SOLUTION: The mean amount purchased by a typical customer at Churchill's Grocery Store is $21.50 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the

Question 1092901: The mean amount purchased by a typical customer at Churchill's Grocery Store is $21.50 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 44 customers, answer the following questions.

a.
What is the likelihood the sample mean is at least $24.50? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability ____

b.
What is the likelihood the sample mean is greater than $19.50 but less than $24.50? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability ____

c.
Within what limits will 90 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!
The mean of the sample distribution is the same as the mean of the population. The standard error is the standard deviation of the population/sqrt(sample size)
Therefore, SE is 7/sqrt (44)=1.055
now use z-test
z>=(x-pop mean)/1.055=(24.50-21.50)/1.055, so z > 2.84 or 0.0023.
---------
z for 19.50 is -2/1.055 or -1.90
z for 24.50 is 2.84.
Want the probability of z being between those two values, and that is 0.9690
---------------------
90% sample means are between z -1.65 and +1.65
z=(x-mean)/sd
-1.65*1.055=x-mean
-1.74+mean=x and that value would be $19.76
+1.74 would be on the other side and that is $23.24
($19.76, $23.24)

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