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Q. 7. Following table gives the height (in
inches) of the employees of an organization: (June
2001)
| Height in inches |
Number of employees |
| 50-60 |
05 |
| 60-70 |
35 |
| 70-80 |
08 |
| 80-90 |
02 |
Calculate the variance of the above class-distribution.
Solution.
| Height in inches |
Mid point
(M) |
Number of employees (f) |
f X M |
x = M - mean
|
x2
|
f X x2
|
| 50-60 |
55
|
05 |
275
|
-11.4 |
129.96
|
649.8
|
| 60-70 |
65
|
35 |
2275
|
-1.4 |
1.96
|
68.6
|
| 70-80 |
75
|
08 |
600
|
8.6 |
73.96
|
591.68
|
| 80-90 |
85 |
02 |
170 |
18.6 |
345.96 |
691.92 |
|
Total
|
|
å N =
50
|
å fM
= 3320
|
|
|
å fx2
= 2002
|
Mean = å fM/å
N = 3320/50 = 66.4
| |
 |
å
fx2
|
| s = |
|
| |
å
N
|
| |
|
2002
|
| or
s = |
|
| |
50
|
or s = 6.32
Variance, s2
= 39.94
Q. 8. The following table gives the height
(in inches) of 100 students, randomly selected from
a college: (Dec.
2002)
| Height in inches |
Number of students |
| 57-60 |
02 |
| 60-63 |
06 |
| 63-66 |
20 |
| 66-69 |
45 |
| 69-72 |
21 |
| 72-75 |
05 |
| 75-78 |
01 |
Calculate the variance of the above frequency
distribution.
Solution.
| Height in inches |
Mid
point
(M) |
Number of students (f) |
f X
M |
x =
M - mean |
x2
|
f X
x2 |
| 57-60 |
58.5 |
02 |
117 |
-8.88 |
78.8544
|
157.7088
|
| 60-63 |
61.5 |
06 |
369 |
-5.88 |
34.5744
|
207.4464
|
| 63-66 |
64.5 |
20 |
1290 |
-2.88 |
8.2944
|
165.888
|
| 66-69 |
67.5 |
45 |
3037.5 |
0.12 |
0.0144 |
0.648 |
| 69-72 |
70.5 |
21 |
1480.5 |
3.12 |
9.7344 |
204.4244 |
| 72-75 |
73.5 |
05 |
367.5 |
6.12 |
37.4544 |
187.272 |
| 75-78 |
76.5 |
01 |
76.5 |
9.12 |
83.1744 |
83.1744 |
|
Total
|
|
å N = 100
|
å fM = 6738
|
|
|
å fx2
= 1006.56 |
Mean = å fM/å
N = 6738/100 = 67.38
| |
|
å
fx2
|
| s = |
|
| |
å
N
|
| |
|
1006.56
|
| or
s = |
|
| |
100
|
or s = 3.17
Variance, s2
= 10.0489
Q. 9. The mean of 5 observations is 4.4 and the
variance is 8.24. If three of the five observations
are 1, 2 and 6, find the other two. (Dec.
2000)
Solution. Given, N = 5, 
= 4.4, and s2
= 8.24
=
å X/N
or å X = N
or å X = 5 X 4.4
= 22
Let the two missing items be p and q
Therefore, 1 + 2 + 6 + p + q = 22
or p + q = 13 ........ (i)
s2 = (å
X2/N) - ()2
or 8.24 = (å X2/5)
- (4.4)2
or å X2
= 138
Also, å X2
= 12 + 22 + 62 +
p2 + q2
or p2 + q2 = 138 - 41
or p2 + q2 = 97
(p + q)2 = p2 + q2
+ 2pq
or (13)2 = 97 + 2pq
or pq = 36
(p - q)2 = p2 + q2
- 2pq
or (p - q)2 = 97 - (2 X 36)
or p - q = 5 ........ (ii)
Adding equation (i) and (ii), we get
p = 9 and q = 4
Thus, the two missing values are 9 and 4.
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