Consider the data sets: X = [-6, 2, -2, 6] , Y = [4, 8, 2, 6] Z = [103, 100, 102, 101] Let SX, SY, SZ be the standard deviations of the sets X, Y and Z respectively. We have the relations,

Option b: SX > SY > SZ. Solution: Compute SX : Data: -6, 2, -2, 6 X = -6+2-2+6/4 = 0 sigmaX2 = (-6)2+(2)2+(-2)2+(6)2/4 = 36+4+4+36/4 = 80/4 = 20 SX = sqrt(20) ≈ 4.47 Compute SY : Data: 4, 8, 2, 6 Y = 4+8+2+6/4 = 5 sigmaY2 = (4-5)2+(8-5)2+(2-5)2+(6-5)2/4 = 1+9+9+1/4 = 20/4 = 5 SY = sqrt(5) ≈ 2.24 Compute SZ : Data: 103, 100, 102, 101 Z = 103+100+102+101/4 = 406/4 = 101.5 sigmaZ2 = (1.5)2+(-1.5)2+(0.5)2+(-0.5)2/4 = 2.25+2.25+0.25+0.25/4 = 5/4 = 1.25 SZ = sqrt(1.25) ≈ 1.12 Comparison: SX ≈ 4.47 > SY ≈ 2.24 > SZ ≈ 1.12 SX > SY > SZ Hence, Option B is the correct answer.

Consider the data sets: $X = [-6, 2, -2, 6]$, $Y = [4, 8, 2, 6]$ $Z = [103, 100, 102, 101]$ Let $S_X, S_Y, S_Z$ be the standard deviations of the sets $X, Y$ and $Z$ respectively. We have the relations,

The correct answer is Option b: $S_X > S_Y > S_Z$. **Compute $S_X$:** Data: $-6, 2, -2, 6$ $$\bar{X} = \frac{-6+2-2+6}{4} = 0$$ $$\sigma_X^2 = \frac{(-6)^2+(2)^2+(-2)^2+(6)^2}{4} = \frac{36+4+4+36}{4} = \frac{80}{4} = 20$$ $$S_X = \sqrt{20} \approx 4.47$$ **Compute $S_Y$:** Data: $4, 8, 2, 6$ $$\bar{Y} = \frac{4+8+2+6}{4} = 5$$ $$\sigma_Y^2 = \frac{(4-5)^2+(8-5)^2+(2-5)^2+(6-5)^2}{4} = \frac{1+9+9+1}{4} = \frac{20}{4} = 5$$ $$S_Y = \sqrt{5} \approx 2.24$$ **Compute $S_Z$:** Data: $103, 100, 102, 101$ $$\bar{Z} = \frac{103+100+102+101}{4} = \frac{406}{4} = 101.5$$ $$\sigma_Z^2 = \frac{(1.5)^2+(-1.5)^2+(0.5)^2+(-0.5)^2}{4} = \frac{2.25+2.25+0.25+0.25}{4} = \frac{5}{4} = 1.25$$ $$S_Z = \sqrt{1.25} \approx 1.12$$ **Comparison:** $$S_X \approx 4.47 > S_Y \approx 2.24 > S_Z \approx 1.12$$ $$\therefore S_X > S_Y > S_Z$$ Hence, **Option B** is the correct answer.

Measures of Central Tendency and DispersionMCQPYQ June 24Question 3146 of 473

All Questions

Consider the data sets: $\displaystyle X = [-6, 2, -2, 6]$ , $\displaystyle Y = [4, 8, 2, 6]$ $\displaystyle Z = [103, 100, 102, 101]$ Let $\displaystyle S_X, S_Y, S_Z$ be the standard deviations of the sets $\displaystyle X, Y$ and $\displaystyle Z$ respectively. We have the relations,

Options

\displaystyle S_X < S_Y < S_Z

\displaystyle S_X > S_Y > S_Z

\displaystyle S_X < S_Z < S_Y

\displaystyle S_X < S_Y = S_Z

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Correct Answer

✅ Option b — $\displaystyle S_X > S_Y > S_Z$

All Options:

A $\displaystyle S_X < S_Y < S_Z$
B $\displaystyle S_X > S_Y > S_Z$
C $\displaystyle S_X < S_Z < S_Y$
D $\displaystyle S_X < S_Y = S_Z$

Detailed Solution & Explanation

**Compute

\displaystyle S_X

:** Data:

\displaystyle -6, 2, -2, 6

\bar{X} = \frac{-6+2-2+6}{4} = 0

\sigma_X^2 = \frac{(-6)^2+(2)^2+(-2)^2+(6)^2}{4} = \frac{36+4+4+36}{4} = \frac{80}{4} = 20

S_X = \sqrt{20} \approx 4.47

**Compute

\displaystyle S_Y

:** Data:

\displaystyle 4, 8, 2, 6

\bar{Y} = \frac{4+8+2+6}{4} = 5

\sigma_Y^2 = \frac{(4-5)^2+(8-5)^2+(2-5)^2+(6-5)^2}{4} = \frac{1+9+9+1}{4} = \frac{20}{4} = 5

S_Y = \sqrt{5} \approx 2.24

**Compute

\displaystyle S_Z

:** Data:

\displaystyle 103, 100, 102, 101

\bar{Z} = \frac{103+100+102+101}{4} = \frac{406}{4} = 101.5

\sigma_Z^2 = \frac{(1.5)^2+(-1.5)^2+(0.5)^2+(-0.5)^2}{4} = \frac{2.25+2.25+0.25+0.25}{4} = \frac{5}{4} = 1.25

S_Z = \sqrt{1.25} \approx 1.12

**Comparison:**

S_X \approx 4.47 > S_Y \approx 2.24 > S_Z \approx 1.12

\therefore S_X > S_Y > S_Z

Hence, **Option B** 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

Standard Deviation

A measure of the dispersion or spread of values in a data set around the mean. It is the square root of Variance. A lower standard deviation indicates data points are clustered close to the mean; a higher value indicates greater spread.

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Correct Answer

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Detailed Solution & Explanation

About This Chapter: Measures of Central Tendency and Dispersion

Key Concepts to Understand

Standard Deviation

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