JEE Main 2023 · 11 April · evening · Single correct · +4 / −1

Let the mean of 6 observations 1,2,4,5, x and y be 5 and their variance be 10 . Then their mean deviation about the mean is equal to :

  1. (a) 10/3
  2. (b) 8/3
  3. (c) 7/3
  4. (d) 3
Answer and solution

Answer (b) 8/3

Solution

Maths sir · 4 steps

The mean gives you x + y and the variance gives x² + y², so pin down x and y before taking deviations.

  1. Use the mean. The six observations add up to 6 × 5 = 30. Since 1 + 2 + 4 + 5 = 12, this gives x + y = 18.
  2. Use the variance. Variance = (Σ xᵢ²)/6 - 5² = 10, so Σ xᵢ² = 6 × 35 = 210. Since 1 + 4 + 16 + 25 = 46, this gives x² + y² = 164.
  3. Find x and y. 2xy = (x+y)² - (x²+y²) = 324 - 164 = 160, so xy = 80. Two numbers with sum 18 and product 80 are 8 and 10.

    M.D. = (|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|)/6 = (4+3+1+0+3+5)/6

  4. So. The absolute deviations add up to 16, so the mean deviation about the mean is 16/6 = 8/3, which is option (b).

The idea to keep. Two unknowns, two conditions: the mean fixes their sum and the variance fixes their sum of squares. Find the actual values, because mean deviation needs each one.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 100 times in JEE Main since 2020. This question is from 2023.