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

Given:

The mean of the given distribution is 1.46.

To do:

We have to find the missing frequencies and the median.

Solution:

Mean $= 1.46$

Let $p_1$ and $p_2$ be the missing frequencies as shown below.

$86+p_1+p_2=200$

$\Rightarrow p_1+p_2=200-86=114$

$p_1=114-p_2$...............(i)

We know that,

Mean $=\frac{\sum f_i x_i}{\sum f_i}$

Therefore,

Mean $=\frac{140+p_1+2 p_2}{200}$

$\Rightarrow 1.46=\frac{140+p_1+2 p_2}{200}$

$\Rightarrow 292=140+p_1+2 p_2$

$\Rightarrow 292-140=114-p_2+2 p_2$                   [From (i)]

$\Rightarrow 152-114= p_2$

$\Rightarrow p_2=38$

$\Rightarrow p_1=114-38=76$

Therefore,

$N = 86+76+38=200$

$\frac{N}{2} = \frac{200}{2} = 100$

The cumulative frequency just greater than $\frac{N}{2}=100$ is 122(=46+76) and the value corresponding to 122 is 1.

This implies,

Median $=1$

The missing frequencies are 76 and 38 and the median is 1.  

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Updated on: 10-Oct-2022

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