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Worked Examples · Example 31

Q.Find the mean deviation about the mean and median for the data: 4, 3, 2, 5, 7, 6, 8.

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✓ Free question

For the data 4,3,2,5,7,6,84,3,2,5,7,6,8, mean == median =5=5, so the mean deviation about the mean and the mean deviation about the median are both 127≈1.71\dfrac{12}{7}\approx1.71.

Mean deviation about the mean: MDxˉ=∑i=1n∣xi−xˉ∣nMD_{\bar x}=\dfrac{\sum_{i=1}^n|x_i-\bar x|}{n}

Mean deviation about the median: MDM=∑i=1n∣xi−M∣nMD_{M}=\dfrac{\sum_{i=1}^n|x_i-M|}{n}

where xˉ\bar x = mean, MM = median, nn = number of observations.

  1. Arrange the data in ascending order: 2,3,4,5,6,7,82,3,4,5,6,7,8; here n=7n=7.
  2. Mean: xˉ=4+3+2+5+7+6+87=357=5\bar x=\dfrac{4+3+2+5+7+6+8}{7}=\dfrac{35}{7}=5.
  3. Median: since n=7n=7 is odd, the median is the (n+12)th=4th\left(\dfrac{n+1}{2}\right)^{th}=4^{th} value of the sorted data, i.e. M=5M=5. So here xˉ=M=5\bar x=M=5, meaning the two mean-deviation calculations use the identical set of deviations.
  4. Absolute-deviation table (from 55):
xix_i2345678
∣xi−5∣\lvert x_i-5\rvert3210123
  1. ∑∣xi−5∣=3+2+1+0+1+2+3=12\sum\lvert x_i-5\rvert = 3+2+1+0+1+2+3=12.
  2. MDxˉ=127≈1.71MD_{\bar x}=\dfrac{12}{7}\approx1.71 and, since M=xˉM=\bar x, MDM=127≈1.71MD_{M}=\dfrac{12}{7}\approx1.71 as well.
  3. Self-check: 12/7=1.7142…12/7 = 1.7142\ldots, rounds to 1.711.71. ✓
✓Final answer

Mean deviation about the mean =1.71=1.71; mean deviation about the median =1.71=1.71 (both equal 127\tfrac{12}{7}, because xˉ=M=5\bar x=M=5)

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