Question

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. Find the probability that the average time waiting in line for these customers is A) Less than 10 minutes B) Between 5 and 10 minutes C) Less than 6 minutes

Answer 1

Answer:

A) 1

B) 1

C) 0

Step-by-step explanation:

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. Find the probability that the average time waiting in line for these customers is A) Less than 10 minutes B) Between 5 and 10 minutes C) Less than 6 minutes

We solve this question using the z score formula

z = (x-μ)/σ/√n, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

n = number of random samples

A) Less than 10 minutes

x < 10

z = 10 - 8.2/ 1.5 / √49

z = 8.4

P-value from Z-Table:

P(x<10) = 1

B) Between 5 and 10 minutes

For x = 5 minutes

z = 5 - 8.2/ 1.5 / √49

z = -14.93333

P-value from Z-Table:

P(x = 5) = 0

For x = 10 minutes

z = 10 - 8.2/ 1.5 / √49

z = 8.4

P-value from Z-Table:

P(x = 10) = 1

The probability that the average time waiting in line for these customers is between 5 and 10 minutes

P(x = 10) - P(x = 5)

= 1 - 0

= 1

C) Less than 6 minutes

x < 6

z = 6 - 8.2/ 1.5 / √49

z = -10.26667

P-value from Z-Table:

P(x<6) = 0