical
progression.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222200000::::: The AM and GM for two observations are 5 and 4 respectively. Find the two
observations.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: If a and b are two positive observations then as given
(cid:30) +
+ (cid:17)
=
(cid:6)
⇒ a+b = 10 …………………………………..(1)
and (cid:30)+ = (cid:19)
⇒ ab = 16 …………………………………..(2)
(cid:22)(cid:30)(cid:24)+(cid:15)(cid:6) (cid:22)(cid:30) +(cid:15)(cid:6) (cid:19)(cid:30)+
∴ = + −
= (cid:7)(cid:21)(cid:6) − (cid:19) × (cid:7)(cid:20)
(cid:10)(cid:10)(cid:11)(cid:12)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
= 36
⇒ a – b = 6 (ignoring the negative sign)……………………….(3)
Adding (1) and (3) We get,
2a = 16
⇒ a = 8
From (1), we get b = 10 – a = 2
Thus, the two observations are 8 and 2.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222211111::::: Find the mean and median from the following data:
Marks : less than 10 less than 20 less than 30
No. of Students : 5 13 23
Marks : less than 40 less than 50
No. of Students : 27 30
Also compute the mode using the approximate relationship between mean, median and mode.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: What we are given in this problem is less than cumulative frequency distribution.
We need to convert this cumulative frequency distribution to the corresponding frequency
distribution and thereby compute the mean and median.
TTTTTaaaaabbbbbllllleeeee 1111111111.....99999
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff MMMMMeeeeeaaaaannnnn MMMMMaaaaarrrrrkkkkksssss fffffooooorrrrr 3333300000 ssssstttttuuuuudddddeeeeennnnntttttsssss
Marks No. of Students Mid - Value fx
i i
Class Interval (f) (x)
i i
(1) (2) (3) (4)= (2)×(3)
0 – 10 5 5 25
10 – 20 13 – 5 = 8 15 120
20 – 30 23 – 13 = 10 25 250
30 – 40 27 – 23 = 4 35 140
40 – 50 30 – 27 = 3 45 135
Total 30 – 670
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:12)(cid:10)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Hence the mean mark is given by
∑fx
x= i i
N
(cid:20)(cid:16)(cid:21)
= (cid:5)(cid:21)
= (cid:6)(cid:6)(cid:3)(cid:5)(cid:5)
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111100000
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff MMMMMeeeeedddddiiiiiaaaaannnnn MMMMMaaaaarrrrrkkkkksssss
Marks No. of Students
(Class Boundary) (Less than cumulative Frequency)
0 0
10 5
20 13
30 23
40 27
50 30
(cid:10) (cid:5)(cid:21)
(cid:7)(cid:17)
Since (cid:6) = (cid:6) = lies between 13 and 23,
we have l = 20, N = 13, N = 23
1 l u
and C = l – l = 30 – 20 = 10
2 1
Thus,
15 −13
Median =20 + ×10
23 −13
= 22
Since Mode =3 Median – 2x approximately, we find that
Mode = 3x22 – 2x22.33
=
(cid:25)(cid:6)(cid:7)(cid:3)(cid:5)(cid:19)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222222222::::: Following are the salaries of 20 workers of a firm expressed in thousand rupees:
5, 17, 12, 23, 7, 15, 4, 18, 10, 6, 15, 9, 8, 13, 12, 2, 12, 3, 15, 14. The firm gave bonus amounting
to Rs. 2000, Rs. 3000, Rs. 4000, Rs.5000 and Rs. 6000 to the workers belonging to the salary
groups 1000 – 5000, 6000 – 10000 and so on and lastly 21000 – 25000. Find the average bonus
paid per employee.
(cid:10)(cid:10)(cid:11)(cid:12)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We first construct frequency distribution of salaries paid to the 20 employees. The
∑fx
average bonus paid per employee is given by i i Where x represents the amount of bonus
N i
paid to the ith salary group and f, the number of employees belonging to that group which
i
would be obtained on the basis of frequency distribution of salaries.
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111111111
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff AAAAAvvvvveeeeerrrrraaaaagggggeeeee bbbbbooooonnnnnuuuuusssss
No of workers Bonus in Rupees
Salary in thousand Rs. Tally Mark (f ) x f x
i i i i
(Class Interval)
(1) (2) (3) (4) (5) = (3) × (4)
1-5 |||| 4 2000 8000
6-10 |||| 5 3000 15000
11-15 |||| ||| 8 4000 32000
16-20 || 2 5000 10000
21-25 | 1 6000 6000
TOTAL – 20 – 71000
Hence, the average bonus paid per employee
(cid:16)(cid:7)(cid:21)(cid:21)(cid:21)
(cid:27)(cid:28)(cid:3)
=
(cid:6)(cid:21)
Rs. = 3550
1111111111.....77777 EEEEEXXXXXEEEEERRRRRCCCCCIIIIISSSSSEEEEE
SSSSSeeeeettttt AAAAA
WWWWWrrrrriiiiittttteeeee dddddooooowwwwwnnnnn ttttthhhhheeeee cccccooooorrrrrrrrrreeeeecccccttttt aaaaannnnnssssswwwwweeeeerrrrrsssss..... EEEEEaaaaaccccchhhhh qqqqquuuuueeeeessssstttttiiiiiooooonnnnn cccccaaaaarrrrrrrrrriiiiieeeeesssss 11111 mmmmmaaaaarrrrrkkkkk.....
1. Measures of central tendency for a given set of observations measures
(i) The scatterness of the observations (ii) The central location of the observations
(iii) Both (i) and (ii) (iv) None of these.
2. While computing the AM from a grouped frequency distribution, we assume that
(i) The classes are of equal length (ii) The classes have equal frequency
(iii) All the values of a class are equal to the mid-value of that class
(iv) None of these.
3. Which of the following statements is wrong?
(i) Mean is rigidly defined
(ii) Mean is not affected due to sampling fluctuations
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:12)(cid:20)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
(iii) Mean has some mathematical properties
(iv) All these
4. Which of the following statements is true?
(i) Usually mean is the best measure of central tendency
(ii) Usually median is the best measure of central tendency
(iii) Usually mode is the best measure of central tendency
(iv) Normally, GM is the best measure of central tendency
5. For open-end classification, which of the following is the best measure of central tendency?
(i) AM (ii) GM (iii) Median (iv) Mode
6. The presence of extreme observations does not affect
(i) AM (ii) Median (iii) Mode (iv)Any of these.
7. In case of an even number of observations which of the following is median ?
(i) Any of the two middle-most value
(ii) The simple average of these two middle values
(iii) The weighted average of these two middle values
(iv) Any of these
8. The most commonly used measure of central tendency is
(i) AM (ii) Median (iii) Mode (iv) Both GM and HM.
9. Which one of the following is not uniquely defined?
(i) Mean (ii) Median (iii) Mode (iv)All of these measures
10. Which of the following measure of the central tendency is difficult to compute?
(i) Mean (ii) Median (iii) Mode (iv)GM
11. Which measure(s) of central tendency is(are) considered for finding the average rates?
(i) AM (ii) GM (iii) HM (iv)Both (ii) and(iii)
12. For a moderately skewed distribution, which of he following relationship holds?
(i) Mean – Mode = 3 (Mean – Median) (ii) Median – Mode = 3 (Mean – Median)
(iii) Mean – Median = 3 (Mean – Mode) (iv) Mean – Median = 3 (Median – Mode)
13. Weighted averages are considered when
(i) The data are not classified
(ii) The data are put in the form of grouped frequency distribution
(iii) All the observations are not of equal importance
(iv) Both (i) and (iii).
