Step 1 :Given a simple random sample of size \(n=150\) from a population of size \(N=15,000\) with a population proportion \(p=0.8\).
Step 2 :We are asked to find the probability of obtaining \(x=123\) or more individuals with the characteristic, i.e., \(P(\hat{p} \geq 0.82)\).
Step 3 :This is a problem of hypothesis testing for proportions. We can solve this problem by first calculating the z-score of the sample proportion and then finding the corresponding probability from the standard normal distribution table.
Step 4 :The z-score is calculated as follows: \(z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p\) is the population proportion, and \(n\) is the sample size.
Step 5 :Substituting the given values, we get \(z = \frac{0.82 - 0.8}{\sqrt{\frac{0.8(1-0.8)}{150}}}\), which simplifies to \(z \approx 0.612\).
Step 6 :This means that the sample proportion of 0.82 is 0.612 standard deviations above the mean.
Step 7 :Now, we need to find the probability corresponding to this z-score from the standard normal distribution table. However, since we want the probability of obtaining a sample proportion greater than or equal to 0.82, we need to find the area to the right of the z-score, which is \(1 - P(Z < z)\).
Step 8 :Calculating this probability, we get \(P(\hat{p} \geq 0.82) = 1 - P(Z < 0.612) \approx 0.2701\).
Step 9 :So, the probability of obtaining a sample proportion greater than or equal to 0.82 is approximately 0.2701.
Step 10 :Final Answer: \(P(\hat{p} \geq 0.82)=\boxed{0.2701}\) (rounded to four decimal places as needed).