Question

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

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(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.)

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

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

Answer 1

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