SOLUTION: 9. Suppose heights of U.S. adults are normally distributed with a mean of 66.3 and a standard deviation of 4.4 inches.
(a) Find the height such that 95% of the population is less
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Question 971287: 9. Suppose heights of U.S. adults are normally distributed with a mean of 66.3 and a standard deviation of 4.4 inches.
(a) Find the height such that 95% of the population is less than this height. (the 9 percentile)
(b) What percent of US adults are taller than 64 inches?
(c) What is the probability that a sample of 18 adults will have a mean height greater than 64 inches?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=( x-66.3)/4.4
1.644 is the z value for the 95th percentile.
Multiply by 4.4 inches
7.24 inches from the mean. That is x
66.3+7.24=73.54 inches.
95% of the population is <= 73.5inches in height.
Taller than 64 inches
z=(64-66.3)/4.4
= -(2.3)/4.4
= -.5227
I want the percentage of the distribution greater than a z-value of -0.5227
This is 0.699 or 70% of the adults are greater than 64 inches tall.
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