R D Sharma Solutions for Chapter: Statistics, Exercise 3: EXERCISE 32.3

Compute the mean deviation from the median of the following distribution:

$\begin{array}{|llllll|}\hline \mathrm{Class}& 0-10& 10-20& 20-30& 30-40& 40-50\\ \mathrm{Frequency}& 5& 10& 20& 5& 10\\ \hline\end{array}$

Find the mean deviation (correct up to three decimal places)from the mean for the following data:

$\begin{array}{|lcccccc|}\hline \mathrm{Classes}& 95-105& 105-115& 115-125& 125-135& 135-145& 145-155\\ \mathrm{Frequencies}& 9& 13& 16& 26& 30& 12\\ \hline\end{array}$

Find the mean deviation from the mean for the following data :

$\begin{array}{|lcccccc|}\hline \mathrm{Classes}& \mathbf{0}\mathbf{-}\mathbf{10}& \mathbf{10}\mathbf{-}\mathbf{20}& \mathbf{20}\mathbf{-}\mathbf{30}& \mathbf{30}\mathbf{-}\mathbf{40}& \mathbf{40}\mathbf{-}\mathbf{50}& \mathbf{50}\mathbf{-}\mathbf{60}\\ \mathrm{Frequencies}& \mathbf{6}& \mathbf{8}& \mathbf{14}& \mathbf{16}& \mathbf{4}& \mathbf{2}\\ \hline\end{array}$

Compute mean deviation (correct up to two decimal places) from mean of the following distribution:

$\begin{array}{|llccccccc|}\hline \mathrm{Marks}& 10-20& 20-30& 30-40& 40-50& 50-60& 60-70& 70-80& 80-90\\ \mathrm{No}.\mathrm{of}students& 8& 10& 15& 25& 20& 18& 9& 5\\ \hline\end{array}$

The age distribution of $100$ life-insurance policy holders is as follows:

$\begin{array}{|lcccccccc|}\hline \begin{array}{l}\mathrm{Age}(\mathrm{on}nearest\\ \mathrm{birth}\mathrm{day})\end{array}& \mathbf{17}\mathbf{-}\mathbf{19}\mathbf{.}\mathbf{5}& \mathbf{20}\mathbf{-}\mathbf{25}\mathbf{.}\mathbf{5}& \mathbf{26}\mathbf{-}\mathbf{35}\mathbf{.}\mathbf{5}& \mathbf{36}\mathbf{-}\mathbf{40}\mathbf{.}\mathbf{5}& \mathbf{41}\mathbf{-}\mathbf{50}\mathbf{.}\mathbf{5}& \mathbf{51}\mathbf{-}\mathbf{55}\mathbf{.}\mathbf{5}& \mathbf{56}\mathbf{-}\mathbf{60}\mathbf{.}\mathbf{5}& \mathbf{61}\mathbf{-}\mathbf{70}\mathbf{.}\mathbf{5}\\ \mathrm{No}.\mathrm{of}persons& 5& \mathbf{16}& \mathbf{12}& \mathbf{26}& \mathbf{14}& \mathbf{12}& \mathbf{6}& \mathbf{5}\\ \hline\end{array}$

Calculate the mean deviation (correct up to two decimal places) from the median age.

Calculate mean deviation about median age for the age distribution of $100$ persons given below:

$\begin{array}{|lcccccccc|}\hline \text{ Age: }& 16-20& 21-25& 26-30& 31-35& 36-40& 41-45& 46-50& 51-55\\ \text{ Number of persons }& 5& 6& 12& 14& 26& 12& 16& 9\\ \hline\end{array}$

Calculate the mean deviation about the mean for the following frequency distribution:

$\begin{array}{|lccccc|}\hline \text{ Class interval: }& 0-4& 4-8& 8-12& 12-16& 16-20\\ \text{ Frequency: }& 4& 6& 8& 5& 2\\ \hline\end{array}$

Calculate mean deviation from the median of the following data:

$\begin{array}{|lccccc|}\hline \text{ Class interval: }& 0-6& 6-12& 12-18& 18-24& 24-30\\ \text{ Frequency: }& 4& 5& 3& 6& 2\\ \hline\end{array}$