(cid:10)(cid:10)(cid:11)(cid:12)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
14. Which of the following results hold for a set of distinct positive observations?
(i) AM ≥ GM ≥ HM (ii) HM ≥ GM ≥ AM
(iii) AM > GM > HM (iv) GM > AM > HM
15. When a firm registers both profits and losses, which of the following measure of central
tendency cannot be considered?
(i) AM (ii) GM (iii) Median (iv) Mode
16. Quartiles are the values dividing a given set of observations into
(i) Two equal parts (ii) Four equal parts(iii) Five equal parts (iv) None of these.
17. Quartiles can be determined graphically using
(i) Histogram (ii) Frequency Polygon (iii) Ogive (iv) Pie chart.
18. Which of the following measure(s) possesses (possess) mathematical properties?
(i) AM (ii) GM (iii) HM (iv) All of these
19. Which of the following measure(s) satisfies (satisfy) a linear relationship between two
variables?
(i) Mean (ii) Median (iii) Mode (iv) All of these
20. Which of he following measures of central tendency is based on only fifty percent of the
central values?
(i) Mean (ii) Median (iii) Mode (iv) Both (i) and(ii)
SSSSSeeeeettttt BBBBB
WWWWWrrrrriiiiittttteeeee dddddooooowwwwwnnnnn ttttthhhhheeeee cccccooooorrrrrrrrrreeeeecccccttttt aaaaannnnnssssswwwwweeeeerrrrrsssss..... EEEEEaaaaaccccchhhhh qqqqquuuuueeeeessssstttttiiiiiooooonnnnn cccccaaaaarrrrrrrrrriiiiieeeeesssss 22222 mmmmmaaaaarrrrrkkkkksssss.....
1. If there are 3 observations 15, 20, 25 then the sum of deviation of the observations from their
AM is
(i) 0 (ii) 5 (iii) –5 (iv) None of these.
2. What is the median for the following observations?
5, 8, 6, 9, 11, 4.
(i) 6 (ii) 7 (iii) 8 (iv) None of these
3. What is the modal value for the numbers 5, 8, 6, 4, 10, 15, 18, 10?
(i) 18 (ii) 10 (iii) 14 (iv) None of these
4. What is the GM for the numbers 8, 24 and 40?
(i) 24 (ii) 12 (iii) 8 15 (iv) 10
5. The harmonic mean for the numbers 2, 3, 5 is
(i) 2.00 (ii) 3.33 (iii) 2.90 (iv) –3 30.
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:12)(cid:22)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
6. If the AM and GM for two numbers are 6.50 and 6 respectively then the two numbers are
(i) 6 and 7 (ii) 9 and 4 (iii) 10 and 3 (iii) 8 and 5.
7. If the AM and HM for two numbers are 5 and 3.2 respectively then the GM will be
(i) 16.00 (ii) 4.10 (iii) 4.05 (iv) 4.00.
8. What is the value of the first quartile for observations 15, 18, 10, 20, 23, 28, 12, 16?
(i) 17 (ii) 16 (iii) 15.75 (iv) 12
9. The third decile for the numbers 15, 10, 20, 25, 18, 11, 9, 12 is
(i) 13 (ii) 10.70 (iii) 11 (iv) 11.50
10. If there are two groups containing 30 and 20 observations and having 50 and 60 as
arithmetic means, then the combined arithmetic mean is
(i) 55 (ii) 56 (iii) 54 (iv) 52.
11. The average salary of a group of unskilled workers is Rs.10000 and that of a group of
skilled workers is Rs.15,000. If the combined salary is Rs.12000, then what is the percentage
of skilled workers?
(i) 40% (ii) 50% (iii) 60% (iv) none of these
12. If there are two groups with 75 and 65 as harmonic means and containing 15 and 13
observation then the combined HM is given by
(i) 65 (ii) 70.36 (iii) 70 (iv) 71.
13. What is the HM of 1,½, 1/3,…………….1/n?
2 n(n+1)
(i) n (ii) 2n (iii) (iv)
(n+1) 2
14. An aeroplane flies from A to B at the rate of 500 km/hour and comes back from B to A at
the rate of 700 km/hour. The average speed of the aeroplane is
(i) 600 km. per hour (ii) 583.33 km. per hour
(iii) 100 35 km. per hour (iv) 620 km. per hour.
15. If a variable assumes the values 1, 2, 3…5 with frequencies as 1, 2, 3…5, then what is the
AM?
11
(i) (ii) 5 (iii) 4 (iv) 4.50
3
16. Two variables x and y are given by y= 2x – 3. If the median of x is 20, what is the median
of y?
(i) 20 (ii) 40 (iii) 37 (iv) 35
(cid:10)(cid:10)(cid:11)(cid:12)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
17. If the relationship between two variables u and v are given by 2u + v + 7 = 0 and if the AM
of u is 10, then the AM of v is
(i) 17 (ii) –17 (iii) –27 (iv) 27.
18. If x and y are related by x–y–10 = 0 and mode of x is known to be 23, then the mode of y
is
(i) 20 (ii) 13 (iii) 3 (iv) 23.
19. If GM of x is 10 and GM of y is 15, then the GM of xy is
(i) 150 (ii) Log 10 × Log 15(iii) Log 150 (iv) None of these.
20. If the AM and GM for 10 observations are both 15, then the value of HM is
(i) Less than 15 (ii) More than 15 (iii) 15 (iv) Can not be determined.
SSSSSeeeeettttt CCCCC
WWWWWrrrrriiiiittttteeeee dddddooooowwwwwnnnnn ttttthhhhheeeee cccccooooorrrrrrrrrreeeeecccccttttt aaaaannnnnssssswwwwweeeeerrrrrsssss..... EEEEEaaaaaccccchhhhh qqqqquuuuueeeeessssstttttiiiiiooooonnnnn cccccaaaaarrrrrrrrrriiiiieeeeesssss 55555 mmmmmaaaaarrrrrkkkkksssss.....
1. What is the value of mean and median for the following data:
Marks : 5–14 15–24 25–34 35–44 45–54 55–64
No. of Student : 10 18 32 26 14 10
(i) 30 and 28 (ii) 29 and 30 (iii) 33.68 and 32.94 (iv)34.21 and 33.18
2. The mean and mode for the following frequency distribution
Class interval : 350–369 370–389 390–409 410–429 430–449 450–469
Frequency : 15 27 31 19 13 6
are
(i) 400 and 390 (ii) 400.58 and 390 (iii) 400.58 and 394.50 (iv) 400 and 394.
3. The median and modal profits for the following data
Profit in ‘000 Rs.: below 5 below 10 below 15 below 20 below 25 below 30
No. of firms: 10 25 45 55 62 65
are
(i) 11.60 and 11.50 (ii) Rs.11556 and Rs.11267
(iii) Rs.11875 and Rs.11667 (iv) 11.50 and 11.67.
4. Following is an incomplete distribution having modal mark as 44
Marks : 0–20 20–40 40–60 60–80 80–100
No. of Students : 5 18 ? 12 5
What would be the mean marks?
