Question

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

Answer 1

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

`P(X<8) = P (X'-u)/(s.d/ √(n)) < (8-8.3)/(1.4/ √(47))]`

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =
`[(8-8.3)/(1.4/ √(47))< (X'-u)/(s.d/ √(n)) < (9-8.3)/(1.4/ √(47))]`

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

`NORMSDIST(6.366)-NORMSDIST(-1.47)`

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) =
`P [Z< (7.5-8.3)/(1.4/ √(47))]`

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000