|
Q. 3. The following table gives weights (in
kilograms) of 100 students, randomly selected from
a college: (June 2002)
|
Weight in Kilograms
|
Number of students
|
|
45-50
|
01
|
|
50-55
|
05
|
|
55-60
|
21
|
|
60-65
|
43
|
|
65-70
|
22
|
|
70-75
|
06
|
|
75-80
|
02
|
Calculate the standard deviation of the above
frequency distribution.
Solution.
| Weight in Kilograms
|
Mid point
(M) |
No. of Students
(f) |
f X M |
x = M - mean
|
x2
|
f X x2
|
| 45-50 |
47.5 |
01 |
47.5 |
-15.3 |
234.09
|
234.09 |
| 50-55 |
52.5 |
05 |
262.5 |
-10.3 |
106.09
|
530.45 |
| 55-60 |
57.5 |
21 |
1207.5 |
-5.3 |
28.09
|
589.89 |
| 60-65 |
62.5 |
43 |
2687.5 |
-0.3 |
0.27
|
11.61 |
| 65-70 |
67.5 |
22 |
1485 |
4.7 |
22.09
|
485.98 |
| 70-75 |
72.5 |
06 |
435 |
9.7 |
94.09
|
564.54 |
| 75-80 |
77.5 |
02 |
155 |
14.7 |
216.09
|
432.18 |
|
Total |
|
å N = 100
|
å fM = 6280
|
|
|
å fx2
= 2848.74 |
Mean = å fM/å
N = 6280/100 = 62.8
| |
 |
å
fx2
|
| s = |
|
| |
å
N
|
| |
|
2848.74
|
| or
s = |
|
| |
100
|
or s = 5.337
Q. 4. Calculate the standard deviation of
the following grouped frequency distribution. (June
2000)
| Height in inches |
Number of students |
| 60-65 |
12 |
| 65-70 |
35 |
| 70-75 |
11 |
| 75-80 |
2 |
| 80-85 |
0 |
Solution.
| Height in inches |
Mid point
(M) |
No. of Students
(f) |
f X M |
x = M - mean
|
x2
|
f X x2
|
| 60-65 |
62.5
|
12 |
750
|
-5.25 |
27.5625
|
330.75
|
| 65-70 |
67.5
|
35 |
2362.5
|
-0.25 |
0.0625
|
2.1875
|
| 70-75 |
72.5
|
11 |
797.5
|
4.75 |
22.5625
|
248.1875
|
| 75-80 |
77.5 |
2 |
155 |
9.75 |
95.0625 |
190.125 |
| 80-85 |
82.5
|
0 |
0
|
14.75 |
217.5625
|
0
|
|
Total
|
|
å N =
60
|
å fM
= 4065
|
|
|
å fx2
= 771.25
|
Mean = å fM/å
N = 4065/60 = 67.75
| |
|
å
fx2
|
| s = |
|
| |
å
N
|
| |
|
771.25
|
| or
s = |
|
| |
60
|
or s = 3.58
Q. 5. For a frequency distribution of marks in
History of 200 candidates, the mean and standard deviation
(s.d.) were found to be 40 and 15 respectively. Later
it was discovered that the score 43 was misread as
53 in distribution. Find the correct mean and standard
deviation corresponding to the correct distribution.
(June 2002, Dec. 99)
Solution. Given, N = 200, 
= 40, and s
= 15
Calculation of Correct Mean
=
å
X/N
or å X = N
or å X = 200 X 40
= 8000
But this is not correct å
X because the score 43 was misread as 53 in the distribution.
Correct å X = 8000
- 53 + 43 = 7990
Correct
= 7990/200 = 39.95
Calculation of Correct Standard Deviation
s2 = (å
X2/N) - ()2
or (15)2 = (å
X2/200) - (40)2
or å X2
= 365000
Correct å X2
= 365000 - (53)2 + (43)2 = 364040
Correct s =[(Correct
å X2/N)
- (Correct )2]
or Correct s =[(364040/200)
- (39.95)2]
= 14.97
Thus, the correct mean is 39.95 and correct standard
deviation is 14.97.
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