SOLUTION: can someone please help me solve this? thank you
Suppose that the heights of adult women in the United States are normally distributed with a mean of 64 inches and a standard de
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Suppose that the heights of adult women in the United States are normally distributed with a mean of 64 inches and a standard de
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Question 185515: can someone please help me solve this? thank you
Suppose that the heights of adult women in the United States are normally distributed with a mean of 64 inches and a standard deviation of 2.2 inches. Jennifer is taller than 70% 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.
You can put this solution on YOUR website! Suppose that the heights of adult women in the United States are normally distributed with a mean of 64 inches and a standard deviation of 2.2 inches. Jennifer is taller than 70% 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.
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The z-score that separates 70% from the top 30% of a normal distribution
is z = 0.52448
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Find the x-score that corresponds to that z-score:
z = (x-u)/s
0.52448 = (x-64)/2.2
x-64 = 1.15368
x = 65.2 inches
Jennifer is at least 65.2 inches tall.
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Cheers,
Stan H.
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