Question

Let $x_1 , x_2,...., x_n$ be n observations, and let $\bar{x}$ be their arithmetic mean and $\sigma^2$ be the variance. Variance of $2x_1, 2x_2, ..., 2x_n$ is $4 \sigma^2$. Arithmetic mean $2x_1, 2x_2, ..., 2x_n $ is 4 $\bar{x}$ .

• A. Statement-1 is false, Statement-2 is true
• B. Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1
• C. Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1
• D. Statement-1 is true, statement-2 is false

Answer 1

The Correct Option is D

If each observation is multiplied by k, mean gets multiplied by k and variance gets multiplied by $k^2$. Hence the new mean should be $2 \bar{x}$ and new variance should be $k^2 \sigma^2$. So statement-1 is true and statement-2 is false.

Answer 2

\(\sigma^2=\sum\frac{x_i^2}{n}-(\sum\frac{x_i}{n})^2\)
and variance of \(2x_1.......2x_n\)
\(\sum\frac{(2x_i)^2}{n}-(\sum\frac{(2x_i)}{n})^2=4\sigma^2\)
\(\therefore\) Statement 1: variance of \(2x_1,2x_2.......2x_n\) is 
\(4\sigma^2 \space{is} \space true\)
\(Then\space A.M\space of 2x_1......2x_n\)
\(=\frac{2x_1+2x_2+......+2x_n}{n}\)
\(=2\bar{x}\)
\(\therefore\space statement\space 2\space is \space false\)