At the last home football game, the senior football players walk through a specially constructed welcoming arch, 2 abreast. It is considered unseemly to bump each other on the way through, so the arch must be wide enough for two players to go through. The distribution of widths of football players with shoulder pads on is approximately normal; the mean padded football player width is 30 inches, and the standard deviation is 5 inches. Let random variable w = width in inches of a randomly selected padded football player. a. Since this is a statistics test, the carpenter who will be constructing the arch has only metric measuring tools, and must convert all the information above to metric measures. 1 inch = 2.54 centimeters, and we define random variable m = width in centimeters of a randomly selected padded football player. What are the mean and standard deviation of m?
b. Suppose the football players are paired randomly to go through the arch. Define random variable v = m1 + m2 to be the collective width (in cm) of two randomly selected football players. What are the mean and standard deviation of v?
c. Suppose that in the original specifications of the arch a 10 inch separation of the football players is specified so that the football players have room to avoid each other. This leads to random variable 1 2 a w w 10 . Describe how this addition of 10 inches of "wiggle room" would change the mean and standard deviation in part (b). Do not recalculate the mean and standard deviation.