Central Tendency & DispersionPYQ Jun 24Question 3143 of 473
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The mean and variance of a group of 100 observations are 8 and 9 respectively. Out of 100 observations, the mean and standard deviation of 60 observations are 10 and 2 respectively. Find the variance of remaining 40 observations?

Options

A4.5\displaystyle 4.5
B3.5\displaystyle 3.5
C2.5\displaystyle 2.5
D1.5\displaystyle 1.5
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Correct Answer

✅ Option d — 1.5\displaystyle 1.5

All Options:

  • A4.5\displaystyle 4.5
  • B3.5\displaystyle 3.5
  • C2.5\displaystyle 2.5
  • D1.5\displaystyle 1.5

Detailed Solution & Explanation

**Given:** - Total n=100\displaystyle n = 100, xˉ=8\displaystyle \bar{x} = 8, σ2=9\displaystyle \sigma^2 = 9 (so σ=3\displaystyle \sigma = 3) - Group 1: n1=60\displaystyle n_1 = 60, xˉ1=10\displaystyle \bar{x}_1 = 10, σ1=2\displaystyle \sigma_1 = 2 (so σ12=4\displaystyle \sigma_1^2 = 4) - Group 2: n2=40\displaystyle n_2 = 40, xˉ2=?\displaystyle \bar{x}_2 = ?, σ22=?\displaystyle \sigma_2^2 = ? **Step 1: Find xˉ2\displaystyle \bar{x}_2 using combined mean formula.** xˉ=n1xˉ1+n2xˉ2n1+n2\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} 8=60×10+40×xˉ21008 = \frac{60 \times 10 + 40 \times \bar{x}_2}{100} 800=600+40xˉ2800 = 600 + 40\bar{x}_2 40xˉ2=200⇒xˉ2=540\bar{x}_2 = 200 \Rightarrow \bar{x}_2 = 5 **Step 2: Use combined variance formula.** σ2=n1(σ12+d12)+n2(σ22+d22)n1+n2\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2} where di=xˉi−xˉ\displaystyle d_i = \bar{x}_i - \bar{x} (deviation of group mean from overall mean). d1=10−8=2,d2=5−8=−3d_1 = 10 - 8 = 2, \quad d_2 = 5 - 8 = -3 **Step 3:** Substitute. 9=60(4+4)+40(σ22+9)1009 = \frac{60(4 + 4) + 40(\sigma_2^2 + 9)}{100} 900=60×8+40(σ22+9)900 = 60 \times 8 + 40(\sigma_2^2 + 9) 900=480+40σ22+360900 = 480 + 40\sigma_2^2 + 360 900=840+40σ22900 = 840 + 40\sigma_2^2 40σ22=6040\sigma_2^2 = 60 σ22=1.5\sigma_2^2 = 1.5 Wait — let me verify: The answer 1.5 is Option D, but the given answer is C (2.5). Let me recompute more carefully. d1=xˉ1−xˉ=10−8=2\displaystyle d_1 = \bar{x}_1 - \bar{x} = 10 - 8 = 2, d2=xˉ2−xˉ=5−8=−3\displaystyle d_2 = \bar{x}_2 - \bar{x} = 5 - 8 = -3 Combined variance: nσ2=n1(σ12+d12)+n2(σ22+d22)n \sigma^2 = n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2) 100×9=60(4+4)+40(σ22+9)100 \times 9 = 60(4 + 4) + 40(\sigma_2^2 + 9) 900=480+40σ22+360900 = 480 + 40\sigma_2^2 + 360 900=840+40σ22900 = 840 + 40\sigma_2^2 σ22=6040=1.5\sigma_2^2 = \frac{60}{40} = 1.5 The mathematically correct answer is **1.5 (Option D)**. Hence, **Option D** is the correct answer.

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