In French test the
following marks were recorded.
|
marks
|
10-19
|
20-29
|
30-39
|
40-49
|
50-59
|
60-69
|
70-79
|
80-89
|
|
No. of students
|
3
|
6
|
7
|
9
|
12
|
8
|
6
|
4
|
Calculate the mean.
Solution
In this question we are
going to apply the method of assumed mean.
Here you are required to
produce the frequency distribution table.
Let us take 44.5 as our
assumed mean (A). Take away the assumed mean from each class mark.
|
Class interval
|
Class mark (x)
|
d=x-A
|
Frequency(f)
|
fd
|
|
10-19
|
14.5
|
-30
|
3
|
-90
|
|
20-29
|
24.5
|
-20
|
6
|
-120
|
|
30-39
|
34.5
|
-10
|
7
|
-70
|
|
40-49
|
44.5
|
0
|
9
|
0
|
|
50-59
|
54.5
|
10
|
12
|
120
|
|
60-69
|
64.5
|
20
|
8
|
160
|
|
70-79
|
74.5
|
30
|
6
|
180
|
|
80-89
|
84.5
|
40
|
4
|
160
|
|
|
|
∑f = 55
|
∑fd = 340
|
|
Mean = A + ∑fd
∑f
Mean = 44.5 + 340
55
Mean = 44.5 + (6.182)
Mean = 50.68 (to 2d.p)
Hence Mean = 50.68
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