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The probability is 0.36 that the amount spent on a randomly selected child will be between what two value equidistant from the mean?

User Wharbio
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Assuming a standard normal distribution, the positive value of interest will be that corresponding to the 50 + 36/2 = 68th percentile. A suitable probability calculator will give that value as 0.46770.


The values that are within ±0.46770 standard deviations from the mean will have a probability of 0.36.

User Bridget The Midget
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