Question

Binomial distribution的问题

One way of checking the effects of undercoverage,
nonresponse, and other sources of error in a sample
survey is to compare the sample with know facts about
the population. About 12% of American adults are black.
What is the probability that a sample of 1500 adults will
contain 170 or fewer blacks?
(a) 0.2148
(b) 0.3241
(c) 0.1890
(d) 0.1771

求答案及过程！

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Answer 1

Firstly, 1700 is big enough so we are inclined to use the central lim theorem. The mean value is given as p=0.12, and we need to estimate the variation.
For a binomial distribution of p vs (1-p) (p already known) and n i.i.d. observations X1,...Xn, the estimate sigma^2=p(1-0.12)^2+(1-p)0.12^2=0.12*0.88, sigma=0.325
So for the bare that there are 170 blacks, (\sum_{i=1,2,...n}(Xi)-pn)/(sqrt(n)*sigma)=-0.7945
From the table of normal distribution, we get that it is about 0.215. So choose A