Suppose a simple random sample of size n 1000 is obtained f

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.)

Solution

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]

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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!