Introduction to Probability and Statistics for Engineers and Scientists

174 Chapter 5: Special Random Variables

moment generating functions and distributions, we can conclude that

∑n
i= 1 Xiis normal
with mean

∑n

i= 1 μiand variance

∑n

i= 1 σi^2.

EXAMPLE 5.5d Data from the National Oceanic and Atmospheric Administration indicate
that the yearly precipitation in Los Angeles is a normal random variable with a mean of
12.08 inches and a standard deviation of 3.1 inches.

(a) Find the probability that the total precipitation during the next 2 years will exceed

25 inches.

(b)Find the probability that next year’s precipitation will exceed that of the

following year by more than 3 inches.

Assume that the precipitation totals for the next 2 years are independent.

SOLUTION LetX 1 andX 2 be the precipitation totals for the next 2 years.

(a) SinceX 1 +X 2 is normal with mean 24.16 and variance 2(3.1)^2 =19.22, it follows

that

P{X 1 +X 2 > 25 }=P

{

X 1 +X 2 −24.16

√

19.22

>

25 −24.16

√

19.22

}

=P{Z>.1916}

≈.4240

(b)Since −X 2 is a normal random variable with mean −12.08 and variance

(−1)^2 (3.1)^2 , it follows thatX 1 −X 2 is normal with mean 0 and variance 19.22.

Hence,

P{X 1 >X 2 + 3 }=P{X 1 −X 2 > 3 }

=P

{

X 1 −X 2

√

19. 22

>

3

√

19. 22

}

=P{Z>.6843}

≈.2469

Thus there is a 42.4 percent chance that the total precipitation in Los Angeles

during the next 2 years will exceed 25 inches, and there is a 24.69 percent chance

that next year’s precipitation will exceed that of the following year by more than

3 inches. ■

Forα∈(0, 1), letzαbe such that

P{Z>zα}= 1 − (zα)=α

That is, the probability that a standard normal random variable is greater thanzαis equal
toα(see Figure 5.9.)