Question:medium

Find the mean deviation about the mean for the data. 

Income per day0-100100-200200-300300-400400-500500-600600-700700-800
Number of persons489107543

Updated On: Jan 22, 2026
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Solution and Explanation

The following table is formed.

Income per dayNumber of persons \(f_i\)Mid-point \(x_i\)\(f_ix_i\)\(|x_i-\bar{x}\)\(f_i|x_i-\bar{x}\)|
0-1004502003081232
100-200815012002081664
200-30092502250108972
300-400103503500880
400-5007450315092644
500-60055502750192960
600-700465026002921168
700-800375022503921176
 50 17900 7896

Here,  \(\sum_{I=1}^{8}f_i=50\sum_{1=i}^8f_ix_i=17900\)

∴ \(\bar{x}=\frac{1}{N}\sum_{i=1}^{8}f_ix_i=\frac{1}{50}×17900=358\)

\(M.D(\bar{x})=\frac{1}{N}\sum_{i=1}^{8}f_i|x_i-\bar{x_i}|=\frac{1}{50}×7896=157.92\)

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