Question

According to the Bureau of Transportation​ Statistics, 81.9​% of American Airlines flights were on time in 2017. Assume this percentage still holds true for American Airlines. For the next

46 flights from American​ Airlines, use the normal approximation to the binomial distribution to complete parts A through D.

A. Determine the probability that fewer than 36 flights will arrive on time. (Round to four decimal places as​ needed.)

B. Determine the probability that exactly 32 flights will arrive on time.

C. Determine the probability that 25​, 26​, 27​, or 28 flights will arrive on time.

D. Determine the probability that 28, 29​, 30, 31, or 32 flights will arrive on time.

Answer 1

n=

46

p=

0.8190

here mean of distribution=μ=np=		37.67
and standard deviation σ=sqrt(np(1-p))=		2.61
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)

  probability that fewer than 36 flights will arrive on time:

probability =P(X<35.5)=(Z<(35.5-37.674)/2.611)=P(Z<-0.83)=0.2033

b)

  probability that exactly 32 flights will arrive on time:

probability =P(31.5<X<32.5)=P((31.5-37.674)/2.611)<Z<(32.5-37.674)/2.611)=P(-2.36<Z<-1.98)=0.0239-0.0091=0.0148

c)

probability that 25​, 26​, 27​, or 28 flights will arrive on time:

probability =P(24.5<X<28.5)=P((24.5-37.674)/2.611)<Z<(28.5-37.674)/2.611)=P(-5.04<Z<-3.51)=0.0002-0=0.0002

d)

probability =P(27.5<X<32.5)=P((27.5-37.674)/2.611)<Z<(32.5-37.674)/2.611)=P(-3.9<Z<-1.98)=0.0239-0=0.0239