Question

A certain flight arrives on time 86 percent of the time. Suppose 112 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 100 flights are on time.

​(b) at least 100 flights are on time.

​(c) fewer than 99 flights are on time.

​(d) between 99 and 106​, inclusive are on time.

Answer 1

a)

n=	112	p=	0.8600
here mean of distribution=μ=np=		96.32
and standard deviation σ=sqrt(np(1-p))=		3.6722
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

P(exactly 100 flights are on time):

probability =

P(99.5<X<100.5)

=

P(0.87<Z<1.14)=

0.8729-0.8078=

0.0651

b)

probability =

P(X>99.5)

=

P(Z>0.87)=

1-P(Z<0.87)=

1-0.8078=

0.1922

c)

probability =

P(X<98.5)

=

P(Z<0.594)=

0.7224

d)

probability =

P(98.5<X<106.5)

=

P(0.59<Z<2.77)=

0.9972-0.7224=

0.2748