SOLUTION: Suppose that the heights of adult women in the United States are normally distributed with a mean of 63.5 inches and a standard deviation of 2.3inches. Jennifer is taller than 75%o

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Question 136842: Suppose that the heights of adult women in the United States are normally distributed with a mean of 63.5 inches and a standard deviation of 2.3inches. Jennifer is taller than 75%of the population of U.S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
Thanks,
Answer by stanbon(75887) (Show Source):
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Suppose that the heights of adult women in the United States are normally distributed with a mean of 63.5 inches and a standard deviation of 2.3inches.
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Jennifer is taller than 75% the population of U.S. women.
Find the z-value corresponding to the lower 75% of the population: z=0.6744897...
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How tall (in inches) is Jennifer?
Find the x-value corresponding to z=0.6744897..
0.6744897 = (x-63.5)/2.3
x-63.5 = 1.551326...
x = 65.05 inches (Jennifer's height)
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Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
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Cheers,
Stan H.