SOLUTION: The mean of a certain random variable, X, is 41 and the standard deviation is 1.3. Find the boundaries that separate the middle 72% of the data.

Algebra ->  Probability-and-statistics -> SOLUTION: The mean of a certain random variable, X, is 41 and the standard deviation is 1.3. Find the boundaries that separate the middle 72% of the data.      Log On


   

Question 1129792: The mean of a certain random variable, X, is 41 and the standard deviation is 1.3.
Find the boundaries that separate the middle 72% of the data.
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Assume X is a normally distributed random variable
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100 - 0.72 = 0.28
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we have 0.14 lower boundary percent and 0.72 + 0.14 = 0.86 upper boundary percent
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assume your problem is looking for the associated X values
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z-score = (X - mean)/standard deviation
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look at table of z-scores for associated z-scores for the lower and upper percents
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z-score(lower) = -2.99
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z-score(upper) = 1.08
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-2.99 = (X - 41)/1.3
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X - 41 = -3.887
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X = 37.113 is approximately 37
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1.08 = (X - 41)/1.3
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X - 41 = 2.34
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X = 43.34 is approximately 43
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lower boundary is 37
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upper boundary is 43
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