Question

Let the Standard deviation of $x_1,x_2$ and $x_3$ be $9$ .Then ,the variance of $3x_1+4,$ $3x_2+4$ and $3x_3+4$ is

• A. $243$
• B. $81$
• C. $729$
• D. $9$
• E. $733$

Answer 1

Answer: (B)

$81$

Answer 2

Given that:

Standard deviation of $x_1,x_2$ and $x_3$ be .

The variance of random variables $3x_1 + 4, 3x_2 + 4$, and $3x_3 + 4,$ to be found ;

For a random variable $X$ with standard deviation $σ$ , the variance of a linear transformation$(aX + b)$, where $'a'$ and $'b'$ are constants and can be calculated as follows:

$Var(aX + b) = a^2 × Var(X)$

applying this to all three given variables we get;

For, $3x_1 + 4: Var(3x_1 + 4)$

$= 3^2 × Var(x1) = 9 × 9 = 81$

For, $3x_2 + 4: Var(3x_2 + 4)$

$= 3^2 × Var(x_2) = 9 × 9 = 81$

For, $3x_3 + 4: Var(3x_3 + 4)$

$= 3^2 × Var(x3) = 9 × 9 = 81$

Hence, the variance for the random variable is $81$ for each of them.(_Ans)