Question

# A certain flight arrives on time 90 percent of the time. Suppose 184 flights are randomly...

A certain flight arrives on time 90 percent of the time. Suppose 184 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

(a) exactly 174 flights are on time. (Round to four decimal places as​ needed.)

(b) at least 174 flights are on time. (Round to four decimal places as​ needed.)

(c) fewer than 170 flights are on time. (Round to four decimal places as​ needed.)

(d) between 170 and 177​, inclusive are on time. (Round to four decimal places as​ needed.)

a)

Mean = n p = 184 * 0.90 = 165.6

Standard deviation = sqrt [ n p (1 - p) ]

= sqrt ( 184 * 0.90 * ( 1 - 0.90) ]

= 4.0694

Using normal approximation,
P(X < x) = P( Z < ( X - mean ) / SD )

P(X = 174) =  P ( 173.5 < X < 174.5 ) (With continuity correction )
P ( 173.5 < X < 174.5 ) = P ( Z < ( 174.5 - 165.6 ) / 4.0694 ) - P ( Z < ( 173.5 - 165.6 ) / 4.0694 )
= P ( Z < 2.19) - P ( Z < 1.94 )
= 0.9857 - 0.9738
= 0.0119

b)

X ~ N ( µ = 165.6 , σ = 4.0694 )
We covert this to standard normal as
P ( X < x) = P ( (Z < X - µ ) / σ )
P(X >= 174) = P(X > 173.5 ) With continuity correction )

P ( X > 173.5 ) = P(Z > (173.5 - 165.6 ) / 4.0694 )
= P ( Z > 1.94 )
= 1 - P ( Z < 1.94 )
= 1 - 0.9738

= 0.0262

c)

P(X < 170)= P(X < 169.5) (With continuity correction )

P ( ( X < 169.5 ) = P ( Z < 169.5 - 165.6 ) / 4.0694 )
= P ( Z < 0.96 )
P ( X < 169.5 ) = 0.8315

d)

P(170 < X < 177) = P(169.5 < X < 177.5) (With continuity correction )

P ( 169.5 < X < 177.5 ) = P ( Z < ( 177.5 - 165.6 ) / 4.0694 ) - P ( Z < ( 169.5 - 165.6 ) / 4.0694 )
= P ( Z < 2.92) - P ( Z < 0.96 )
= 0.9982 - 0.8315
= 0.1668