Question

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
Total Frequency (No. of days): 46 ? ? 25 10 5 200

Answer 1

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 = x + 2y + 140

Given

N = 200

46 + x + y + 25 + 10 + 5 = 200

x + y = 200 – 46 – 25 – 10 – 5

x + y = 114 —- (1)

And, Mean = 1.46

Sum/ N = 1.46

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

x + 2y = 292 – 140

x + 2y = 152 —- (2)

Subtract equation (1) from equation (2)

x + 2y – x – y = 152 – 114

y = 38

Putting the value of y in equation (1), we have 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 then the median is 1