Question

A certain flight arrives on time 89 percent of the time. Suppose 148 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 134 flights are on time.

​(b) at least 134 flights are on time.

​(c) fewer than 120 flights are on time.

​(d) between 120 and 143​, inclusive are on time.

Answer 1

X ~ Bin ( n , p)

Where n = 148 , p = 0.89

Mean = n p = 148 * 0.89 = 131.72

Standard deviation = sqrt ( n p ( 1 - p) ) = sqrt ( 148 * 0.89 ( 1 - 0.89)) = 3.8065

Uisng normal approximation,

P(X < x) = P(Z < (X - mean) / SD)

a)

With continuity correction,

P(X = 134) = P(133.5 < X < 134.5)

P ( 133.5 < X < 134.5 ) = P ( Z < ( 134.5 - 131.72 ) / 3.8065 ) - P ( Z < ( 133.5 - 131.72 ) / 3.8065 )
= P ( Z < 0.73) - P ( Z < 0.47 )
= 0.7673 - 0.6808
= 0.0865

b)

With continuity correction

P(X >= 134) = P(X > 133.5)

=  P ( X > 133.5 ) = P(Z > (133.5 - 131.72 ) / 3.8065 )
= P ( Z > 0.47 )
= 1 - P ( Z < 0.47 )
= 1 - 0.6808

= 0.3192

c)

With continuity correction

P(X < 120) = P(X < 119.5)

P ( ( X < 119.5 ) = P ( Z < 119.5 - 131.72 ) / 3.8065 )
= P ( Z < -3.21 )
P ( X < 119.5 ) = 0.0007

d)

With continuity correction

P(120 < X < 143) = P(119.5 < X< 143.5)

=  P ( 119.5 < X < 143.5 ) = P ( Z < ( 143.5 - 131.72 ) / 3.8065 ) - P ( Z < ( 119.5 - 131.72 ) / 3.8065 )
= P ( Z < 3.09) - P ( Z < -3.21 )
= 0.999 - 0.0007
= 0.9983