a casino and play a certain game. The chance the person will win the game is 0.3

Question

a casino and play a certain game. The chance the person will win the game is 0.32. Once they play the first game, win or lose, they are to play the game 3 more times for a total of 4 games. A random variable X is to count how many of the 4 games the gambler wins (a) Finish the probablity distribution of X below. Use tour decimals in each of your entries. P(X = x) (b) From the distribution you found in part (a), what can you say about the distribution of X? The distribution of X is ? with an mean of games won and a standard deviation of ll games won. Enter your answers to two decimals.)

Explanation / Answer

p = 0.32

n = 4

a) P(X = 0) = 4C0 * 0.320 * (1 - 0.32)4 = 0.2138

P(X = 1) = 4C1 * 0.321 * (1 - 0.32)3 = 0.4025

P(X = 2) = 4C2 * 0.322 * (1 - 0.32)2 = 0.2841

P(X = 3) = 4C3 * 0.323 * (1 - 0.32)1 = 0.0891

P(X = 4) = 4C4 * 0.324 * (1 - 0.32)0 = 0.0105

b) Mean = n * p = 4 * 0.32 = 1.28

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(4 * 0.32 * 0.68) = 0.93

The distrubution of X is binomial distribution with a mean of 1.28 games won and standard deviation of 0.93 games won

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