SOLUTION: This is a multiple choice proble that I am not getting any of the answers to. The average amount customers at a certain grocery store spend yearly is $636.55. Assume the variable

Question 150420: This is a multiple choice proble that I am not getting any of the answers to.
The average amount customers at a certain grocery store spend yearly is $636.55. Assume the variable is normally distributed. If the standard deviation is $89.46, find the probability that a randomly selected customer spends between $550.67 and $836.94.

0.144 = 14.4%

0.820 = 82.0%

0.156 = 15.6%

0.943 = 94.3%
Please help,
Ellen
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
find the z values for the upper and bounds of the range and then find the portion of the distribution represented

lower __ z=(550.67-636.55)/89.46 __ z=-.96 (approx)

upper __ z=(836.94-636.55)/89.46 __ z=2.24 (approx)

this range represents about 82% of the distribution
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