Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Calculate the mean deviation from the median of the following data: Class inerval 0to6 Frequency 4

Calculate the mean deviation from the median of the following data:

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

Answers (1)

A frequency distribution table is given and we have to find the mean deviation about the median

 Let us make a table from the given data and fill out the other columns after calculation

\begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & \begin{array}{l} \text { Cumulative } \\ \text { frequency } \\ \left(\mathrm{c}_{\mathrm{f}} f\right) \end{array} \\ \hline 0-6 & 3 & 4 & 4 \\ \hline 6-12 & 9 & 5 & 9 \\ \hline 12-18 & 15 & 3 & 12 \\ \hline 18-24 & 21 & 6 & 18 \\ \hline 24-30 & 27 & 2 & 20 \\ \hline & \text { Total } & \mathrm{N}=20 & \\ \hline \end{array}

 

Now here N=20, which is even

Here \text{Median Class}= \frac{N}{2}=10^{th}~term~ \\\\

 \\ \text{This observation lie in the class interval 12-18, so median can be written as} \\ M=l+\frac{\frac{N}{2}-cf}{f}*h \\ \text{ Here l=12, cf=9, f=3, h=6 and N=20} \\\text{Substituting these values, the above equation becomes}, M=12+\frac{\frac{20}{2}-9}{3}*6\\ M=12+\frac{10-9}{3}*6

\\ =12+\frac{1\ast6}{3}~~ \\\\ \\ ~ M=12+2=14 \\\\

\begin{array}{|l|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & d_{i}=\left|x_{i}-M\right| & f_{i} d_{i} \\ \hline 0-6 & 3 & 4 & 11 & 44 \\ \hline 6-12 & 9 & 5 & 5 & 25 \\ \hline 12-18 & 15 & 3 & 1 & 3 \\ \hline 18-24 & 21 & 6 & 7 & 42 \\ \hline 24-30 & 27 & 2 & 13 & 26 \\ \hline & \text { Total } & \mathrm{N}=20 & & =140 \\ \hline \end{array}

\\ mean Deviation=\frac{ \Sigma f_{i}d_{i}}{ \Sigma f_{i}}=\frac{140}{20}=7 \\\\

 

 

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question