(i) 45 (ii) 46 (iii) 47 (iv) 48
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:12)(cid:24)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
5. The data relating to the daily wage of 20 workers are shown below:
Rs.50, Rs.55, Rs.60, Rs.58, Rs.59, Rs.72, Rs.65, Rs.68, Rs.53, Rs.50, Rs.67, Rs.58, Rs.63,
Rs.69, Rs.74, Rs.63, Rs.61, Rs.57, Rs.62, Rs.64.
The employer pays bonus amounting to Rs.100, Rs.200, Rs.300, Rs.400 and Rs.500 to the
wage earners in the wage groups Rs. 50 and not more than Rs. 55 Rs. 55 and not more
than Rs. 60 and so on and lastly Rs. 70 and not more than Rs. 75, during the festive month
of October.
What is the average bonus paid per wage earner?
(i) Rs.200 (ii) Rs.250 (iii) Rs.285 (iv) Rs.300
6. The third quartile and 65th percentile for the following data
Profits in ‘000 Rs.: les than 10 10–19 20–29 30–39 40–49 50–59
No. of firms : 5 18 38 20 9 2
are
(i) Rs.33500 and Rs.29184 (ii) Rs.33000 and Rs.28680
(iii) Rs.33600 and Rs.29000 (iv) Rs.33250 and Rs.29250.
7. For the following incomplete distribution of marks of 100 pupils, median mark is known
to be 32.
Marks : 0–10 10–20 20–30 30–40 40–50 50–60
No. of Students : 10 – 25 30 – 10
What is the mean mark?
(i) 32 (ii) 31 (iii) 31.30 (iv) 31.50
8. The mode of the following distribution is Rs. 66. What would be the median wage?
Daily wages (Rs.) : 30–40 40–50 50–60 60–70 70–80 80–90
No of workers : 8 16 22 28 – 12
(i) Rs.64.00 (ii) Rs.64.56 (iii) Rs.62.32 (iv) Rs.64.25
(cid:10)(cid:10)(cid:11)(cid:12)(cid:25) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
AAAAANNNNNSSSSSWWWWWEEEEERRRRRSSSSS
SSSSSeeeeettttt AAAAA
1 (ii) 2 (iii) 3 (ii) 4 (i) 5 (iii) 6 (ii)
7 (ii) 8 (i) 9 (iii) 10 (iv) 11 (iv) 12 (i)
13 (iii) 14 (iii) 15 (ii) 16 (ii) 17 (iii) 18 (iv)
19 (iv) 20 (ii)
SSSSSeeeeettttt BBBBB
1 (i) 2 (ii) 3 (ii) 4 (iii) 5 (iii) 6 (ii)
7 (iv) 8 (iii) 9 (ii) 10 (ii) 11 (i) 12 (ii)
13 (iii) 14 (ii) 15 (i) 16 (iii) 17 (iii) 18 (ii)
19 (i) 20 (iii)
SSSSSeeeeettttt CCCCC
1 (iii) 2 (iii) 3 (iii) 4 (iv) 5 (iv) 6 (i)
7 (iii) 8 (iii)
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:12)(cid:26)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
1111111111.....88888 DDDDDEEEEEFFFFFIIIIINNNNNIIIIITTTTTIIIIIOOOOONNNNN OOOOOFFFFF DDDDDIIIIISSSSSPPPPPEEEEERRRRRSSSSSIIIIIOOOOONNNNN
The second important characteristic of a distribution is given by dispersion. Two distributions
may be identical in respect of its first important characteristic i.e. central tendency and yet
they may differ on account of dispersion. The following figure shows a number of distributions
having identical measure of central tendency and yet varying measure of scatterness. Obviously,
distribution is having the maximum amount of dispersion.
A
B
C
FFFFFiiiiiggggguuuuurrrrreeeee 1111111111.....11111
Showing distributions with identical measure of central tendency
and varying amount of dispersion.
Dispersion for a given set of observations may be defined as the amount of deviation of the
observations, usually, from an appropriate measure of central tendency. Measures of dispersion
may be broadly classified into
1. Absolute measures of dispersion. 2. Relative measures of dispersion.
Absolute measures of dispersion are classified into
(i) Range (ii) Mean Deviation
(iii) Standard Deviation (iv) Quartile Deviation
Likewise, we have the following relative measures of dispersion :
(i) Coefficient of range. (ii) Coefficient of Mean Deviation
(iii) Coefficient of Variation (iv) Coefficient of Quartile Deviation.
We may note the following points of distinction between the absolute and relative measures of
dispersion :
I Absolute measures are dependent on the unit of the variable under consideration whereas
the relative measures of dispersion are unit free.
(cid:10)(cid:10)(cid:11)(cid:20)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
II For comparing two or more distributions, relative measures and not absolute measures of
dispersion are considered.
III Compared to absolute measures of dispersion, relative measures of dispersion are difficult
to compute and comprehend.
CCCCChhhhhaaaaarrrrraaaaacccccttttteeeeerrrrriiiiissssstttttiiiiicccccsssss fffffooooorrrrr aaaaannnnn iiiiidddddeeeeeaaaaalllll mmmmmeeeeeaaaaasssssuuuuurrrrreeeee ooooofffff dddddiiiiissssspppppeeeeerrrrrsssssiiiiiooooonnnnn
As discussed in section 11.2 an ideal measure of dispersion should be properly defined, easy to
comprehend, simple to compute, based on all the observations, unaffected by sampling
fluctuations and amenable to some desirable mathematical treatment.
1111111111.....99999 RRRRRAAAAANNNNNGGGGGEEEEE
For a given set of observations, range may be defined as the difference between the largest and
smallest observation. Thus if L and S denote the largest and smallest observations respectively
then we have
Range = L – S
The corresponding relative measure of dispersion, known as coefficient of range, is given by
- ,
− (cid:7)(cid:21)(cid:21)
Coefficient of range = - , ×
+
For a grouped frequency distribution, range is defined as the difference between the two extreme
class boundaries. The corresponding relative measure of dispersion is given by the ratio of the
difference between the two extreme class boundaries to the total of these class boundaries,
expressed as a percentage.
WWWWWeeeee mmmmmaaaaayyyyy nnnnnooooottttteeeee ttttthhhhheeeee fffffooooollllllllllooooowwwwwiiiiinnnnnggggg iiiiimmmmmpppppooooorrrrrtttttaaaaannnnnttttt rrrrreeeeesssssuuuuulllllttttt iiiiinnnnn cccccooooonnnnnnnnnneeeeeccccctttttiiiiiooooonnnnn wwwwwiiiiittttthhhhh rrrrraaaaannnnngggggeeeee:::::
RRRRReeeeesssssuuuuulllllttttt:::::
Range remains unaffected due to a change of origin but affected in the same ratio due to
a change in scale i.e., if for any two constants a and b, two variables x and y are related by
y = a + bx,
Then the range of y is given by
(cid:27) (cid:25)(cid:25)+(cid:25) (cid:27)
(cid:26) = × (cid:2)…………………………………………… (11.23)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222233333::::: Following are the wages of 8 workers expressed in rupees:
82, 96, 52, 75, 70, 65, 50, 70. Find the range and also it’s coefficient.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: The largest and the smallest wages are L = Rs.96 and S= Rs.50
Thus range = Rs.96 – Rs.50 = Rs.46
(cid:14)(cid:20) (cid:17)(cid:21)
− (cid:7)(cid:21)(cid:21)
Coefficient of range = (cid:14)(cid:20) (cid:17)(cid:21) ×
+
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:20)(cid:10)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
= 31.51
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222244444 ::::: What is the range and its coefficient for the following distribution of weights?
