If the mean and SD of X are a and b respectively, then the S.D of X = X-a/b is

Option c: 1. Solution: Given: Mean of X = a , SD of X = b Let Z = X - a/b (this is the standard normal transformation / z-score) Step 1: Find mean of Z . E(Z) = E(X-a/b) = E(X) - a/b = a - a/b = 0 Step 2: Find SD of Z . Using the property: if Z = X - a/b = 1/bX - a/b , this is a linear transformation Z = c + kX where k = 1/b : SD(Z) = |k| × SD(X) = 1/|b| × b = b/b = 1 (assuming b > 0 ) Result: SD of Z = 1 Hence, Option C is the correct answer.

If the mean and SD of $X$ are $a$ and $b$ respectively, then the S.D of $X = \frac{X-a}{b}$ is

The correct answer is Option c: $1$. **Given:** Mean of $X = a$, SD of $X = b$ **Let** $Z = \frac{X - a}{b}$ (this is the standard normal transformation / z-score) **Step 1: Find mean of $Z$.** $$E(Z) = E\left(\frac{X-a}{b}\right) = \frac{E(X) - a}{b} = \frac{a - a}{b} = 0$$ **Step 2: Find SD of $Z$.** Using the property: if $Z = \frac{X - a}{b} = \frac{1}{b}X - \frac{a}{b}$, this is a linear transformation $Z = c + kX$ where $k = \frac{1}{b}$: $$\text{SD}(Z) = |k| \cdot \text{SD}(X) = \frac{1}{|b|} \times b = \frac{b}{b} = 1$$ (assuming $b > 0$) **Result:** SD of $Z = 1$ Hence, **Option C** is the correct answer.

Measures of Central Tendency and DispersionMCQMTP May 18Question 3150 of 473

All Questions

If the mean and SD of $\displaystyle X$ are $\displaystyle a$ and $\displaystyle b$ respectively, then the S.D of $\displaystyle X = \frac{X-a}{b}$ is

Options

\displaystyle a/b

\displaystyle -1

\displaystyle 1

\displaystyle ab

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

Correct Answer

✅ Option c — $\displaystyle 1$

All Options:

A $\displaystyle a/b$
B $\displaystyle -1$
C $\displaystyle 1$
D $\displaystyle ab$

Detailed Solution & Explanation

**Given:** Mean of

\displaystyle X = a

, SD of

\displaystyle X = b

**Let**

\displaystyle Z = \frac{X - a}{b}

(this is the standard normal transformation / z-score) **Step 1: Find mean of

\displaystyle Z

.**

E(Z) = E\left(\frac{X-a}{b}\right) = \frac{E(X) - a}{b} = \frac{a - a}{b} = 0

**Step 2: Find SD of

\displaystyle Z

.** Using the property: if

\displaystyle Z = \frac{X - a}{b} = \frac{1}{b}X - \frac{a}{b}

, this is a linear transformation

\displaystyle Z = c + kX

where

\displaystyle k = \frac{1}{b}

\text{SD}(Z) = |k| \cdot \text{SD}(X) = \frac{1}{|b|} \times b = \frac{b}{b} = 1

(assuming

\displaystyle b > 0

) **Result:** SD of

\displaystyle Z = 1

Hence, **Option C** 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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If the mean and SD of X\displaystyle XX are a\displaystyle aa and b\displaystyle bb respectively, then the S.D of X=X−ab\displaystyle X = \frac{X-a}{b}X=bX−a​ is