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.

Central Tendency & DispersionMTP 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

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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.

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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