Question

A certain flight arrives on time 88 percent of the time. Suppose 145 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 128 flights are on time.

​(b) at least 128 flights are on time.

​(c) fewer than 124 flights are on time.

​(d) between 124 and 125​, inclusive are on time.

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

Answer 1

SOLUTION:

From given data,

A certain flight arrives on time 88 percent of the time. Suppose 145 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

Where,

p = 88% = 88/100 = 0.88

q = 1-p = 1-0.88 =0.12

n = 145

Mean = = n*p = 145*0.88 = 127.6

Standard deviation = = sqrt(n*p*q) = sqrt(145*0.88 *0.12) = 3.91305

Z = X- / = X-127.6 / 3.91305

​(a) exactly 128 flights are on time.

exactly 128 means P(X=128) = P(127.5 < X < 128.5) (By using continues correlation )

At X = 127.5

Z =(127.5-127.6) / 3.91305​​​​​​​ = -0.02

At X = 128.5

Z =(128.5-127.6) / 3.91305​​​​​​​ = 0.22

P(127.5 < X < 128.5) = P(-0.02 < Z < 0.22)

P(127.5 < X < 128.5) = P(Z < 0.22) -  P(Z < -0.02)

P(127.5 < X < 128.5) =0.58706 - 0.49202

P(127.5 < X < 128.5) =0.0950

​(b) at least 128 flights are on time.

P(X > 128) = P(X > 127.5) (by using continues correction)

P(X > 127.5) = P(Z > (127.5-127.6)/3.91305)

P(X > 127.5) = P( Z > -0.02)

P(X > 127.5) = 1- P( Z< -0.02)

P(X > 127.5) = 1- 0.49202

P(X > 127.5) = 0.5079

​(c) fewer than 124 flights are on time.

P(X < 124 ) =P(X<124.5) (by using continouse correction)

P(X<124.5) = P(Z < (124.5-127.6)/3.91305)

P(X<124.5) = P(Z < -0.79)

P(X<124.5) = 0.2147

​(d) between 124 and 125​, inclusive are on time.

At X = 124

Z =(124-127.6) / 3.91305​​​​​​​ = -0.91

At X = 125

Z =(125-127.6) / 3.91305​​​​​​​ = -0.66

P(124 < X < 125) = P(-0.91 < Z < -0.66)

P(124 < X < 125) = P(Z < -0.66) -  P(Z < -0.91)

P(124 < X < 125) =0.25463 - 0.18141

P(124 < X < 125) =0.0732