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Question: If X is a random variable such that is \[\sigma ({\text{X}})\] = 2.6, then \[\sigma (1 - 4{\text{X}}...

If X is a random variable such that is σ(X)\sigma ({\text{X}}) = 2.6, then σ(1−4X)\sigma (1 - 4{\text{X}}) is equal to ?
A. 7.8
B. -10.4
C. 13
D. 10.4

Explanation

Solution

Hint: As we know that variance σ2(X{\sigma ^2}({\text{X}}) is equal to the square of the standard deviation σ(X{\sigma ({\text{X}}}). So, let us find variance for random variable X.

Complete step-by-step answer:
As we know that, if Y is any random variable then σ(Y)\sigma ({\text{Y}}) denotes standard deviation of Y. And σ2(Y){\sigma ^2}({\text{Y}}) denotes variance of Y.
So, Variance(Y) = (S.D.)2{({\text{S}}{\text{.D}}{\text{.}})^2}
So, Variance(X) = (σ(X))2{\left( {\sigma ({\text{X}})} \right)^2} = (2.6)2{(2.6)^2} = 6.76
As, we know that, if Y is any random variable then,
Variance (aY + b) = {{\text{a}}^2}$$$\times$ variance(Y), where a and b are constant. So, we write (1 – 4X) as ( – 4X + 1). So, a = –4 and b = 1. So, Variance (1 – 4X) = Variance ( – 4X + 1) = {\left( { - 4} \right)^2}$$$\timesVariance(X)=16Variance (X) = 16\times$ 6.76 = 108.16
Now we know that variance is the square of standard deviation.
So, σ(1−4X)\sigma (1 - 4{\text{X}}) will be the square root of Variance (1 – 4X).
So, σ(1−4X)\sigma (1 - 4{\text{X}}) = 108.16\sqrt {108.16} = 10.4
Hence, the correct option will be D.

Note: Whenever we come up with this type of problem then first, we have to find the variance of the given standard deviation Y, then we had to use that to find the variance of (aY + b). And after that we will square root the variance to find the required standard deviation and remember that standard deviation is always positive. This will be the easiest and efficient way to find the solution of the problem.