Question

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

Answer 1

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

Step-by-step 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:

`_nC_k* p^n* (1-p)^(k-n) \\ \\=_(70)C_0* ((1)/(38))^0* (37)/(38)^(70) \\ \\=0.15`

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