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:
[tex]_nC_k\times p^n\times (1-p)^{k-n}
\\
\\=_{70}C_0\times (\frac{1}{38})^0\times \frac{37}{38}^{70}
\\
\\=0.15[/tex]
Since this is not below 0.05, we do not consider this unusual.
