Question

If `X` is a random variable such that `sigma(X)=2.6 ` then `sigma(1-4X)` is equal to ` 4$ 4 5 21/3 (A) 7/2 97 (A) 7.8 (C) 13 `

Answer 1

Answer:

The value of $(\sigma_{1-4X} )$ is 10.4.

Step-by-step explanation:

It is provided that 'X' is random variable with standard deviation $(\sigma_{X} )$ = 2.6

Then the variance of the random variable X is,

$Variance (X) = (Standard \ Deviation\ (X))^{2} \\\sigma^{2}_{X} = (\sigma_{X} )^{2} \\=(2.6)^{2}\\ =6.76$

Compute the variance of (1 - 4X) using the formula $Var(aX-b)=a^{2} Var(X)$ as follows:

$Var(aX-b)=a^{2} Var(X)\\Var(1-4X)=4^{2} Var(X)\\=16\times6.76\\=108.16$

Then the value of $(\sigma_{1-4X} )$ is:

$Var(1-4X)=108.16\\\sigma_{1-4X} =\sqrt{Var(1-4X)} \\=\sqrt{108.16}\\ =10.4$

Thus, the value of $(\sigma_{1-4X} )$ is 10.4.