The answer has "the division by n" part missing.

The Gateway to Computer Science Excellence

+10 votes

Which of the following assertions are **CORRECT**?

- P: Adding $7$ to each entry in a list adds $7$ to the mean of the list
- Q: Adding $7$ to each entry in a list adds $7$ to the standard deviation of the list
- R: Doubling each entry in a list doubles the mean of the list
- S: Doubling each entry in a list leaves the standard deviation of the list unchanged

- $P$, $Q$
- $Q$, $R$
- $P$, $R$
- $R$, $S$

+22 votes

Best answer

Suppose we double each entry of a list

Initial Mean $(M_I)=\frac{ \sum_{i=1}^{n} x_i}{n}$

New Mean $(M_N)= \frac{\sum_{i=1}^{n}2\times x_i}{n}$

$\quad \quad \quad =\frac{2}{n} \sum_{i=1}^{n}x_i$

So, when each entry in the list is doubled, mean also gets doubled.

Standard Deviation $\sigma_I = \sqrt {\frac{1}{N}\sum_{i=1}^{n} \left(M_I - x_i\right)^2}$

New Standard Deviation $\sigma_N = \sqrt {\frac{1}{N}\sum_{i=1}^{n} \left(M_N - 2 \times x_i\right)^2}$

$\quad \quad \quad= \sqrt {\frac{1}{N}\sum_{i=1}^{n} \left(2 \times \left(M_I - x_i\right)\right)^2}$

$\quad \quad \quad= 2 \sigma_I$

So, when each entry is doubled, standard deviation also gets doubled.

When we add a constant to each element of the list, it gets added to the mean as well. This can be seen from the formula of mean.

When we add a constant to each element of the list, the standard deviation (or variance) remains unchanged. This is because, the mean also gets added by the same constant and hence the deviation from the mean remains the same for each element.

So, here P and R are correct.

Correct Answer: $C$

52,223 questions

59,811 answers

201,020 comments

118,087 users