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. 
---------------------------------
Draw a normal curve with mean = 63.5 in. and standard deviation of 2.2 in
---------------------------------
Put a point on the horizontal axis where 70% of the area under the curve
would be to the left.  Label the point "x".
---------------------
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
----------------
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
========================
Cheers,
Stan H.