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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

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 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 \(\sum\)fx = x + 2y + 140

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

\(\frac{\sum fx}{N}=1.46\)

\(=\frac{x+2y+140}{200}=1.46\)

= 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

\(\frac{N}{2}=\frac{200}{2}=100\)

The cumulative frequency just more than \(\frac{N}{2}\) is 122 so the median is 1

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