Question

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 a mean 14.1 pounds and a 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 a mean 6.7 pounds and a standard deviation of 0.5 pounds. 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((1.3)^2/10 + (0.5)^2/15

Answer 1

Answer:

i think its D. C. or E. I'm not that great in math im kind of struggling thru it

Answer 2

Answer:

D)

Step-by-step explanation:

Let's start writing the data from the exercise.

The random variables are :

X : '' Weight of fleece produced by a sheep from Northern Farm ''

Y : '' Weight of fleece produced by a sheep from Western Farm ''

W : '' Total weight of fleece from 10 randomly selected sheep from Northern Farm and 15 randomly selected sheep from Western Farm ''

Let's use μ to denote mean, σ to denote standard deviation and VAR to denote variance.

In the exercise :

μ(X) = 14.1 pounds

σ(X) = 1.3 pounds

μ(Y) = 6.7 pounds

σ(Y) = 0.5 pounds

The equation that represents W is

$W=10X+15Y$

Let's suppose that we have two random variables X1 and X2. Let's also assume that X1 and X2 are independent. If we have the following linear combination of the random variables :

aX1 + bX2

The variance of the linear combination is :

$VAR(aX1+bX2)=a^{2}VAR(X1)+b^{2}VAR(X2)$

Applying this to the exercise :

$VAR(W)=VAR(10X+15Y)=10^{2}VAR(X)+15^{2}VAR(Y)$ (I)

We know that VAR(X) = σ²(X) ⇒

$VAR(X)=(1.3pounds)^{2}$

And also

$VAR(Y)=(0.5pounds)^{2}$

If we replace in (I) ⇒

$VAR(W)=(10)^{2}(1.3)^{2}+(15)^{2}(0.5)^{2}$

Given that we find the variance of W, If we want to obtain the standard deviation we need to apply square root to the variance of W ⇒

σ(W) = $\sqrt{(10)^{2}(1.3)^{2}+(15)^{2}(0.5)^{2}}$

Finally, the correct option is D)