Question:
If the mean of the data $p,6,6,7,8,11,15,16$ is 3 times $p$, then the mean deviation of the data from its mean is
If the mean of the data $p,6,6,7,8,11,15,16$ is 3 times $p$, then the mean deviation of the data from its mean is
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Mean deviation uses absolute differences from mean; compute mean first precisely.
Updated On: May 19, 2025
- $2.25$
- $3.75$
- $4.4$
- $2.5$
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The Correct Option is B
Solution and Explanation
Let total number of terms = 8 (including $p$)
Mean = $\dfrac{p + 6 + 6 + 7 + 8 + 11 + 15 + 16}{8} = \dfrac{p + 69}{8}$
Given: $\dfrac{p + 69}{8} = 3p \Rightarrow p + 69 = 24p \Rightarrow 23p = 69 \Rightarrow p = 3$
Data becomes: $3,6,6,7,8,11,15,16$
Mean = $\dfrac{72}{8} = 9$
Mean deviation = $\dfrac{1}{n} \sum |x_i - \bar{x}|$
$= \dfrac{1}{8} (|3 - 9| + |6 - 9| + |6 - 9| + |7 - 9| + |8 - 9| + |11 - 9| + |15 - 9| + |16 - 9|)$
$= \dfrac{1}{8} (6 + 3 + 3 + 2 + 1 + 2 + 6 + 7) = \dfrac{30}{8} = 3.75$
Mean = $\dfrac{p + 6 + 6 + 7 + 8 + 11 + 15 + 16}{8} = \dfrac{p + 69}{8}$
Given: $\dfrac{p + 69}{8} = 3p \Rightarrow p + 69 = 24p \Rightarrow 23p = 69 \Rightarrow p = 3$
Data becomes: $3,6,6,7,8,11,15,16$
Mean = $\dfrac{72}{8} = 9$
Mean deviation = $\dfrac{1}{n} \sum |x_i - \bar{x}|$
$= \dfrac{1}{8} (|3 - 9| + |6 - 9| + |6 - 9| + |7 - 9| + |8 - 9| + |11 - 9| + |15 - 9| + |16 - 9|)$
$= \dfrac{1}{8} (6 + 3 + 3 + 2 + 1 + 2 + 6 + 7) = \dfrac{30}{8} = 3.75$
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