SOLUTION: Scores on a test have a mean of 71 and Q3 is 82. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.)
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-> SOLUTION: Scores on a test have a mean of 71 and Q3 is 82. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.)
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Question 1123352: Scores on a test have a mean of 71 and Q3 is 82. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.) Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! I assume that Q3 means the third quartile
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the percentile associated with Q3 is 75%
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the z-score associated with 75% is approximately 0.67
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0.67 = (82 - 71)/standard deviation
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standard deviation = 11/0.67 is approximately 16.42
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z-score = (90 - 71)/16.42 is approximately 1.16
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P ( X < 90 ) is approximately 0.8770 of 87.70 percentile
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