Data Management - Stats Help?
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Data Management - Stats Help? The owner of a convenience store has copies of the local newspaper delivered early each morning. The demand for papers is normally distributed with a mean of 72 and a standard deviation of 14. a) What is the probability that there will not be enough newspapers to meet demand if the owner orders 67 copies? b) How many copies should be ordered so that the probability of "selling out" is at most 12% ? papers. Thank you in advance!
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Answer:
a) Calculate the z value for 67 z0 = (67-72)/14 Then look up P(z<z0) in the standard normal table. You want P(z>z0) which is 1-P(z<z0) (It will be more that 0.50 since 67<72 and you're finding the area to the right) b) You want the area under the standard normal density such that P(z>z0) = 0.12 So, look up the z0 value where P(z<z0) = 0.88 then solve for the x value from z0 = (x-72)/14 (It will be more than 72 since the area to the right, 0.12, is less than .50)
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Other answers
a) Calculate the z value for 67 z0 = (67-72)/14 Then look up P(z<z0) in the standard normal table. You want P(z>z0) which is 1-P(z<z0) (It will be more that 0.50 since 67<72 and you're finding the area to the right) b) You want the area under the standard normal density such that P(z>z0) = 0.12 So, look up the z0 value where P(z<z0) = 0.88 then solve for the x value from z0 = (x-72)/14 (It will be more than 72 since the area to the right, 0.12, is less than .50)
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