At the last home football game of the season, the senior football players walk t

Question

At the last home football game of the season, the senior football players walk through a specially constructed welcoming arch, 2 abreast. The arch is 50” wide. 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 is approximately normal with a mean of 30 inches, and standard deviation of 5 inches. Let random variable w = width in inches of a randomly selected padded football player. Suppose the football players are paired randomly to go through the arch. Define random variable to be the collective width -- in inches -- of two randomly selected football players. What are the mean and standard deviation of v? The original specifications of the arch specified a 10 inch separation of the players so that they have room to avoid each other. Define the random variable, “amount” of room needed for two football players: . How does the mean and standard deviation in part (b) differ from the mean and standard deviation of the random variable, a? Do not recalculate the mean and standard deviation. d) The school safety committee is concerned about the amount of “wiggle” room in the arch. The wiggle room is the amount of space left over when two players go through the arch. Using your results from (b) and (c) above, define a random variable, g, which the committee could use in their analysis of the amount of wiggle room.

Explanation / Answer

Solution:

a) m= (2.54w) = 2.54w = 76.2 cm

m = (2.54w) = 2.54 w = 12.7 cm

b) v = 2 w = 60''

v = sqrt( 2w + 2w) = sqrt(52 + 52) = 7.07''

c) The mean would increase by 10'' tp 70''

The standard deviation would remain unchanged.

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