Calculate the mean deviation from the median of the following frequency distribution :

$\begin{matrix} Heights in inches & 58 & 59 & 60 & 61 & 62 & 63 & 64 & 65 & 66 No. of students & 15 & 20 & 32 & 35 & 22 & 20 & 10 & 8 \end{matrix}$

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Solution

In order to find the mean deviation from the median, we well first have to calculate the median. M is the value of x corresponding to the cumulative frequency just greater than or equal to $\frac{N}{2}$ .

$\begin{matrix} x_{i} & f_{i} & Cumulative frequency = | x_{i} - 61 | & | d_{i} | & f_{i} | d_{i} | 58 & 15 & 15 & 3 & 45 59 & 20 & 35 & 2 & 40 60 & 32 & 67 & 1 & 32 61 & 35 & 102 & 0 & 0 62 & 35 & 137 & 1 & 35 63 & 22 & 159 & 2 & 44 64 & 20 & 179 & 3 & 60 65 & 10 & 189 & 4 & 40 66 & 8 & 197 & 5 & 40 N = \sum f_{i} = 197 & \sum_{i = 1}^{n} f_{i} | d_{i} | = 336 \end{matrix}$

Here,
$\frac{N}{2} = \frac{197}{2} = 98.5$
The cumulative frequency just greater than 98.5 is 102. The corresponding value of x is 61. Therefore, the median is 61.
$M . D . = \frac{1}{N} \sum_{i = 1}^{n} f_{i} | d_{i} | = \frac{1}{197} \times 336 = 1.7055$

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