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 2.2inches. Jennifer is taller than 70
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Question 141200: 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 2.2inches. 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 63.5 inches and a standard deviation of 2.2inches. 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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Draw a normal curve with mean = 63.5 in. and standard deviation of 2.2 in
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Put a point on the horizontal axis where 70% of the area under the curve
would be to the left. Label the point "x".
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Find the z-value of the point which has 70% of the area to the left. Use your chart to find that.
Ans: z = 0.5244
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Find the value on your normal-curve sketch corresponding to z=0.5244
Use z = (x-u)/sigma
0.5244 = (x-63.5)/2.2
x-63.5 = 1.1537
x = 64.65 inches
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Cheers,
Stan H.
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