Binomial distribution的问题
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
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