If all the observations are increased by 6, then the variance of the series will be

Option c: Unchanged. Solution: Concept: Variance is independent of change of origin. Adding any constant to all observations does NOT change variance. Proof: If xi' = xi + 6, then x-bar' = x-bar + 6. xi' - x-bar' = xi - x-bar sigma'^2 = frac{sum(xi' - x-bar')^2}{n} = frac{sum(xi - x-bar)^2}{n} = sigma^2 Variance is Unchanged. Hence, Option C is the correct answer.

If all the observations are increased by $6$, then the variance of the series will be

The correct answer is Option c: Unchanged. **Concept:** Variance is independent of change of origin. Adding any constant to all observations does NOT change variance. **Proof:** If $x_i' = x_i + 6$, then $\bar{x}' = \bar{x} + 6$. $$x_i' - \bar{x}' = x_i - \bar{x}$$ $$\sigma'^2 = \frac{\sum(x_i' - \bar{x}')^2}{n} = \frac{\sum(x_i - \bar{x})^2}{n} = \sigma^2$$ **Variance is Unchanged.** Hence, **Option C** is the correct answer.

Central Tendency & DispersionMTP Sep 24 Series IIQuestion 3215 of 473

All Questions

If all the observations are increased by $\displaystyle 6$ , then the variance of the series will be

Options

AIncreased

BDecreased

CUnchanged

DNone of these

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Correct Answer

✅ Option c — Unchanged

All Options:

AIncreased
BDecreased
CUnchanged
DNone of these

Detailed Solution & Explanation

**Concept:** Variance is independent of change of origin. Adding any constant to all observations does NOT change variance. **Proof:** If

\displaystyle x_i' = x_i + 6

, then

\displaystyle \bar{x}' = \bar{x} + 6

x_i' - \bar{x}' = x_i - \bar{x}

\sigma'^2 = \frac{\sum(x_i' - \bar{x}')^2}{n} = \frac{\sum(x_i - \bar{x})^2}{n} = \sigma^2

**Variance is Unchanged.** Hence, **Option C** is the correct answer.

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If all the observations are increased by 6\displaystyle 66, then the variance of the series will be