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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Statistics

Let the six numbers a1,a2,a3,a4,a5,a6a_1, a_2, a_3, a_4, a_5, a_6, be in A.P. and a1+a3=10a_1 + a_3 = 10. If the mean of these six numbers is 192\dfrac{19}{2} and their variance is σ2\sigma^2, then 8σ28\sigma^2 is equal to :
  1. A.
    200
  2. B.
    210
  3. C.
    220
  4. D.
    105
Show answer and solution

Correct option: B

Solution

Let the AP have first term a and common difference d. From a₁ + a₃ = 10: 2a + 2d = 10, so a + d = 5, meaning a₂ = 5.

Mean of six terms = a + (5/2)d = 19/2. Since a = 5 - d: (5 - d) + (5/2)d = 19/2 → 5 - d/2 = 19/2 → d/2 = 5 - 19/2 = 1/2, so d = 1... wait let me redo.

a + (5/2)d = 19/2. With a = 5 - d: 5 - d + (5/2)d = 19/2 → 5 + (3/2)d = 19/2 → (3/2)d = 9/2 → d = 3.

So a = 2, and the six terms are: 2, 5, 8, 11, 14, 17.

Mean = 57/6 = 19/2 ✓

Variance σ² = (1/6)Σ(aᵢ - mean)². Deviations from 19/2: (2 - 19/2) = -15/2, (5 - 19/2) = -9/2, (8 - 19/2) = -3/2, (11 - 19/2) = 3/2, (14 - 19/2) = 9/2, (17 - 19/2) = 15/2

[JEE Main 2023 · 24 Jan · Shift 2]

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