Measures of Central Tendency and DispersionMCQPYQ Jun 24Question 3143 of 473
All Questions

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
For any discrepancies in this question, email contact@cadada.in

Correct Answer

Option d1.5\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

Ad

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=200xˉ2=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=xˉixˉ\displaystyle d_i = \bar{x}_i - \bar{x} (deviation of group mean from overall mean). d1=108=2,d2=58=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=xˉ1xˉ=108=2\displaystyle d_1 = \bar{x}_1 - \bar{x} = 10 - 8 = 2, d2=xˉ2xˉ=58=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.

About This Chapter: Measures of Central Tendency and Dispersion

Paper

Paper 3: Quantitative Aptitude

Weightage

12-15 Marks

Key Topics

Mean, Median, Mode, Range, Mean Deviation, Standard Deviation

The core foundation of Statistics. This chapter covers Mean (Arithmetic, Geometric, Harmonic), Median, Mode, and their properties. It also explores measures of spread like Range, Mean Deviation, Standard Deviation, and Quartile Deviation.

View Official ICAI Syllabus

Exam Strategy Tip

Do not just memorize formulas; ICAI loves asking about the mathematical properties (e.g., 'sum of deviations from the AM is always zero'). You can usually eliminate 2 options just by knowing the properties.

Key Concepts to Understand

Related Comparison Tables

More Questions from Measures of Central Tendency and Dispersion

Ready to Master Measures of Central Tendency and Dispersion?

Practice all 473 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free