Question:medium

Find the mean deviation about the median for the data

xi1521273035
fi35678

Updated On: Jan 22, 2026
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Solution and Explanation

The given observations are already in ascending order. 

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

\(x_i\)\(f_i\)\(c.f\)
1533
2158
27614
30721
35829

Here, N = 29, which is odd.

∴ Median= \((\frac{29+1}{2})^{th}\)  observation = 15th observation.

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30. 

∴ Median = 30

The absolute values of the deviations from median, i.e \(|x_i-M|,\) are 

\(|x_i,M|\)159305
\(f_i\)35678
\(f_i|x-M|\)454518040

\(\sum_{I=1}^{5}f_i=29\) ,  \(\sum_{I=1}^{5}f_i|x_i-M|=148\)

\(=M.D.(M)=\frac{1}{N}\sum_{i=1}^{5}f_i|x_i-M|=\frac{1}{29}×148=5.1\)

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