A. Suppose the heights of 10-year-old girls are Normally distributed with mean =54.8 inches and =2.8 inches. (They aren't; I made these numbers up.)

1. Draw (or have a computer generate) a Normal curve with the parameters labeled. In your picture, shade in the region that represents the proportion of 10-year-old girls who are more than 60.4 inches.
2. Find the area of the region you shaded, and interpret that area in two ways (like I did at the bottom of this page)
   1. as aproportion ofall 10-year-old girls
   2. as a probability aboutone such girl
3. If we select one 10-year-old girl at random, what is the probability that she is more than 60 inches tall?
4. If we select one 10-year-old girl at random, what is the probability that she is more than 5 inches away from the mean? (She might be very tall, or she might be very short.)
5. What percentage of 10-year-old girls will be within 1 inch of the mean?
6. Find the 95th percentile of height a value that separates the shortest 95% of 10-year-old girls from the tallest 5% of 10-year-old girls. Round your answer to the nearest tenth of an inch.

B. Ball bearings are manufactured with a mean diameter of 4 millimeters (mm). Because of variability in the manufacturing process, the diameters of the ball bearings are approximately Normally distributed, with a standard deviation 0.012 mm.

1. What proportion of ball bearings have a diameter more than 4.02 mm?
2. Any ball bearings that have a diameter less than 3.97 mm or greater than 4.03 mm are discarded. What proportion of ball bearings will be discarded?
3. Using the results of the last problem, if 20,000 ball bearings are manufactured in a day, how many should the plant manager expect to discard?
4. An order comes in for 40,000 ball bearings, requesting that all the ball bearings delivered must be between 3.99 mm and 4.01 mm. How many bearings should the plant manager manufacture in order to satisfy the order? (You must explain how you are getting your answer to receive full credit.)