The table shows the distribution of masks obtained by students in an examination.
Marks
50 - 54
55 - 59
60 - 64
65 - 69
70 - 74
75 - 79
80 - 84
85 - 89
Frequency
5
15
20
28
12
9
7
4
Using an assumed mean of 67, calculate, correct to one decimal place. the a) Mean b) Standard deviation of the distribution
Explanation
x
f
d
fd
\(\delta\)
f\(\delta\)
52
5
-15
-75
225
1125
57
15
-10
-150
100
1500
62
20
-5
-100
25
500
67
28
0
0
0
0
72
12
5
60
25
300
77
9
10
90
100
900
82
7
15
105
225
1575
87
4
20
80
400
1600
Total
100
10
7500
(a) To find mean; = 67 + \(\frac{10}{100}\) = 67.1 (b) Substitute and find the standard deviation as \(\sqrt{\frac{7500}{100} - (\frac{10}{100})^2}\) = \(\sqrt{74.99}\) = 8.6597 = 8.7, correct to one decimal place
