# 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

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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