SOLUTION: Suppose the wing lengths of houseflies are normally distributed with a mean of 45.5 millimeters and a
standard deviation of 3.92 millimeters. Use the standard normal distribution
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-> SOLUTION: Suppose the wing lengths of houseflies are normally distributed with a mean of 45.5 millimeters and a
standard deviation of 3.92 millimeters. Use the standard normal distribution
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Question 1180807: Suppose the wing lengths of houseflies are normally distributed with a mean of 45.5 millimeters and a
standard deviation of 3.92 millimeters. Use the standard normal distribution and a calculator or table to
estimate the following.
a. The percent of houseflies with wing lengths over 35 millimeters
b. The percent of houseflies with wing lengths over 50 millimeters
c. The percent of houseflies with wing lengths over 55 millimeters Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z(35)>(35-45.5)/3.92=-2.68. Probability is 0.9963 or 99.63%
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over 50 is z>(50-45.5)/3.92 or 1.148. Probability is 0.1255 or 12.55%
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over 55 is z>(55-45.5)/3.92 or 2.423. Probability is 0.0077 or 0.77%
