(1)
The sample distribution taken from a binomial distribution is approximately normal because sample/population size < 0.05 ("small" sample). Also a binomial distribution approximates a normal distribution when np>K and np(1-p)>K, where K ranges between 5 & 10, depending on preference.
(2)
The expected value of p̂, or E[p̂] is the mean of the population. The mean of a binomial population is np.
(4)
P(X≥123)=ΣP(Xi),i=123...n
where P(Xi)=C(n,i)pi(1-p)(n-i)
and C(n,i)=n!/(i!(n-i)!) is combination of i objects from n.
It's a relative long process to do the summation. You can use binomial tables, or you can approximate the value using the normal approximation to binomial distribution, as hinted in part (1).
(5) see (4)
Suppose a simple random sample of size n=150 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p=0.8. Please answer questions (1) through (5) below.
(1) Describe the sampling distribution of p^ (choose the correct phrase that best describes the shape of the sampling below)
a) Not normal because n <, 0.05N and np (1-p) <10.
b) Approximately normal because n <, 0.05N and np (1-p) >10.
c) Approximately normal because n <, 0.05N and np (1-p) <10.
d) Not normal because n <, 0.05N and np (1-p) >10.
2) Determine the mean of the sampling distribution of p^.
p^ = ______ (round to one decimal place as needed)
3) Determine the standard deviation of the sampling distribution of p^.
p^ = ______ (round six decimal places as needed)
4) What is the probability of obtaining x=123 or more individuals with the characteristic? that is , what P(p^ > 0.82)?
P( p^ > 0.82) = _____ (Round four decimals places)
5) What is the probability of obtaining x = 111 of fewer individuals with the characteristic ? that is what is P(p^ < 0.74)?
P(p^ < 0.74) = _______ (Round to four decimal places as needed)
1 answer
1 answer