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
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	\(\sum\)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