Question

Show all steps. 1. Th an airport check-in counter 4minutes. e amount of time that a customer spends waiting at is a random variable with mean 18.5 minutes and standard deviation 1 Suppose that a random sample of 49 customers is observed. Find probability that the average time waiting in the line is a) less than 20minutes 6p b) between 15 and 21minutes

Answer 1

$X \sim N ( \mu = 18.5 , \sigma = 14 )$

P ( X < 20 )

Standardizing the value

$Z = ( X - \mu ) / (\sigma/\sqrt{n})$

$Z = ( 20 - 18.5 ) / ( 14 /\sqrt{ 49 })$

Z = 0.75

$P ( ( X - \mu ) / (\sigma/\sqrt{n}) < ( 20 - 18.5 ) / ( 14 /\sqrt{ 49 } )$

P ( X < 20 ) = P ( Z < 0.75 )

P ( X < 20 ) = 0.7734

Part b)

$X \sim N ( \mu = 18.5 , \sigma = 14 )$

P ( 15 < X < 21 )

Standardizing the value

$Z = ( X - \mu ) / (\sigma/\sqrt{n})$

$Z = ( 15 - 18.5 ) / ( 14 /\sqrt{ 49 })$

Z = -1.75

$Z = ( 21 - 18.5 ) / ( 14 /\sqrt{ 49 })$

Z = 1.25

P ( -1.75 < Z < 1.25 )

P ( 15 < X < 21 ) = P ( Z < 1.25 ) - P ( Z < -1.75 )

P ( 15 < X < 21 ) = 0.8944 - 0.0401

P ( 15 < X < 21 ) = 0.8543