344 Consistency and Limiting DistributionsExample 5.3.2.LetXdenote the mean of a random sample of size 75 from the
distribution that has the pdff(x)={
10 <x< 1
0elsewhere.For this situation, it can be shown that the pdf ofX,g(x), has a graph when
0 <x<1 that is composed of arcs of 75 different polynomials of degree 74. The
computation of such a probability asP(0. 45 <X< 0 .55) would be extremely
laborious. The conditions of the theorem are satisfied, sinceM(t) exists for all real
values oft.Moreover,μ=^12 andσ^2 = 121 , so that using R we have approximately
P(0. 45 <X< 0 .55) = P[√
n(0. 45 −μ)
σ
<√
n(X−μ)
σ
<√
n(0. 55 −μ)
σ]= P[− 1. 5 <30(X− 0 .5)< 1 .5]
≈ pnorm(1.5)−pnorm(− 1 .5) = 0. 8663.Example 5.3.3(Normal Approximation to the Binomial Distribution). Suppose
thatX 1 ,X 2 ,...,Xnis a random sample from a distribution that isb(1,p). Here
μ=p, σ^2 =p(1−p), andM(t) exists for all real values oft.IfYn=X 1 +···+Xn,
it is known thatYnisb(n, p). Calculations of probabilities forYn,whenwedo
not use the Poisson approximation, are simplified by making use of the fact that
(Yn−np)/
√
np(1−p)=√
n(Xn−p)/√
p(1−p)=√
n(Xn−μ)/σhas a limiting
distribution that is normal with mean zero and variance 1.
Frequently, statisticians say thatYn,ormoresimplyY, has an approximate
normal distribution with meannpand variancenp(1−p). Even withnas small
as 10, withp=^12 so that the binomial distribution is symmetric aboutnp=5,
we note in Figure 5.3.1 how well the normal distribution,N(5,^52 ), fits the binomial
distribution,b(10,^12 ), where the heights of the rectangles represent the probabilities
of the respective integers 0, 1 , 2 ,...,10. Note that the area of the rectangle whose
base is (k− 0. 5 ,k+0.5) and the area under the normal pdf betweenk− 0 .5and
k+0.5areapproximatelyequal for eachk=0, 1 , 2 ,...,10, even withn= 10. This
example should help the reader understand Example 5.3.4.
Example 5.3.4.With the background of Example 5.3.3, letn= 100 andp=^12 ,
and suppose that we wish to computeP(Y=48, 49 , 50 , 51 ,52). SinceYis a random
variable of the discrete type,{Y =48, 49 , 50 , 51 , 52 }and{ 47. 5 <Y < 52. 5 }are
equivalent events. That is,P(Y =48, 49 , 50 , 51 ,52) =P(47. 5 <Y < 52 .5). Since
np=50andnp(1−p) = 25, the latter probability may be written
P(47. 5 <Y < 52 .5) = P(
47. 5 − 50
5<Y− 50
5<52. 5 − 50
5)= P(
− 0. 5 <Y− 50
5
< 0. 5)
.