Question

A certain flight arrives on time 83 percent of the time. Suppose 125 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 92 flights are on time.

​(b) at least 92 flights are on time.

​(c) fewer than 112 flights are on time.

​(d) between 112 and 116 inclusive are on time.

Answer 1

x=A certain flight arrives on time

p=.83

n=155

Using normal approximation of binomial:
μ = n*p = 125 * .83 = 103.75

σ = sqrt(n * p * (1-p) ) = sqrt(17.6375) ≈ 4.199702

X~

a) exactly 92 flights are on time.

ie P(X=92)=0

if X is a continuous random variable hence the probability of getting exactly is zero

b)(b) at least 92 flights are on time.

ie

Standardize the value of X using z score formula

hence Z-score for 92 ie

P(X≥92)=P(z>-2.79)

The probability is calculated as follows using z tables:

P(z>-2.79)=1−P(z<-2.79)

P(z<-2.79)=0.0026

P(z>-2.79)=1−0.0026

P(X≥92)=0.9974

probability at least 92 flights are on time. is 0.9974

(c) fewer than 112 flights are on time.

P(x<112)

hence Z-score for 112 ie

=1.9644

=P(z<−1.9644)

P(z<1.9644)=0.9753

∴P(X<112)=0.9753

fewer than 112 flights are on time 0.9753

​(d) between 112 and 116 inclusive are on time.

hence Z-score for 112 and 116

Z112=1.9644.

Z116=2.9169

P(112≤x≤116)=P(1.9644<z<2.9169)

P(1.9644<z<2.9169)=P(z<<2.9169)−P(z<1.9644)

P(1.9644<z<2.9169)=0.9982-0.9753=0.0230

P(112≤x≤116)=0.0230

between 112 and 116 inclusive are on time. is 0.0230