Weights in kgs. : 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74
No. of Students : 12 18 23 10 3
SSSSSooooollllluuuuutttttiiiiiooooonnnnn ::::: The lowest class boundary is 49.50 kgs. and the highest class boundary is 74.50 kgs.
Thus we have
Range = 74.50 kgs. – 49.50 kgs.
= 25 kgs.
74.50- 49.50
´ 100
Also, coefficient of range =
74.50+ 49.50
25
= ´ 100
124
= 20.16
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222255555 ::::: If the relationship between x and y is given by 2x+3y=10 and the range of
x is Rs. 15, what would be the range of y?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since 2x+3y=10
(cid:7)(cid:21) (cid:6)
(cid:24) (cid:2)
Therefore, y = (cid:5) (cid:5)
Applying (11.23) , the range of y is given by
(cid:27) (cid:25)+(cid:25) (cid:27)
= ×
(cid:26) (cid:2)
= 2/3 × Rs. 15
= Rs.10.
1111111111.....1111100000 MMMMMEEEEEAAAAANNNNN DDDDDEEEEEVVVVVIIIIIAAAAATTTTTIIIIIOOOOONNNNN
Since range is based on only two observations, it is not regarded as an ideal measure of dispersion.
A better measure of dispersion is provided by mean deviation which, unlike range, is based on
all the observations. For a given set of observation, mean deviation is defined as the arithmetic
mean of the absolute deviation of the observations from an appropriate measure of central
tendency. Hence if a variable x assumes n values x , x , x …x , then the mean deviation of x
1 2 3 n
about an average A is given by
(cid:10)(cid:10)(cid:11)(cid:20)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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1
MD = ∑ x −A ……………………………………….(11.24)
A n i
For a grouped frequency distribution, mean deviation about A is given by
1
MD = ∑ x −A f …………………………………....(11.25)
A n i i
Where x and f denote the mid value and frequency of the i-th class interval and
i i
N = ∑f
i
In most cases we take A as mean or median and accordingly, we get mean deviation about
mean or mean deviation about median.
A relative measure of dispersion applying mean deviation is given by
’!(cid:30)(cid:1)(cid:25)(cid:25)(cid:13)!(cid:29)(cid:8)(cid:30)"(cid:8)((cid:1)(cid:25)(cid:25)(cid:30)+( "(cid:25)(cid:25)(cid:12)
(cid:7)(cid:21)(cid:21)
Coefficient of mean deviation = (cid:12) × …………….(11.26)
MMMMMeeeeeaaaaannnnn dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn tttttaaaaakkkkkeeeeesssss iiiiitttttsssss mmmmmiiiiinnnnniiiiimmmmmuuuuummmmm vvvvvaaaaallllluuuuueeeee wwwwwhhhhheeeeennnnn ttttthhhhheeeee dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnnsssss aaaaarrrrreeeee tttttaaaaakkkkkeeeeennnnn fffffrrrrrooooommmmm ttttthhhhheeeee mmmmmeeeeedddddiiiiiaaaaannnnn.....
AAAAAlllllsssssooooo mmmmmeeeeeaaaaannnnn dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn rrrrreeeeemmmmmaaaaaiiiiinnnnnsssss uuuuunnnnnccccchhhhhaaaaannnnngggggeeeeeddddd ddddduuuuueeeee tttttooooo aaaaa ccccchhhhhaaaaannnnngggggeeeee ooooofffff ooooorrrrriiiiigggggiiiiinnnnn bbbbbuuuuuttttt ccccchhhhhaaaaannnnngggggeeeeesssss iiiiinnnnn ttttthhhhheeeee sssssaaaaammmmmeeeee
rrrrraaaaatttttiiiiiooooo ddddduuuuueeeee tttttooooo aaaaa ccccchhhhhaaaaannnnngggggeeeee iiiiinnnnn ssssscccccaaaaallllleeeee iiiii.....eeeee..... iiiiifffff yyyyy ===== aaaaa +++++ bbbbbxxxxx,,,,, aaaaa aaaaannnnnddddd bbbbb bbbbbeeeeeiiiiinnnnnggggg cccccooooonnnnnssssstttttaaaaannnnntttttsssss,,,,,
then MD of y = |b| × MD of x ………………………(11.27)
Example. 11.26 : What is the mean deviation about mean for the following numbers?
5, 8, 10, 10, 12, 9.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
The mean is given by
(cid:17) (cid:18) (cid:7)(cid:21) (cid:7)(cid:21) (cid:7)(cid:6) (cid:14)
(cid:4) + + + + +
= (cid:20) = 9
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111122222
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff MMMMMDDDDD aaaaabbbbbooooouuuuuttttt AAAAAMMMMM
x (cid:2) (cid:2)
−
i (cid:8)
5 4
8 1
10 1
10 1
12 3
9 0
Total 10
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:20)(cid:20)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Thus mean deviation about mean is given by
∑ x −x 10
i = =1.67
n 6
Example. 11.27: Find mean deviations about median and also the corresponding coefficient
for the following profits (‘000 Rs.) of a firm during a week.
82, 56, 75, 70, 52, 80, 68.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
The profits in thousand rupees is denoted by x. Arranging the values of x in an ascending
order, we get
52, 56, 68, 70, 75, 80, 82.
Therefore, Me = 70. Thus, Median profit = Rs. 70,000.
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111133333
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff MMMMMeeeeeaaaaannnnn dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn aaaaabbbbbooooouuuuuttttt mmmmmeeeeedddddiiiiiaaaaannnnn
x |x–Me|
i i
52 18
56 14
68 2
70 0
75 5
80 10
82 12
Total 61
∑ x −Median
Thus mean deviation about median = i
n
(cid:20)(cid:7)
(cid:27)(cid:28)(cid:3) (cid:7)(cid:21)(cid:21)(cid:21)
= (cid:16) ×
= (cid:27)(cid:28)(cid:3)(cid:18)(cid:16)(cid:7)(cid:19)(cid:3)(cid:6)(cid:18)
(cid:10)(cid:10)(cid:11)(cid:20)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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’&(cid:25)(cid:25)(cid:30)+( "(cid:25)(cid:25).!(cid:13)(cid:8)(cid:30)(cid:1)
(cid:7)(cid:21)(cid:21)
Coefficient of mean deviation = ’!(cid:13)(cid:8)(cid:30)(cid:1) ×
(cid:18)(cid:16)(cid:7)(cid:19)(cid:3)(cid:6)(cid:18)
(cid:7)(cid:21)(cid:21)
= (cid:16)(cid:21)(cid:21)(cid:21)(cid:21) ×
= 12.45
Example 11.28 : Compute the mean deviation about the arithmetic mean for the following data:
x : 1 3 5 7 9
f : 5 8 9 2 1
Also find the coefficient of the mean deviation about the AM.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We are to apply formula (11.25) as these data refer to a grouped frequency distribution
the AM is given by
∑fx
x = i i
N
(cid:17) (cid:7) (cid:18) (cid:5) (cid:14) (cid:17) (cid:6) (cid:16) (cid:7) (cid:14)
× + × + × + × + × (cid:5)(cid:3)(cid:18)(cid:18)
= (cid:17) (cid:18) (cid:14) (cid:6) (cid:7) =
+ + + +
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111144444
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff MMMMMDDDDD aaaaabbbbbooooouuuuuttttt ttttthhhhheeeee AAAAAMMMMM
x f (cid:2) − (cid:2) (cid:9)(cid:2) − (cid:2)
(cid:1)
(4) = (2) (3)
(1) (2) (3)
1 5 2.88 14.40
3 8 0.88 7.04
5 9 1.12 10.08
7 2 3.12 6.24
9 1 5.12 5.12
Total 25 – 42.88
Thus, MD about AM is given by
∑f x−x
N
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:20)(cid:22)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
(cid:19)(cid:6)(cid:3)(cid:18)(cid:18)
= (cid:6)(cid:17)
=1.72
’&(cid:25)(cid:25)(cid:30)+( "(cid:25)(cid:25)(cid:12)’
(cid:7)(cid:21)(cid:21)
Coefficient of MD about its AM = (cid:12)’ ×
1.72
= ×100
3.88
= 44.33
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....2222299999 ::::: Compute the coefficient of mean deviation about median for the following
distribution:
Weight in kgs. : 40-50 50-60 60-70 70-80
No. of persons : 8 12 20 10
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We need to compute the median weight in the first stage
TTTTTaaaaabbbbbllllleeeee 1111111111..... 1111155555
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff mmmmmeeeeedddddiiiiiaaaaannnnn wwwwweeeeeiiiiiggggghhhhhttttt
Weight in kg No. of Persons
(CB) (Cumulative Frequency)
40 0
50 8
60 20
70 40
80 50
(cid:10)(cid:10)(cid:11)(cid:20)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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N
−N
Hence, M =l + 2 l ×C
1 N −N
u l
[ (cid:6)(cid:17) (cid:6)(cid:21) ]
(cid:20)(cid:21) − (cid:7)(cid:21) /0(cid:3)
= + (cid:19)(cid:21) (cid:6)(cid:21) × = 62.50 Kg.
