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Q.Find the mean deviation about the median of the data : 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2025Subjective· 2mImportance★★★★★
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Median =47.5=47.5; averaging ∣xi−47.5∣|x_i-47.5| over all 10 values gives mean deviation =7=7.

Arrange the data in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.36,\ 42,\ 45,\ 46,\ 46,\ 49,\ 51,\ 53,\ 60,\ 72.

With n=10n=10 (even), the median is the average of the 5th and 6th values:

Median=46+492=47.5.\text{Median} = \dfrac{46+49}{2} = 47.5.

Compute absolute deviations ∣xi−47.5∣|x_i - 47.5|:

| xix_i | 36 | 42 | 45 | 46 | 46 | 49 | 51 | 53 | 60 | 72 |

|---|---|---|---|---|---|---|---|---|---|---| …

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