SOLUTION: #8-13
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 fol
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-> SOLUTION: #8-13
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 fol
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Question 1202886: #8-13
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.
a. 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.)
b. 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.)
c. Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.) ____ and _____.
You can put this solution on YOUR website! m = mean = 23.5
d = standard devition = 6
n = 52 = sample size
s = standard error = d / sqrt(n) = .83205
a. 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.)
z = (x-m)/s = (25.5 - 23.5) / .83205 = 2.4
area to the left of that = .9918
area to the right of that = 1 - .9918 = .0082.
the probability of getting a sample mean greater than 25.5 = .0082.
b. 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.)
z-score for lower bound = (22.5 - 23.5) / .83205 = -1.2
area to the left of that z-score = .1151
z-score for upper gound = (25.5 - 23.5) / .83205 = 2.4
area to the left of that z-score = .9918
area in between = .99918 minus .1151 = .8767
the probability of getting a sample mean between 22.5 and 25.5 is equal to .8767.
c. Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.)
z-score formula for lower bound is -1.96 = (x - 23.5) / .83205.
solve for x to get x = -1.96 * .83205 + 23.5 = 21.87.
z-score formula for upper bound is 1.96 = (x - 23.5) / .83205.
solve for x to get x = 1.96 * .83205 + 23.5 = 25.13
95% confidence interval is from 21.87 to 25.13
here's what the solutions look like on the z-score graphing calculator.
when using z-score, the mean is 0 and the standard deviation is 1.
when using raw scores, the mean is 23.5 and the standard deviation is .83205.
note that what is titled standard deviation is the standare error.
the standard error is another name for the standard deviation of the distribution of sample means.