−
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111166666
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff mmmmmeeeeeaaaaannnnn dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn ooooofffff wwwwweeeeeiiiiiggggghhhhhttttt aaaaabbbbbooooouuuuuttttt mmmmmeeeeedddddiiiiiaaaaannnnn
weight mid-value No. of persons x −Me f x −Me
i i i
(kgs.) (x) kgs. (f) (kgs.) (kgs.)
i i
(1) (2) (3) (4) (5)=(3)×(4)
40–50 45 8 17.50 140
50–60 55 12 7.50 90
60–70 65 20 2.50 50
70–80 75 10 12.50 125
Total – 50 – 405
∑f x −Median
Mean deviation about median = i i
N
(cid:19)(cid:21)(cid:17)
/0(cid:3)
= (cid:17)(cid:21)
(cid:18)(cid:3)(cid:7)(cid:21)(cid:25)10(cid:3)
=
’!(cid:30)(cid:1)(cid:25)(cid:25)(cid:13)!(cid:29)(cid:8)(cid:30)"(cid:8)((cid:1)(cid:25)(cid:25)(cid:30)+( "(cid:25)(cid:25).!(cid:13)(cid:8)(cid:30)(cid:1)
(cid:7)(cid:21)(cid:21)
Coefficient of mean deviation about median= ’!(cid:13)(cid:8)(cid:30)(cid:1) ×
(cid:18)(cid:3)(cid:7)(cid:25)(cid:21)
(cid:7)(cid:21)(cid:21)
= (cid:25)(cid:20)(cid:6)(cid:3)(cid:17)(cid:21) ×
= (cid:7)(cid:6)(cid:3)(cid:14)(cid:20)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333300000::::: If x and y are related as 4x+3y+11 = 0 and mean deviation of x is 5.40, what
is the mean deviation of y?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since 4x + 3y + 11 = 0
−11 −4
Therefore, y = + x
3 3
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:20)(cid:24)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Hence MD of y= |b| × MD of x
(cid:19)
(cid:25)(cid:17)(cid:3)(cid:19)(cid:21)(cid:25)
= (cid:5) ×
= 7.20
1111111111.....1111111111 SSSSSTTTTTAAAAANNNNNDDDDDAAAAARRRRRDDDDD DDDDDEEEEEVVVVVIIIIIAAAAATTTTTIIIIIOOOOONNNNN
Although mean deviation is an improvement over range so far as a measure of dispersion is
concerned, mean deviation is difficult to compute and further more, it cannot be treated
mathematically. The best measure of dispersion is, usually, standard deviation which does not
possess the demerits of range and mean deviation.
Standard deviation for a given set of observations is defined as the root mean square deviation
when the deviations are taken from the AM of the observations. If a variable x assumes n
values x , x , x ………..x then its standard deviation(s) is given by
1 2 3 n
∑(x −x)2
s= i ……………………….(11.28)
n
FFFFFooooorrrrr aaaaa gggggrrrrrooooouuuuupppppeeeeeddddd fffffrrrrreeeeeqqqqquuuuueeeeennnnncccccyyyyy dddddiiiiissssstttttrrrrriiiiibbbbbuuuuutttttiiiiiooooonnnnn,,,,, ttttthhhhheeeee ssssstttttaaaaannnnndddddaaaaarrrrrddddd dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn iiiiisssss gggggiiiiivvvvveeeeennnnn bbbbbyyyyy
∑f (x −x)2
s= i i .……………………… (11.29)
N
(11.28) and (11.29) can be simplified to the following forms
∑x2
s= i −x2 for unclassified data
n
∑fx2
= i i −x2 for a grouped frequency distribution.
N
..……………………… (11.30)
Sometimes the square of standard deviation, known as variance, is regarded as a measure of
dispersion. We have, then,
∑(x −x)2
Variance = s2 = i for unclassified data
n
∑f(x −x)2
= i i for a grouped frequency distribution……………..(11.31)
N
A relative measure of dispersion using standard deviation is given by coefficient of variation
(v) which is defined as the ratio of standard deviation to the corresponding arithmetic mean,
expressed as a percentage.
,&
(cid:7)(cid:21)(cid:21)
Coefficient of Variation (CV) = (cid:12)’ × ..……………………… (11.32)
(cid:10)(cid:10)(cid:11)(cid:20)(cid:25) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
IIIIIlllllllllluuuuussssstttttrrrrraaaaatttttiiiiiooooonnnnn
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333311111::::: Find the standard deviation and the coefficient of variation for the following
numbers: 5, 8, 9, 2, 6
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: We present the computation in the following table.
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111177777
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff ssssstttttaaaaannnnndddddaaaaarrrrrddddd dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn
x x 2
i i
5 25
8 64
9 81
2 4
6 36
30 ∑x2 = 210
i
Applying (11.30), we get the standard deviation as
∑x2
s= i −x2
n
210 30 2 (cid:28)(cid:8)(cid:1)(cid:23)!(cid:2) Σ (cid:2) (cid:8)
= − = (cid:1)
5 5
= 42−36
= 6
= (cid:6)(cid:3)(cid:19)(cid:17)
The coefficient of variation is
,&
(cid:7)(cid:21)(cid:21)
CV = × (cid:12)’
(cid:6)(cid:3)(cid:19)(cid:17)
(cid:7)(cid:21)(cid:21)
= × (cid:20)
= (cid:19)(cid:21)(cid:3)(cid:18)(cid:5)
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:20)(cid:26)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333322222::::: Show that for any two numbers a and b, standard deviation is given
(cid:30) +
−
by .
