If the standard deviation of 1, 2, 3, 4, ..., 10 is sigma, then the SD of 11, 12, 13, 14, ..., 20 is:

Option c: sigma. Solution: Key insight: The series 11, 12, 13, ..., 20 is obtained by adding 10 to each term of 1, 2, 3, ..., 10. Property: Adding a constant to all observations does NOT change the Standard Deviation. Proof: If xi' = xi + 10, then x-bar' = x-bar + 10. xi' - x-bar' = (xi + 10) - (x-bar + 10) = xi - x-bar Deviations are unchanged, hence SD is unchanged. Conclusion: SD of {11, 12, ..., 20} = SD of {1, 2, ..., 10} = sigma Hence, Option C is the correct answer.

If the standard deviation of $1, 2, 3, 4, ..., 10$ is $\sigma$, then the SD of $11, 12, 13, 14, ..., 20$ is:

The correct answer is Option c: $\sigma$. **Key insight:** The series $11, 12, 13, ..., 20$ is obtained by adding 10 to each term of $1, 2, 3, ..., 10$. **Property:** Adding a constant to all observations does NOT change the Standard Deviation. **Proof:** If $x_i' = x_i + 10$, then $\bar{x}' = \bar{x} + 10$. $$x_i' - \bar{x}' = (x_i + 10) - (\bar{x} + 10) = x_i - \bar{x}$$ Deviations are unchanged, hence SD is unchanged. **Conclusion:** SD of $\{11, 12, ..., 20\}$ = SD of $\{1, 2, ..., 10\}$ = $\sigma$ Hence, **Option C** is the correct answer.

Central Tendency & DispersionMTP Dec 22 - Series IQuestion 3188 of 473

All Questions

If the standard deviation of $\displaystyle 1, 2, 3, 4, ..., 10$ is $\displaystyle \sigma$ , then the SD of $\displaystyle 11, 12, 13, 14, ..., 20$ is:

Options

\displaystyle 10\sigma

\displaystyle 10+\sigma

\displaystyle \sigma

DNone of these

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Correct Answer

✅ Option c — $\displaystyle \sigma$

All Options:

A $\displaystyle 10\sigma$
B $\displaystyle 10+\sigma$
C $\displaystyle \sigma$
DNone of these

Detailed Solution & Explanation

**Key insight:** The series

\displaystyle 11, 12, 13, ..., 20

is obtained by adding 10 to each term of

\displaystyle 1, 2, 3, ..., 10

. **Property:** Adding a constant to all observations does NOT change the Standard Deviation. **Proof:** If

\displaystyle x_i' = x_i + 10

, then

\displaystyle \bar{x}' = \bar{x} + 10

x_i' - \bar{x}' = (x_i + 10) - (\bar{x} + 10) = x_i - \bar{x}

Deviations are unchanged, hence SD is unchanged. **Conclusion:** SD of

\displaystyle \{11, 12, ..., 20\}

= SD of

\displaystyle \{1, 2, ..., 10\}

\displaystyle \sigma

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

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If the standard deviation of 1,2,3,4,...,10\displaystyle 1, 2, 3, 4, ..., 101,2,3,4,...,10 is σ\displaystyle \sigmaσ, then the SD of 11,12,13,14,...,20\displaystyle 11, 12, 13, 14, ..., 2011,12,13,14,...,20 is: