Question

A certain flight arrives on time

90

percent of the time. Suppose

171

flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly

152

flights are on time.

​(b) at least

152

flights are on time.

​(c) fewer than

143

flights are on time.

​(d) between

143

and

153

​,

inclusive are on time.

​(a)

​P(152

​)equalsnothing

​(Round to four decimal places as​ needed.)

​(b)

​P(Xgreater than or equals

152​)equalsnothing

​(Round to four decimal places as​ needed.)

​(c)

​P(Xless than

143​)equalsnothing

​(Round to four decimal places as​ needed.)

​(d)

​P(143

less than or equalsXless than or equals153​)equalsnothing​(Round to four decimal places as​ needed.)

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 171 * 0.9 ) = 153.9
Variance = n * P * Q = ( 171 * 0.9 * 0.1 ) = 15.39
Standard deviation = √(variance) = √(15.39) = 3.923

Part a)
P ( X = 152 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 152 - 0.5 < X < 152 + 0.5 ) = P ( 151.5 < X < 152.5 )

X ~ N ( µ = 153.9 , σ = 3.923 )
P ( 151.5 < X < 152.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 151.5 - 153.9 ) / 3.923
Z = -0.61
Z = ( 152.5 - 153.9 ) / 3.923
Z = -0.36
P ( -0.61 < Z < -0.36 )
P ( 151.5 < X < 152.5 ) = P ( Z < -0.36 ) - P ( Z < -0.61 )
P ( 151.5 < X < 152.5 ) = 0.3594 - 0.2709
P ( 151.5 < X < 152.5 ) = 0.0885

Part b)
P ( X >= 152 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 152 - 0.5 ) =P ( X > 151.5 )

X ~ N ( µ = 153.9 , σ = 3.923 )
P ( X > 151.5 ) = 1 - P ( X < 151.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 151.5 - 153.9 ) / 3.923
Z = -0.61
P ( ( X - µ ) / σ ) > ( 151.5 - 153.9 ) / 3.923 )
P ( Z > -0.61 )
P ( X > 151.5 ) = 1 - P ( Z < -0.61 )
P ( X > 151.5 ) = 1 - 0.2709
P ( X > 151.5 ) = 0.7291

Part c)
P ( X < 143 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 143 - 0.5 ) = P ( X < 142.5 )

X ~ N ( µ = 153.9 , σ = 3.923 )
P ( X < 142.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 142.5 - 153.9 ) / 3.923
Z = -2.91
P ( ( X - µ ) / σ ) < ( 142.5 - 153.9 ) / 3.923 )
P ( X < 142.5 ) = P ( Z < -2.91 )
P ( X < 142.5 ) = 0.0018

Part d)
P ( 143 <= X <= 153 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 143 - 0.5 < X < 153 + 0.5 ) = P ( 142.5 < X < 153.5 )

X ~ N ( µ = 153.9 , σ = 3.923 )
P ( 142.5 < X < 153.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 142.5 - 153.9 ) / 3.923
Z = -2.91
Z = ( 153.5 - 153.9 ) / 3.923
Z = -0.1
P ( -2.91 < Z < -0.1 )
P ( 142.5 < X < 153.5 ) = P ( Z < -0.1 ) - P ( Z < -2.91 )
P ( 142.5 < X < 153.5 ) = 0.4602 - 0.0018
P ( 142.5 < X < 153.5 ) = 0.4584