SOLUTION: The amount of Jenns phone bill is normally distributed with a mean of $65 and a standard deviation of $9. What percentage of her phone bills are between 32$ and 92?
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Question 1155679: The amount of Jenns phone bill is normally distributed with a mean of $65 and a standard deviation of $9. What percentage of her phone bills are between 32$ and 92? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
low end z=(32-65)/9=-33/9
high end z=(92-65)/9=3
The probability of z between -3.67 and 3 is 0.9985
