SOLUTION: Suppose that the heights of adult women in the United States are normally distributed with a mean of 64.5 inches and a standard deviation of 2.4 inches. Jennifer is taller than 70%

Question 293222: Suppose that the heights of adult women in the United States are normally distributed with a mean of 64.5 inches and a standard deviation of 2.4 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.
I can work it out to the point where P(Z>(x-64.5/2.4)=0.7
I don't really get the explanation that I have in front of me so I tried a websearch. I found a lot of answers to similar problems, but most of the answers require graphing calculators and I'm working by hand. I would appreciate some advice on what to do next.
Answer by stanbon(75887) (Show Source): 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 64.5 inches and a standard deviation of 2.4 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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You have to have some technology that gives you z-scores or you
have to know how to use a z-chart (which you can find on-line).
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Procedure.
Draw a normal curve.
Let "z" be the horizontal axis.
Mark a point on the axis that has 70% of the area to its left
and 30% to its right.
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Find the z-value of that point.
I get z = 0.5244
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Draw another normal curve.
Let "x" be the horizontal axis.
Put a point on the x-axis that corresponds to z=0.5244.
That point will have 70% of the heights to the left and
30% of the heights to the right.
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Now find the x-value of that point using x = zs + u
I get x = 0.5244*2.4 + 64.5
x = 65.76 inches.
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Cheers,
Stan H.
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