(cid:6)
(cid:30) +
(cid:2) +
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: For two numbers a and b, AM is given by = (cid:6)
The variance is
∑(x −x)2
s2 = i
2
( (cid:30) +)(cid:6) ( (cid:30) +)(cid:6)
(cid:30)(cid:24) + +(cid:24) +
+
(cid:6) (cid:6)
=
(cid:6)
(a−b)2 (a−b)2
+
= 4 4
2
(cid:22)(cid:30) +(cid:15)(cid:6)
−
=
(cid:19)
(cid:30) +
−
(cid:28)
⇒ =
(cid:6)
(The absolute sign is taken, as SD cannot be negative).
(cid:1)(cid:6) (cid:7)
−
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333333333::::: Prove that for the first n natural numbers, SD is .
(cid:7)(cid:6)
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: for the first n natural numbers AM is given by
(cid:7) (cid:6) (cid:5) (cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)
(cid:2) + + + +
=
(cid:1)
(cid:1)(cid:22)(cid:1) (cid:7)(cid:15)
+
= (cid:6)(cid:1)
(cid:1) (cid:7)
+
= (cid:6)
∑x2
∴ SD = i −x2
n
(cid:7)(cid:6) (cid:6)(cid:6) (cid:5)(cid:6)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3)(cid:3) (cid:1)(cid:6) ((cid:1) (cid:7))(cid:6)
+ + + +
= −
(cid:1) (cid:6)
(cid:1)(cid:22)(cid:1) (cid:7)(cid:15)(cid:22)(cid:6)(cid:1) (cid:7)(cid:15) (cid:22)(cid:1) (cid:7)(cid:15)(cid:6)
+ + +
= −
(cid:20)(cid:1) (cid:19)
(cid:10)(cid:10)(cid:11)(cid:21)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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(cid:22)(cid:1) + (cid:7)(cid:15)(cid:22)(cid:19)(cid:1) + (cid:6) − (cid:5)(cid:1) − (cid:5)(cid:15) (cid:1)(cid:6) − (cid:7)
= =
(cid:7)(cid:6) (cid:7)(cid:6)
(cid:1)(cid:6) (cid:7)
−
Thus, SD of first n natural numbers. SD =
(cid:7)(cid:6)
We consider the following formula for computing standard deviation from grouped frequency
distribution with a view to saving time and computational labour:
∑fd2 ∑fd 2
S = i i − i i ……………………………..(11.33)
N N
x −A
Where d = i
i C
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333344444::::: Find the SD of the following distribution:
Weight (kgs.) : 50-52 52-54 54-56 56-58 58-60
No. of Students : 17 35 28 15 5
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111177777
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff SSSSSDDDDD
Weight No. of Students Mid-value d=x – 55 fd fd2
i i i i i i
(kgs.) (f) (x) 2 (5)=(2)×(4) (6)=(5)×(4)
i i
(1) (2) (3) (4)
50-52 17 51 –2 –34 68
52-54 35 53 –1 –35 35
54-56 28 55 0 0 0
56-58 15 57 1 15 15
58-60 5 59 2 10 20
Total 100 – – – 44 138
Applying (11.33), we get the SD of weight as
∑fd2 ∑fd 2
= i i − i i ×C
N N
(cid:7)(cid:5)(cid:18) (cid:22) − (cid:19)(cid:19)(cid:15)(cid:6)
= − × (cid:6)10(cid:28)(cid:3)
(cid:7)(cid:21)(cid:21) (cid:7)(cid:21)(cid:21)
= (cid:7)(cid:3)(cid:5)(cid:18) − (cid:21)(cid:3)(cid:7)(cid:14)(cid:5)(cid:20) × (cid:6)(cid:25)10(cid:28)(cid:3)
(cid:6)(cid:3)(cid:7)(cid:18)(cid:25)10(cid:28)(cid:3)(cid:25)
=
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:21)(cid:10)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
PPPPPrrrrrooooopppppeeeeerrrrrtttttiiiiieeeeesssss ooooofffff ssssstttttaaaaannnnndddddaaaaarrrrrddddd dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn
I. If all the observations assumed by a variable are constant i.e. equal, then the SD is zero.
This means that if all the values taken by a variable x is k, say , then s = 0. This result
applies to range as well as mean deviation.
II. SD remains unaffected due to a change of origin but is affected in the same ratio due to
a change of scale i.e., if there are two variables x and y related as y = a+bx for any two
constants a and b, then SD of y is given by
(cid:28) (cid:25)+(cid:25)(cid:25)(cid:28)
(cid:26) = (cid:2) ………………………..(11.34)
III. If there are two groups containing n and n observations, x and x as respective AM’s,
1 2 1 2
s and s as respective SD’s , then the combined SD is given by
1 2
(cid:1) (cid:28) (cid:6) (cid:1) (cid:28) (cid:6) (cid:1) (cid:13) (cid:6) (cid:1) (cid:13) (cid:6)
+ + +
(cid:7) (cid:7) (cid:6) (cid:6) (cid:7) (cid:7) (cid:6) (cid:6)
s = (cid:1) (cid:1) ………………………..(11.35)
+
(cid:7) (cid:6)
(cid:13) (cid:2) (cid:2)
where, = −
(cid:7) (cid:7)
(cid:13) (cid:2) (cid:2)
= −
(cid:6) (cid:6)
n x +n x
x = 1 1 2 2
and = combined AM
n +n
1 2
This result can be extended to more than 2 groups. For x(72) groups, we have
∑ns2+∑nd2
s= i i i i
……………………….. (11.36)
∑n
i
With d = x −x
i i
∑nx
x = i i
and
∑n
i
Where x =x (11.35) is reduced to
1 2
(cid:1) (cid:28) (cid:6) (cid:1) (cid:28) (cid:6)
+
(cid:7) (cid:7) (cid:6) (cid:6)
s = (cid:1) (cid:1)
+
(cid:7) (cid:6)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333355555::::: If AM and coefficient of variation of x are 10 and 40 respectively, what is the
variance of (15–2x)?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: let y = 15 – 2x
Then applying (11.34), we get,
s = 2 × s ………………………………… (1)
y x
As given cv = coefficient of variation of x = 40 andx= 10
x
s
This cv = x×100
x x
(cid:10)(cid:10)(cid:11)(cid:21)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
,
(cid:19)(cid:21) (cid:2) (cid:7)(cid:21)(cid:21)
⇒ = (cid:7)(cid:21) ×
⇒ S =4
x
From (1),S =2×4=8
y
Therefore, variance of (15−2x)=S 2 =64
y
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333366666::::: Compute the SD of 9, 5, 8, 6, 2.
