Suppose a simple random sample of size n = 1000 is obtained from a population wh
ID: 3379342 • Letter: S
Question
Suppose a simple random sample of size n = 1000 is obtained from a population whose size is N =2,000.000 and whose population portion with a specified characteristic is p = 0.53. Complete parts (a) through (c) below Describe the sampling distribution of p. Approximately normal. mup = 0.53 and sigma^ p 0.0002 Approximately normal. mup = 0.53 and sigma^ p 0.0158 Approximately normal. mup = 0.53 and sigma^ p0.0004 What is the probability of obtaining x = 570 or more individuals with the characteristic? P(x GE 570) = (Round to four decimal places as needed.)Explanation / Answer
a)
By central limit theorem, it will have the same mean, u(p^) = 0.53.
However, the standard deviation is
sigma(p^) = sqrt(p(1-p)/n) = sqrt(0.53*(1-0.53)/1000)
= 0.015782902
Thus, it is OPTION B. [ANSWER, B]
***********************
b)
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 1000
p = the probability of a success = 0.53
x = our critical value of successes = 570
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 569 ) = 0.993914889
Thus, the probability of at least 570 successes is
P(at least 570 ) = 0.006085111 [ANSWER]
Note that I used the binomial distribution here. Should you have another approach to this problem, please resubmit this question, stating the method you like, so we can help you! Thanks!