Question 1112696: Hello , Cant Solve This , Someone help me ?
The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 52 customers, answer the following questions.
Q1)What is the likelihood the sample mean is at least $25.50? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
Probability ?
Q2)What is the likelihood the sample mean is greater than $22.50 but less than $25.50? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
Probability ?
Q3) Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.)
Sample mean ....... and ......
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x bar-mean)/sigma/sqrt(n)
=(25.50-23.50)/6/sqrt(52)
=2*sqrt(52)/6
=2.40
probability z is greater than that is 0.0082
---------for 22.50
z=-1*sqrt(52)/6
= -1.20
probability z is between these two values, -1.20 and 2.40, is 0.8767
---------------
95% will occur between +/- 1.96*6/sqrt(52)
This is 1.63 on each side of the given mean of 23.50
(21.87, 25.13)
|
|
|