Without any more computation, obtain the SD of
Sample I –1, –5, –2, –4, –8,
Sample II 90, 50, 80, 60, 20,
Sample III 23, 15, 21, 17, 9.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111188888
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff SSSSSDDDDD
x x 2
i i
9 81
5 25
8 64
6 36
2 4
30 210
The SD of the original set of observations is given by
∑x2 ∑x 2
s = i - i
n n
2
210 30
= −
5 5
= (cid:19)(cid:6) − (cid:5)(cid:20)
= 6
= (cid:6)(cid:3)(cid:19)(cid:17)
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:21)(cid:20)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
If we denote the original observations by x and the observations of sample I by y, then we have
y = –10 + x
y = (–10) + (1) x
, (cid:7) ,
∴ (cid:26) = × (cid:2)
= (cid:7)(cid:25) × (cid:25)(cid:6)(cid:3)(cid:19)(cid:17)(cid:25)(cid:25)
=(cid:6)(cid:3)(cid:19)(cid:17)(cid:25)(cid:25)
In case of sample II, x and y are related as
Y = 10x
= 0 + (10)x
(cid:28) (cid:7)(cid:21) (cid:28)
∴ = ×
(cid:26) (cid:2)
= (cid:7)(cid:21) × (cid:6)(cid:3)(cid:19)(cid:17)
= 24.50
And lastly, y= (5)+(2)x
⇒ s
y
= 2 ×2.45
= 4.90
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333377777::::: For a group of 60 boy students, the mean and SD of stats. marks are 45 and
2 respectively. The same figures for a group of 40 girl students are 55 and 3 respectively. What
is the mean and SD of marks if the two groups are pooled together?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: As given n = 60, x = 45, s = 2 n = 40, x = 55, s = 3
1 1 1 2 2 2
Thus the combined mean is given by
(cid:1) (cid:2) (cid:1) (cid:2)
+
(cid:2) (cid:7) (cid:7) (cid:6) (cid:6)
=
(cid:1) (cid:1)
+
(cid:7) (cid:6)
(cid:20)(cid:21) (cid:19)(cid:17) (cid:19)(cid:21) (cid:17)(cid:17)
× + ×
=
(cid:20)(cid:21) (cid:19)(cid:21)
+
=49
(cid:13) (cid:2) (cid:2) (cid:19)(cid:17) (cid:19)(cid:14) (cid:24)(cid:19)
Thus = − = − =
(cid:7) (cid:7)
(cid:13) (cid:2) (cid:2) (cid:17)(cid:17) (cid:19)(cid:14) (cid:20)
= − = − =
(cid:6) (cid:6)
Applying (11.35), we get the combined SD as
(cid:1) (cid:28) (cid:6) (cid:1) (cid:28) (cid:6) (cid:1) (cid:13) (cid:6) (cid:1) (cid:13) (cid:6)
+ + +
(cid:7) (cid:7) (cid:6) (cid:6) (cid:7) (cid:7) (cid:6) (cid:6)
s = (cid:1) (cid:1)
+
(cid:7) (cid:6)
(cid:20)(cid:21) (cid:6)(cid:6) (cid:19)(cid:21) (cid:5)(cid:6) (cid:20)(cid:21) (cid:22) (cid:19)(cid:15)(cid:6) (cid:19)(cid:21) (cid:20)(cid:6)
× + × + × − + ×
s =
(cid:20)(cid:21) (cid:19)(cid:21)
+
(cid:10)(cid:10)(cid:11)(cid:21)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
= 30
= (cid:17)(cid:3)(cid:19)(cid:18)
Example 11.38: The mean and standard deviation of the salaries of the two factories are provided
below:
Factory No. of Employees Mean Salary SD of Salary
A 30 Rs.4800 Rs.10
B 20 Rs. 5000 Rs.12
i) Find the combined mean salary and standard deviation of salary.
ii) Examine which factory has more consistent structure so far as satisfying its employees are
concerned.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Here we are given
n = 30, x = Rs.4800, s = Rs.10,
1 1 1
n = 20, x = Rs.5000, s = Rs.12
2 2 2
30×Rs.4800+20×Rs.5000
i) =Rs.4800
30+20
d =x −x = Rs.4,800 - Rs.4880 = - Rs.80
1 1
d =x −x= Rs.5,000 - Rs.4880 = Rs.120
2 2
hence, the combined SD in rupees is given by
30×102+20×122+30×(−80)2+20×1202
s=
30+20
= 9717.60
= 98.58
thus the combined mean salary and the combined standard deviation of salary are Rs.4880
and Rs.98.58 respectively.
ii) In order to find the more consistent structure, we compare the coefficients of variation of
, ,
(cid:7)(cid:21)(cid:21) (cid:12) (cid:7)(cid:21)(cid:21) 2
the two factories. Letting CV = × (cid:2) and CV = × (cid:2)
A (cid:12) B 2
We would say factory A is more consistent
if CV < CV . Otherwise factory B would be more consistent.
A B
s s 100×10
Now CV =100× A =100× 1 = =0.21
A x x 4800
A 1
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:21)(cid:22)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
s s 100×12
CV =100× B =100× 2 = =0.24
and B x x 5000
B 2
Thus we conclude that factory A has more consistent structure.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....3333399999::::: A student computes the AM and SD for a set of 100 observations as 50 and
5 respectively. Later on, she discovers that she has made a mistake in taking one observation as
60 instead of 50. What would be the correct mean and SD if
i) The wrong observation is left out?
ii) The wrong observation is replaced by the correct observation?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: As given, n = 100, x=50, S = 5
Wrong observation = 60(x), correct observation = 50(V)
∑x
x = i
n
⇒ ∑x = nx = 100 × 50 = 5000
i
∑x2
and s2 = i −x2
n
⇒ ∑x2 =n(x2+s2)=100(502+52)=252500
i
i) Sum of the 99 observations = 5000 – 60 = 4940
AM after leaving the wrong observation = 4940/99 = 49.90
Sum of squares of the observation after leaving the wrong observation
= 252500 – 602 = 248900
Variance of the 99 observations = 248900/99 – (49.90)2
= 2514.14 – 2490.01
= 24.13
∴ SD of 99 observations = 4.91
ii) Sum of the 100 observations after replacing the wrong observation by the correct observation
= 5000 – 60 + 50 =4990
4990
AM = = 49.90
100
Corrected sum of squares = 252500 + 502 – 602 = 251400
251400
Corrected SD =
–(49.90)2
100
= 45.99
= 6.78
(cid:10)(cid:10)(cid:11)(cid:21)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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1111111111.....1111122222 QQQQQUUUUUAAAAARRRRRTTTTTIIIIILLLLLEEEEE DDDDDEEEEEVVVVVIIIIIAAAAATTTTTIIIIIOOOOONNNNN
Another measure of dispersion is provided by qqqqquuuuuaaaaarrrrrtttttiiiiillllleeeee dddddeeeeevvvvviiiiiaaaaatttttiiiiiooooonnnnn or ssssseeeeemmmmmiiiii ----- iiiiinnnnnttttteeeeerrrrr –––––qqqqquuuuuaaaaarrrrrtttttiiiiillllleeeee
range which is given by
$ $
−
$ (cid:5) (cid:7)
(cid:13) = (cid:6) ……………………………..(11.37)
A relative measure of dispersion using quartiles is given by coefficient of quartile deviation
which is
$ $
−
(cid:5) (cid:7) (cid:7)(cid:21)(cid:21)
Coefficient of quartile deviation =$ $ × …………………………….(11.38)
+
(cid:5) (cid:7)
Quartile deviation provides the best measure of dispersion for open-end classification. It is also
less affected due to sampling fluctuations. Like other measures of dispersion, quartile deviation
remains unaffected due to a change of origin but is affected in the same ratio due to change in
scale.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....4444400000 ::::: Following are the marks of the 10 students : 56, 48, 65, 35, 42, 75, 82, 60, 55,
50. Find Quartile deviation and also its coefficient.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn:::::
After arranging the marks in an ascending order of magnitude, we get 35, 42, 48, 50, 55, 56,
60, 65, 75, 82
(n +1)
First Quartile (Q ) = th observation
1 4
(10+1)
= th observation
4
= 2.75th observation
= 2nd observation + 0.75 × difference between the third and the 2nd observation.
