Question

A certain flight arrives on time 82 percent of the time. Suppose 171 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 156 flights are on time. ​(b) at least 156 flights are on time. ​(c) fewer than 151 flights are on time. ​(d) between 151 and 152​, inclusive are on time.

Answer 1

n=	171	p=	0.8200
here mean of distribution=μ=np=		140.22
and standard deviation σ=sqrt(np(1-p))=		5.0239
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

a)

P(exactly 156 flights are on time):

probability =

P(155.5<X<156.5)

=

P(3.04<Z<3.24)=

0.9994-0.9988=

0.0006

b)

P(at least 156 flights are on time):

probability =

P(X>155.5)

=

P(Z>3.041)=

1-P(Z<3.04)=

1-0.9988=

0.0012

c)

P( fewer than 151 flights are on time):

probability =

P(X<150.5)

=

P(Z<2.05)=

0.9798

d)

P(between 151 and 152)

probability =

P(150.5<X<152.5)

=

P(2.05<Z<2.44)=

0.9927-0.9798=

0.0129