Question on AP Stats variance and standard deviation [closed]

$\begingroup$

This was a question from the 2018 EXAM:

A company that makes fleece clothing uses fleece produced from two farms, Northern Farm and Western Farm. Let the random variable $X$ represent the weight of fleece produced by a sheep from Northern Farm. The distribution of $X$ has mean 14.1 pounds and standard deviation 1.3 pounds. Let the random variable $Y$ represent the weight of fleece produced by a sheep from Western Farm. The distribution of $Y$ has mean 6.7 pounds and standard deviation 0.5 pound. Assume $X$ and $Y$ are independent. Let $W$ equal the total weight of fleece from 10 randomly selected sheep from Northern Farm and 15 randomly selected sheep from Western Farm. Which of the following is the standard deviation, in pounds, of $W$?

(A) $1.3+0.5$

(B)$\sqrt{1.3^{2}+0.5^{2}}$

(C) $\sqrt{10(1.3)^2+15(0.5)^2}$

(D) $\sqrt{10^2(1.3)^2+15^2(0.5)^2}$

(E) $\sqrt{\frac{1.3^2}{10}+\frac{0.5^2}{15}}$

The answer is (C) and I wanted a more in depth solution.

I know that $W=10X+15Y$ But I wanted to know what to do from here.

$\endgroup$ 1

1 Answer

$\begingroup$

So, the standard of deviation is the square root of the variance. The question is asking you to compute:

$$\sigma(W) = \sqrt{\text{Var}(W)}$$

It's a straightforward computation from here, we will use the fact that $W = 10X + 15Y$ and the properties of the variance of independent random variables. \begin{align*} \sigma(W)^2 = \text{Var}(W) &= \text{Var}(10X + 15Y)\\ &= \text{Var}(10X) + \text{Var}(15Y) = 10^2\text{Var(X)} + 15^2\text{Var(Y)}\\ &= 10^2(1.3)^2 + 15^2(0.5)^2\\ \sigma(W) &= \sqrt{10^2(1.3)^2 + 15^2(0.5)^2} \end{align*}

$\endgroup$ 1