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After a touchdown in American football, a team can elect to kick the ball for 1

ID: 3042272 • Letter: A

Question

After a touchdown in American football, a team can elect to kick the ball for 1 extra point (called a point after try, PAT) if the kick is successful or advance the ball into the end zone for 2 points (called a two-point conversion) if the advance is successful. In each scenario, 0 points are awarded if the attempt is unsuccessful. During the 2016 season, NFL teams converted 93.6% of PAT kicks and 48.6% of two-point conversions. Answer the following questions:

a) Write out the probability distributions for PATs and two-point conversions. Let X be the random variable that represents the number of points for a PAT attempt and Y be the random variable that represents the number of points scored for a two-point conversion attempt.

b) Compute the expected number of points for each scenario.

c) Which strategy is better in terms of the expected number of points scored?

d) Compute the variance for two-point conversions, V ar (Y)

e) Which strategy, PATs or two-point conversions, is more variable in terms of number of points scored? Hint: Think about which strategy produces point values that deviate from the expected value more often.

With work please!

Explanation / Answer

(a) Probability distributions for PATs and two pooint conversions are the example of binomial distributions. So, X have binomial distribution here for probability = 0.936 and similary, Y have binomial distribution here for probability = 0.486.

(b) Expected number of Points in PAT = 0.936 * 1 = 0.936

Expected number of pointss in two point conversions = 0.486 * 2 = 0.972

(c) Here the two point conversion strategy is better in terms of expected number of points scored.

(d) Var(Y) = sqrt (2 * 0.486 * 0.514) = 0.707

(e) Here the strategy two point conversions is more variables in terms of number of points scored.

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