Question

A certain flight arrives on time 86 percent of the time. Suppose 169 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that (a) exactly 140 flights are on time. (b) at least 140 flights are on time. (c) fewer than 148 flights are on time. (d) between 148 and 159, inclusive are on time. (a) P( 140)- (Round to four decimal places as needed.) (b) PIX z 140)(Round to four decimal places as needed.) (c)P 148)-(Round to four decimal places as needed) (d) P(148X159)(Round to four decimal places as needed.)

Answer 1

n=	169	p=	0.8600
here mean of distribution=μ=np=		145.34
and standard deviation σ=sqrt(np(1-p))=		4.5108
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

a)

probability =

P(139.5<X<140.5)

=

P(-1.29<Z<-1.07)=

0.1423-0.0985=

0.0438

b)

probability =

P(X>140.5)

=

P(Z>-1.07)=

1-P(Z<-1.07)=

1-0.1423=

0.8577

c)

probability =

P(X<147.5)

=

P(Z<0.48)=

0.6844

d)

probability =

P(147.5<X<159.5)

=

P(0.48<Z<3.14)=

0.9992-0.6844=

0.3148