= 42 + 0.75 × (48 – 42)
= 46.50
(cid:5)(cid:22)(cid:1) (cid:7)(cid:15)
+
Third Quartile (Q ) = (cid:19) th observation
3
= 8.25 th observation
= 65 + 0.25 × 10
= 67.50
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:21)(cid:24)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
Thus applying (11.37), we get the quartile deviation as
$ (cid:5) − $ (cid:7) (cid:20)(cid:16)(cid:3)(cid:17)(cid:21) − (cid:19)(cid:20)(cid:3)(cid:17)(cid:21) (cid:7)(cid:21)(cid:3)(cid:17)(cid:21)
= =
(cid:6) (cid:6)
Also, using (11.38), the coefficient of quartile deviation is
Q –Q
3 1
×100
= Q +Q
3 1
(cid:20)(cid:16)(cid:3)(cid:17)(cid:21) (cid:19)(cid:20)(cid:3)(cid:17)(cid:21)
− (cid:7)(cid:21)(cid:21)
= (cid:20)(cid:16)(cid:3)(cid:17)(cid:21) (cid:19)(cid:20)(cid:3)(cid:17)(cid:21) ×
+
= 18.42
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....4444411111 ::::: If the quartile deviation of x is 6 and 3x + 6y = 20, what is the quartile deviation
of y?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: 3x + 6y = 20
20 −3
⇒ y = + x
6 6
(cid:5)
−
(cid:26) 3 (cid:30)4"(cid:8)(cid:31)!(cid:25)(cid:25)(cid:13)!(cid:29)(cid:8)(cid:30)"(cid:8)((cid:1)(cid:25)((cid:9)(cid:25)(cid:2)
Therefore, quartile deviation of = ×
(cid:20)
1
= x 6
2
= 3.
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....4444422222::::: Find an appropriate measures of dispersion from the following data:
Daily wages (Rs.) : upto 20 20-40 40-60 60-80 80-100
No. of workers : 5 11 14 7 3
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Since this is an open-end classification, the appropriate measure of dispersion would
be quartile deviation as quartile deviation does not taken into account the first twenty five
percent and the last twenty five per cent of the observations.
TTTTTaaaaabbbbbllllleeeee 1111111111.....1111199999
CCCCCooooommmmmpppppuuuuutttttaaaaatttttiiiiiooooonnnnn ooooofffff QQQQQuuuuuaaaaarrrrrtttttiiiiillllleeeee
Daily wages in Rs. No. of workers
(Class boundary) (less than cumulative frequency)
a 0
20 5
40 16
60 30
80 37
100 40
(cid:10)(cid:10)(cid:11)(cid:21)(cid:25) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
Here a denotes the first Class Boundary
[ (cid:7)(cid:21)(cid:24)(cid:17) ]
(cid:6)(cid:21) (cid:6)(cid:21)
Q = Rs. + (cid:7)(cid:20)(cid:24)(cid:17) × = Rs. 29.09
1
Q = Rs. 60
3
Thus quartile deviation of wages is given by
Q –Q
3 1
2
Rs. 60–Rs. 29.09
=
2
= Rs. 15.46
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....4444433333::::: The mean and variance of 5 observations are 4.80 and 6.16 respectively. If
three of the observations are 2,3 and 6, what are the remaining observations?
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: Let the remaining two observations be a and b, then as given
2+3+6+a+b
= 4.80
5
⇒ 11+a+b =24
⇒ a+b =13 .................(1)
(cid:6)(cid:6) (cid:30)(cid:6) +(cid:6) (cid:5)(cid:6) (cid:20)(cid:6)
+ + + + (cid:24)(cid:22)(cid:19)(cid:3)(cid:18)(cid:21)(cid:15)(cid:6)
and
(cid:17)
(cid:19)(cid:14) (cid:30)(cid:6) +(cid:6)
+ + (cid:24)(cid:6)(cid:5)(cid:3)(cid:21)(cid:19) (cid:20)(cid:3)(cid:7)(cid:20)
⇒ =
(cid:17)
⇒ 49 + a2 + b2 =146
⇒ a2 + b2 =97 .................(2)
From (1), we get a = 13 – b ...........(3)
Eliminating a from (2) and (3), we get
(13 – b)2 + b2 =97
⇒ 169 – 26b + 2b2 =97
⇒ b2 – 13 b + 36 = 0
⇒ (b–4)(b–9) =0
⇒ b = 4 or 9
From (3), a= 9 or 4
Thus the remaining observations are 4 and 9.
(cid:19)(cid:5)(cid:3)(cid:5)(cid:17)(cid:19)(cid:5)(cid:17)(cid:1)(cid:19) (cid:10)(cid:10)(cid:11)(cid:21)(cid:26)
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(cid:1)(cid:2)(cid:3)(cid:4)(cid:5)(cid:6)(cid:2)(cid:4)(cid:7)(cid:8)(cid:9)(cid:7)(cid:10)(cid:2)(cid:11)(cid:12)(cid:6)(cid:3)(cid:13)(cid:7)(cid:12)(cid:2)(cid:11)(cid:14)(cid:2)(cid:11)(cid:10)(cid:15)(cid:7)(cid:3)(cid:11)(cid:14)(cid:7)(cid:14)(cid:16)(cid:4)(cid:17)(cid:2)(cid:6)(cid:4)(cid:16)(cid:8)(cid:11)
EEEEExxxxxaaaaammmmmpppppllllleeeee 1111111111.....4444444444 ::::: After shift of origin and change of scale, a frequency distribution of a continuous
variable with equal class length takes the following form of the changed variable (d):
d : –2 –1 0 1 2
frequency : 17 35 28 15 5
If the mean and standard deviation of the original frequency distribution are 54.12 and 2.1784
respectively, find the original frequency distribution.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: we need find out the origin A and scale C from the given conditions.
(cid:2) (cid:24)(cid:12)
(cid:8)
Since d = (cid:11)
i
⇒ x = A + Cd
i i
once A and C are known, the mid- values x’s would be known. Finally, we convert the mid-
i
values to the corresponding class boundaries by using the formula:
LCB = x – C/2
i
and UCB = x + C/2
i
On the basis of the given data, we find that
∑fd = –44, ∑fd2 = 138 and N = 100
i i i i
∑fd2 ∑fd 2
Hence s = i i − i i ×C
N N
2
138 −44
⇒ 2.1784= − ×C
100 100
⇒ (cid:6)(cid:3)(cid:7)(cid:16)(cid:18)(cid:19) = (cid:7)(cid:3)(cid:5)(cid:18) − (cid:21)(cid:3)(cid:7)(cid:14)(cid:5)(cid:20) × (cid:11)
⇒ (cid:6)(cid:3)(cid:7)(cid:16)(cid:18)(cid:19) = (cid:7)(cid:3)(cid:21)(cid:18)(cid:14)(cid:6) × (cid:11)
⇒ C = 2
∑fd
Further, x =A+ i i ×C
N
(cid:19)(cid:19)
(cid:12) − (cid:6)
⇒ 54.12 = + (cid:7)(cid:21)(cid:21) ×
⇒ 54.12 = A – 0.88
⇒ A = 55
Thus x = A + Cd
i i
⇒ x = 55 + 2d
i i
(cid:10)(cid:10)(cid:11)(cid:22)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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