CBS televised a recent Superbowl football game between the New Orleans Saints an

Question

CBS televised a recent Superbowl football game between the New Orleans Saints and the Indiannapolis Colts. The game indicates that the probability of a household tuning to the game is .45 among US households. Let the random variable X be the number of households tuned to the game and it is a binomial random variable. Suppose 10 households were randomly selected. A. What is the probability that no more than 3 households tuned to the game ( keep 3 decimal places) B. Suppose 10 households were randomly selected. What is the mean number of households tuned to the game? C. What is the standard deviation tuned to the game? (keep two decimal places) D. Based on your calculations, is one household tuned to the game unusually low? Why?

Explanation / Answer

p = 0.45

n = 10

a) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                  = 10C0 * 0.450 * 0.5510 + 10C1 * 0.451 * 0.559 + 10C2 * 0.452 * 0.558 + 10C3 * 0.453 * 0.557

                  = 0.266

b) mean = n * p

               = 10 * 0.45

               = 4.5 or 5 (approx.)

c) standard deviation = sqrt(n * p * (1 - p))

                                  = sqrt(10 * 0.45 * 0.55)

                                  = 1.57

d) P(X = 1) = 10C1 * 0.451 * 0.559

                   = 0.0207

Since the probability is less than 0.05 , this is unusual.

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