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1 vote
Suppose that X is a random variable with mean 20 and standard deviation 5. Also

suppose that Y is a random variable with mean 40 and standard deviation 10. Find the
variance and standard deviation of the random variable Z for each of the following
cases. Be sure to show your work.
(a) Z = 2 + 10X.
(b) Z = 10X − 2.
(c) Z = X + Y.
(d) Z = X − Y.
(e) Z = −3X − 2Y.

User Rod Lima
by
7.1k points

1 Answer

5 votes
Disregard my earlier question.

Recall that the variance of a random variable
X, denoted
\mathbb V(X), is given by


\mathbb E((X-\mu)^2)

where
\mathbb E(X) denotes the expected value/mean of
X, and
\mu=\mathbb E(X) is the actual mean of
X.

Now, recall that


\mathbb V(X)=\mathbb E((X-\mu)^2)=\mathbb E(X^2-2\mu X+\mu^2)

\mathbb V(X)=\mathbb E(X^2)-2\mu^2+\mu^2

\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2

(a)
Z=2+10X


\mathbb V(Z)=\mathbb E((2+10X)^2)-\mathbb E(2+10X)^2

\mathbb V(Z)=\mathbb E(4+40X+100X^2)-(\mathbb E(2)+\mathbb E(10X))^2

\mathbb V(Z)=\mathbb E(4)+40\mathbb E(X)+100\mathbb E(X^2)-(\mathbb E(2)+10\mathbb E(X))^2

\mathbb V(Z)=100\mathbb E(X^2)-40000

\mathbb V(Z)=100\left(\mathbb E(X^2)-\mathbb E(X)^2+\mathbb E(X)^2\right)-40000

\mathbb V(Z)=100\left(\mathbb V(X)+\mathbb E(X)^2\right)-40000

\mathbb V(Z)=100\left(5^2+400\right)-40000

\mathbb V(Z)=2500

The standard deviation is the square root of the variance, so for (a) you have


√(\mathbb V(Z))=√(2500)=50

That should give you an idea as to how to figure out the rest.
User Brad Gardner
by
7.3k points
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