Question 185515
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.