Class interval | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
Frequency | 17 | 35 | 43 | 40 | 21 | 24 |
Answer:
Here, Let the assumed mean be
(2)
Thus, we have
and(1)
Using step deviation method
Mean,
(1)
Class interval
Frequency
Mid-value
0-50
50-100
100-150
150-200
200-250
250-300
17
35
43
40
21
24
25
75
125=A
175
225
275
-2
-1
0
1
2
3
-34
-35
0
40
42
72
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