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The mean deviation of the following data from
Class interval : 0-5 5-10 10-15 15-20 20-25
Frequency : 3 4 8 10 5

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Detailed Solution

Computation of mean deviation about mean:

Class interval	Mid-values $\|x_{i}>$	Frequency $(f_{i})$	$Σ f_{i} x_{i}$	$\|x_{i} - \bar{X}\|$	$Σ f_{i} \|x_{i} - \bar{X}\|$
	$\begin{aligned} N = Σ f_{i} \\ = 30 \end{aligned}$	$Σ f_{i} x_{i} = 425$	$\begin{matrix} Σ f_{i} \|x_{i} - \bar{X}\| \\ = 150 \end{matrix}$

Class interval

Mid-values

|x_{i}>

Frequency

(f_{i})

Σ f_{i} x_{i}

|x_{i} - \bar{X}|

Σ f_{i} |x_{i} - \bar{X}|

0-5

5-10

10-15

15-20

20-25

2.5

7.5

12.5

17.5

22.5

7.5

100

175

112.5

\begin{aligned} \frac{35}{3} \\ \frac{20}{3} \\ \frac{5}{3} \\ \frac{10}{3} \\ \frac{25}{3} \end{aligned}

$\begin{aligned} \frac{80}{3} \\ \frac{40}{3} \\ \frac{100}{3} \\ \frac{125}{3} \end{aligned}$

\begin{aligned} N = Σ f_{i} \\ = 30 \end{aligned}

Σ f_{i} x_{i} = 425

\begin{matrix} Σ f_{i} |x_{i} - \bar{X}| \\ = 150 \end{matrix}

$\bar{X} = \frac{Σ f_{i} x_{i}}{Σ f_{i}} = \frac{425}{30} = \frac{85}{6}$

$∴$ Mean deviation about mean $= \frac{Σ f_{i} |x_{i} - \bar{X}|}{Σ f_{i}} = \frac{150}{30} = 5$

Following are some definitions related to mean deviation:

Mean coefficient of dispersion

$= \frac{Mean deviation from the mean}{Mean}$

Median coefficient of dispersion

$= \frac{Mean deviation from the median}{Median}$

Mode coefficient of dispersion

$= \frac{Mean deviation from the mode}{Mode}$

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