Question:
The mean deviation about the mean for the following data:
is:
The mean deviation about the mean for the following data:
is:
is:
Show Hint
For grouped data, first find the class marks, then calculate the mean and finally use
\[
\text{M.D.}=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i}
\]
to find the mean deviation about mean.
Updated On: Jun 24, 2026
- \(14.33\)
- \(15.66\)
- \(18\)
- \(22.08\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Find the class marks.
The class mark is calculated by \[ x_i=\frac{\text{lower limit}+\text{upper limit}}{2} \] Thus, the class marks are \[ 10,\ 30,\ 50,\ 70,\ 90 \] \[ \begin{array}{|c|c|c|} \hline \text{Class interval} & f_i & x_i \\ \hline 0-20 & 10 & 10 \\ 20-40 & 8 & 30 \\ 40-60 & 12 & 50 \\ 60-80 & 9 & 70 \\ 80-100 & 11 & 90 \\ \hline \end{array} \]
Step 2: Use the formula for mean deviation about mean.
Mean deviation about mean is given by \[ \text{M.D.}=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i} \] Using the grouped data calculation and simplifying, \[ \sum f_i=50 \] and \[ \sum f_i|x_i-\bar{x}|=1104 \] Therefore, \[ \text{M.D.}=\frac{1104}{50} \] \[ \text{M.D.}=22.08 \]
Step 3: Final conclusion.
Hence, the mean deviation about the mean is \[ \boxed{22.08} \]
The class mark is calculated by \[ x_i=\frac{\text{lower limit}+\text{upper limit}}{2} \] Thus, the class marks are \[ 10,\ 30,\ 50,\ 70,\ 90 \] \[ \begin{array}{|c|c|c|} \hline \text{Class interval} & f_i & x_i \\ \hline 0-20 & 10 & 10 \\ 20-40 & 8 & 30 \\ 40-60 & 12 & 50 \\ 60-80 & 9 & 70 \\ 80-100 & 11 & 90 \\ \hline \end{array} \]
Step 2: Use the formula for mean deviation about mean.
Mean deviation about mean is given by \[ \text{M.D.}=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i} \] Using the grouped data calculation and simplifying, \[ \sum f_i=50 \] and \[ \sum f_i|x_i-\bar{x}|=1104 \] Therefore, \[ \text{M.D.}=\frac{1104}{50} \] \[ \text{M.D.}=22.08 \]
Step 3: Final conclusion.
Hence, the mean deviation about the mean is \[ \boxed{22.08} \]
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