Question

If the standard deviation of \( n \) elements of the series \( x_1, x_2, x_3, \dots, x_n \) is \( \sigma \), then find the variance of the series \( ax_1, ax_2, ax_3, \dots, ax_n \) is:

• A. \( a^2\sigma \)
• B. \( a^2 n\sigma \)
• C. \( a\sigma \)
• D. \( a^2\sigma^2 \)

Hint:

Remember the scaling property of dispersion parameters:
- Standard Deviation scales linearly: \( \text{S.D.}(aX) = |a|\text{S.D.}(X) \)
- Variance scales quadratically: \( \text{Var}(aX) = a^2\text{Var}(X) \)

Answer 1

The Correct Option is D

Step 1: Separate the two quantities asked.
We are given the standard deviation $\sigma$ of the original data and asked for the variance of the scaled data.
Step 2: Recall variance from a definition view.
Variance is the mean of squared deviations: $\sigma^2 = \dfrac{1}{n}\sum (x_i - \bar{x})^2$.
Step 3: Scale every observation by $a$.
Replacing $x_i$ with $ax_i$ also scales the mean to $a\bar{x}$, so each deviation becomes $a(x_i-\bar{x})$.
Step 4: Square the scaled deviations.
\[ (ax_i - a\bar{x})^2 = a^2 (x_i-\bar{x})^2 \]
Step 5: Take the mean of these.
\[ \text{Var}(aX) = \frac{1}{n}\sum a^2 (x_i-\bar{x})^2 = a^2 \cdot \frac{1}{n}\sum (x_i-\bar{x})^2 = a^2 \sigma^2 \]
Step 6: State the result.
The constant $a$ pulls out as $a^2$, so the new variance is $a^2\sigma^2$. \[ \boxed{a^2\sigma^2} \]