Question

A certain flight arrives on time 90 percent of the time. Suppose 185 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 170 flights are on time. ​(b) at least 170 flights are on time. ​(c) fewer than 174 flights are on time. ​(d) between 174 and 178​, inclusive are on time.

Answer 1

P(flight arrives on time), p = 0.90

q = 1 - p = 0.10

Sample size, n = 185

Mean = np

= 185 x 0.90

= 166.5

Standard deviation =

=

= 4.08

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(exactly 170 flights are on time) = P(169.5 < X < 170.5)

= P(X < 170.5) - P(X < 169.5)

= P(Z < (170.5 - 166.5)/4.08) - P(Z < (169.5 - 166.5)/4.08)

= P(Z < 0.98) - P(Z < 0.74)

= 0.8365 - 0.7704

= 0.0661

b) P(at least 170 flights are on time) = P(X > 169.5)                       (continuity correction applied)

= 1 - P(X < 169.5)

= 1 - 0.7704

= 0.2296

c) P(fewer than 174 flights are on time) = P(X < 173.5)                       (continuity correction applied)

= P(Z < (173.5 - 166.5)/4.08)

= P(Z < 1.72)

= 0.9573

d) P(between 174 and 178​, inclusive are on time)

= P(174 X 178)

P(X < 178.5) - P(X < 173.5)

= P(Z < (178.5 - 166.5)/4.08) - P(Z < (173.5 - 166.5)/4.08)

= P(Z < 2.94) - P(Z < 1.72)

= 0.9984 - 0.9573

= 0.0411