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Question 897:



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  1. Q897_1_q897-zscoreareas.xls
For this question we need to use the properties of the normal curve. I've attached a spreadsheet which will lookup the percentile values from the normal distribution for each of the z-scores provided.

A z-score just refers to the number of standard deviations in the normal curve and the probability is the percent of the area.

a. We need the area below 1.35 standard deviations which is .9115
b. To find the area in between two z-scores we subtract the smaller percentage from the larger and get .9772-.0329 = .9444.
c. This last one is tricky. We need to find the area below 1.02 which is .8461 and add this to the area above 2.77. The area below 2.77 is .9972 so the area above this is 1-.9972 = .0028. So we add .0028 to .8461 and get .8489.

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