Find the missing frequencies and the median for the following distribution if the mean is 1.46.
No. of accidents: | 0 | 1 | 2 | 3 | 4 | 5 | Total |
Frequency (No. of days): | 46 | ? | ? | 25 | 10 | 5 | 200 |
No. of accidents (x) | No. of days (f) | Fx |
0 | 46 | 0 |
1 | X | X |
2 | Y | 2y |
3 | 25 | 75 |
4 | 10 | 40 |
5 | 5 | 25 |
N = 200 |
|
Given, N = 200
= 46 + x + y + 25 + 10 + 5 = 200
= x + y = 200 – 46 – 25 – 10 – 5
= x + y = 114 (i)
And Mean = 1.46
= 1.46
= 
= 1.46
= x + 2y + 140 = 292
= x + 2y = 292 – 140
= x + 2y = 152 (ii)
Subtract (i) from (ii), we get
X + 2y – x – y = 152 – 114
y = 38
Put the value of y in (i), we get
x = 114 – 38 = 76
No. of accidents | No. of days | Cumulative frequency |
0 | 46 | 46 |
1 | 76 | 122 |
2 | 38 | 160 |
3 | 25 | 185 |
4 | 10 | 195 |
5 | 5 | 200 |
N = 200 |
We have, N = 200
= 
= 100
The cumulative frequency just more than 
is 122 so the median is 1
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