A slot machine at a casino pays out an average of $0.90, with a standard deviation of $120. It costs a dollar to play. If gamblers play this machine 5,760,000 times in a month, what is the probability that the casino will come out ahead? Assume that the casino's total profit follows a Normal model.
a. 0.036
b. 0.964
c. 0.977
d. 0.933
e. 0.023
2007-07-19 16:33:32 · 1 answers · asked by publix_589 2 in Science & Mathematics ➔ Mathematics
Let X be the total net payout for the month.
We know X has a mean of 5,760,000 (0.90 - 1.00) = -$576,000. The variance of X is 5,760,000 times the variance of a single play, so the std dev of X is √5,760,000 = 2400 times the std dev of a single play, i.e. $288,000. So
P(casino wins) = P(X < 0) = P(Z < (0 - (-576000)) / 288000) = P(Z < 2) = 0.977 to 3 d.p. So answer (c) is correct.
2007-07-19 16:42:30 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