Question

A commonly used practice of airline companies is to sell more tickets than actual seats to a particular flight because customers who buy tickets do not always show up for the flight. Suppose that the percentage of no shows at flight time is 2%. For a particular flight with 380 seats, a total of 384 tickets were sold. Use normal approximation to find the probability that

(a) at most 375 passengers will show up.

(b) the airline overbooked this flight.

(c) between 4 and 8 passengers (both inclusive) will not show up.

Answer 1

n=

384

p=

0.0200

here mean of distribution=μ=np=		7.68
and standard deviation σ=sqrt(np(1-p))=		2.74
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)P( at most 375 passengers will show up ) =P(At least 5 did not show up)

probability =P(X>4.5)=P(Z>(4.5-7.68)/2.743)=P(Z>-1.16)=1-P(Z<-1.16)=1-0.123=0.8770

b)

the airline overbooked this flight =P(at most 3 did not show up)

probability =P(X<3.5)=(Z<(3.5-7.68)/2.743)=P(Z<-1.52)=0.0643

c)

probability =P(3.5<X<8.5)=P((3.5-7.68)/2.743)<Z<(8.5-7.68)/2.743)=P(-1.52<Z<0.3)=0.6179-0.0643=0.5536