If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning. a. find the mean and standard deviation for the number of wins of gamblers who bet on the number 7 seventy seventy times. b. would 0 wins in seventy seventy bets be an unusually low number of wins

A) The mean would be 1.842 and the standard deviation would be 1.339.
b) It would not be unusually low.

Explanation
a) This can be modeled as a binomial distribution, since there are two outcomes (7 or not 7), the events are independent (getting a 7 on one spin does not influence the chance of a 7 on the next spin) and there are a fixed number of trials (70).  In a binomial distribution, the mean is given by n*p, where n is the number of trials and p is the probability.

μ=n*p=70*(1/38) = 1.842.

The standard deviation of a binomial distribution is given by
σ=√(n*p*(1-p))=√(70(1/38)(37/38)=1.339.

b) To determine if this is unusually low, we find the probability:


Since this is not below 0.05, we do not consider this unusual.

Try a or d, hope that