Question

A certain flight arrives on time 86 percent of the time. Suppose 178 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 161 flights are on time.

​(b) at least 161 flights are on time.

​(c) fewer than 147 flights are on time.

​(d) between 147 and 160​, inclusive are on time.

Answer 1

Solution:

We are given

n = 178

p = 0.86

q = 1 – p = 1 – 0.86 = 0.14

Mean = np = 178*0.86 = 153.08

SD = sqrt(npq) = sqrt(178*0.86*0.14) = 4.629384

Part a

We have to find P(X=161) = P(160.5<X<161.5) (continuity correction)

P(160.5<X<161.5) = P(X<161.5) – P(X<160.5)

Z = (X – mean)/SD

Z = (161.5 - 153.08) / 4.629384 = 1.818816

P(Z<1.818816) = 0.96553

(by using z-table)

Z = (160.5 - 153.08) / 4.629384 =1.602805

P(Z<1.602805) = 0.945511

(by using z-table)

P(160.5<X<161.5) = P(X<161.5) – P(X<160.5)

P(160.5<X<161.5) = 0.96553 - 0.945511

P(160.5<X<161.5) = 0.020019

Required probability = 0.020019

Part b

Here, we have to find P(X≥161) = P(X>160.5) (continuity correction)

P(X>160.5) = 1 – P(X<160.5) = 1 - 0.945511 = 0.054489

[We have P(X<160.5) = 0.945511 from above part a]

Required probability = 0.054489

Part c

Here, we have to find P(X<147) = P(X<146.5) (continuity correction)

Z = (X – mean)/SD

Z = (146.5 - 153.08) / 4.629384 =-1.42136

P(Z<-1.42136) = 0.077607

(by using z-table)

Required probability = 0.077607

Part d

We have to find P(147<X<160) = P(146.5≤X<160.5) (continuity correction)

P(146.5≤X<160.5) = P(X<160.5) – P(X<146.5) = 0.945511 - 0.077607 = 0.867904

[By using part b and c]

Required probability = 0.867904