Measures of Central Tendency and DispersionPYQ Jan 26Question 4537 of 473
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If two samples of size 30 and 20 have means as 55 and 60 and standard deviation 5 and 6 respectively then what would be the standard deviation of combined sample of size 50?

Options

A5.95
B5.90
C5.85
D5.80
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Correct Answer

✅ Option a — 5.95

All Options:

  • A5.95
  • B5.90
  • C5.85
  • D5.80

Detailed Solution & Explanation

The formula for the combined standard deviation (σ12\displaystyle \sigma_{12}) of two samples is:
σ12=n1(σ12+d12)+n2(σ22+d22)n1+n2\sigma_{12} = \sqrt{\frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}}
where:
- n1=30\displaystyle n_1 = 30, xˉ1=55\displaystyle \bar{x}_1 = 55, σ1=5\displaystyle \sigma_1 = 5
- n2=20\displaystyle n_2 = 20, xˉ2=60\displaystyle \bar{x}_2 = 60, σ2=6\displaystyle \sigma_2 = 6
- d1=xˉ1−xˉ12\displaystyle d_1 = \bar{x}_1 - \bar{x}_{12}
- d2=xˉ2−xˉ12\displaystyle d_2 = \bar{x}_2 - \bar{x}_{12}
- xˉ12\displaystyle \bar{x}_{12} is the combined mean.

**Step 1: Calculate the combined mean (xˉ12\displaystyle \bar{x}_{12})**
xˉ12=n1xˉ1+n2xˉ2n1+n2\bar{x}_{12} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}
xˉ12=30(55)+20(60)30+20=1650+120050=285050=57\bar{x}_{12} = \frac{30(55) + 20(60)}{30 + 20} = \frac{1650 + 1200}{50} = \frac{2850}{50} = 57

**Step 2: Calculate the deviations (d1\displaystyle d_1 and d2\displaystyle d_2)**
d1=xˉ1−xˉ12=55−57=−2  ⟹  d12=4d_1 = \bar{x}_1 - \bar{x}_{12} = 55 - 57 = -2 \implies d_1^2 = 4
d2=xˉ2−xˉ12=60−57=3  ⟹  d22=9d_2 = \bar{x}_2 - \bar{x}_{12} = 60 - 57 = 3 \implies d_2^2 = 9

**Step 3: Calculate the combined standard deviation (σ12\displaystyle \sigma_{12})**
Substitute the values into the formula:
σ122=30(52+4)+20(62+9)30+20\sigma_{12}^2 = \frac{30(5^2 + 4) + 20(6^2 + 9)}{30 + 20}
σ122=30(25+4)+20(36+9)50\sigma_{12}^2 = \frac{30(25 + 4) + 20(36 + 9)}{50}
σ122=30(29)+20(45)50\sigma_{12}^2 = \frac{30(29) + 20(45)}{50}
σ122=870+90050=177050=35.4\sigma_{12}^2 = \frac{870 + 900}{50} = \frac{1770}{50} = 35.4
σ12=35.4≈5.9498\sigma_{12} = \sqrt{35.4} \approx 5.9498
Thus, the standard deviation of the combined sample is approximately 5.95\displaystyle 5.95.

Hence, **Option A** 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.

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