Assume the credit card balances of younger college educated employed persons are normally distributed with a mean of $ 6,358 and a standard deviation of $1,907 – assume these are population values. What is the probability of a randomly selected credit card holder has a credit card balance less than $5750? Use the standard normal table for this, using the population mean and standard deviation. Now you randomly select 81 credit card holders. What is the probability that their mean credit card balance is less than $5750? Use the standard normal table for this, but this time use the population mean and standard error (standard deviation/SQRT(81)).
Assume the credit card balances of younger college educated employed persons are normally distributed with a mean of $ 6,358 and a standard deviation of $1,907 – assume these are population values. What is the probability of a randomly selected credit card holder has a credit card balance less than $5750? Use the standard normal table for this, using the population mean and standard deviation. Now you randomly select 81 credit card holders. What is the probability that their mean credit card balance is less than $5750? Use the standard normal table for this, but this time use the population mean and standard error (standard deviation/SQRT(81)).
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 19PFA
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- Assume the credit card balances of younger college educated employed persons are
normally distributed with a mean of $ 6,358 and a standard deviation of $1,907 – assume these are population values.
-
- What is the probability of a randomly selected credit card holder has a credit card balance less than $5750? Use the standard normal table for this, using the population mean and standard deviation.
- Now you randomly select 81 credit card holders. What is the probability that their mean credit card balance is less than $5750? Use the standard normal table for this, but this time use the population mean and standard error (standard deviation/SQRT(81)).
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