There is a 0.18910.1891 probability that a best-of-seven contest will last four

Question

There is a

0.18910.1891

probability that a best-of-seven contest will last four games, a

0.21250.2125

probability that it will last five games, a

0.27330.2733

probability that it will last six games, and a

0.32510.3251

probability that it will last seven games. Verify that this is a probability distribution. Find its mean and standard deviation. Is it unusual for a team to "sweep" by winning in four games?

1. What is the mean of the probability distribution?

2. Find the standard deviation of the probability distribution

Explanation / Answer

P(X = 4) = 0.1891

P(X = 5) = 0.2125

P(X = 6) = 0.2733

P(X = 7) = 0.3251

a) mean, E(X) = 4 * 0.1891 + 5 * 0.2125 + 6 * 0.2733 + 7 * 0.3251 = 5.7344

b) E(X2) = 42 * 0.1891 + 52 * 0.2125 + 62 * 0.2733 + 72 * 0.3251 = 34.1068

Var(X) = E(X2) - (E(X))2 = 34.1068 - 5.73442 = 1.2235

Standard deviation = sqrt(1.2235) = 1.1061

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