Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
