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	45	0
1	x	x
2	y	2y
3	25	75
4	10	40
5	5	25
	N = 200	`sumf_1x_1=x+2y+140`

Given, N = 200

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ 86 + x + y = 200

⇒ x + y = 200 - 86

⇒ x+ y = 114                   ..............(1)

And Mean = 1.46

`rArr(sumfx)/N=1.46`

`rArr(x+2y+140)/200=1.46`

⇒ x + 2y + 140 = 1.46 x 200

⇒ x + 2y + 140 = 292

⇒ x + 2y = 292 - 140

⇒ x + 2y = 152                 ................(2)

Subtract equation (1) from equation (2)

⇒ x + 2y - (x+ y) = 152 - 114

⇒ x + 2y - x - y = 38

⇒ y = 38

Put the value of y in (1),

⇒ x+ y = 114

⇒ x + 38 = 114

⇒ x = 114 - 38

⇒ x = 76

Hence, the missing frequencies are 38 and 76.

(2) Calculation of median.

No. of accidents (x)	No. of days (f)	Cumulative Frequency
0	45	46
1	76	122
2	38	160
3	25	185
4	10	195
5	5	200
	N = 200

We have

N = 200

So, `N/2=200/2=100`

Thus, the cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.

Hence, the median is 1.