Introduction to Probability and Statistics for Engineers and Scientists

5.5Normal Random Variables 173

whereris a constant. Ifr=3, andVcan be assumed (to a very good approximation) to
be a normal random variable with mean 6 and standard deviation 1, find

(a)E[W];

(b)P{W> 120 }.

SOLUTION
(a) E[W]=E[ 3 V^2 ]
= 3 E[V^2 ]
=3(Var[V]+E^2 [V])
=3(1+36)= 111

(b) P{W> 120 }=P{ 3 V^2 > 120 }

=P{V>

√

40 }

=P{V− 6 >

√

40 − 6 }

=P{Z>.3246}

= 1 − (.3246)

=.3727 ■

Another important result is that the sum of independent normal random variables is
also a normal random variable. To see this, suppose thatXi,i=1,...,n, are independent,
withXibeing normal with meanμiand varianceσi^2. The moment generating function
of

∑n

i= 1 Xiis as follows.

E

[

exp

{

t

∑n

i= 1

Xi

}]

=E

[

etX^1 etX^2 ···etXn

]

=

∏n

i= 1

E

[

etXi

]

by independence

=

∏n

i= 1

eμit+σ

i^2 t^2 /2

=eμt+σ

(^2) t (^2) /2
where
μ=
∑n
i= 1
μi, σ^2 =
∑n
i= 1
σi^2
Therefore,
∑n
i= 1 Xihas the same moment generating function as a normal random variable
having meanμand varianceσ^2. Hence, from the one-to-one correspondence between