JEE Main 2021 · 27 July · evening · Single correct · +4 / −1

Let the mean and variance of the frequency distribution
x:, x₁ = 2, x₂ = 6, x₃ = 8, x₄ = 9; f:, 4, 4, α, β
be 6 and 6.8 respectively. If x₃ is changed from 8 to 7, then the mean for the new data will be :

  1. (a) 4
  2. (b) 5
  3. (c) 17/3
  4. (d) 16/3
Answer and solution

Answer (c) 17/3

Solution

Maths sir · 4 steps

Mean and variance give two equations in α and β. Solve them, then shift only the x₃ terms.

  1. Use the mean. Total frequency is N = 8 + α + β, and Σ f x = 8 + 24 + 8α + 9β = 32 + 8α + 9β. Setting Σ f x = 6N gives 32 + 8α + 9β = 48 + 6α + 6β, so 2α + 3β = 16.
  2. Use the variance. Measure deviations from the mean 6. The squared deviations are 16, 0, 4, 9, so Σ f(x-6)² = 64 + 4α + 9β.

    (64 + 4α + 9β)/(8 + α + β) = 6.8 ⇒ 14α - 11β = 48

  3. Solve the pair. From the first equation, α = (16 - 3β)/2. Putting this into 14α - 11β = 48 gives 112 - 21β - 11β = 48, so β = 2 and α = 5. Then N = 15 and Σ f x = 6 × 15 = 90.
  4. So. Changing x₃ from 8 to 7 lowers each of its α = 5 values by 1, so the sum drops by 5 to 85. The new mean is 85/15 = 17/3, option (c).

The idea to keep. When one value changes, do not recompute everything. The new sum is the old sum plus (change in x) times its frequency, and N stays the same.

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

Study Statistics

Asked about 100 times in JEE Main since 2020. This question is from 2021.