The number of hours of classes attended by students in a semester and the corresponding number of students is tabulated below. Find the mean deviation about median.

$\begin{matrix} N u m b e r o f h o u r s & 100 - 120 & 120 - 140 & 140 - 160 & 160 - 180 & 180 - 200 N u m b e r o f s t u d e n t s & 19 & 28 & 30 & 16 & 7 \end{matrix}$

15.86

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18.02

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19.68

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21.88

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Solution

The correct option is C 19.68
$\begin{matrix} C l a s s I n t e r v a l & F r e q u e n c y & C u m u l a t i v e F r e q u e n c y 100 - 120 & 19 & 19 120 - 140 & 28 & 47 140 - 160 & 30 & 77 160 - 180 & 16 & 93 180 - 200 & 7 & 100 \end{matrix}$

$\frac{N}{2} = 50$
Median class is 140-160

$M = L + \frac{\frac{N}{2} - c . f}{f} \times h = 140 + \frac{50 - 47}{30} \times 20 = 142$

Deviations from the median are tabulated below

$\begin{matrix} X (c l a s s m i d p o i n t) & F & D = | X - M | & F D 110 & 19 & 32 & 608 130 & 28 & 12 & 336 150 & 30 & 8 & 240 170 & 16 & 28 & 448 190 & 7 & 48 & 336 \sum F = 100 & \sum F D = 1968 \end{matrix}$

$M . D (M) = \frac{1968}{100} = 19.68$

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