Given a population proportion \(p=0.8\) and a simple random sample of size \(N=200\) from a population of size \(n=20,000\), the probability of obtaining \(x \geq 166\) individuals with the characteristic is found using the normal approximation to the binomial distribution.
To find the probability of obtaining \(x \geq 166\) individuals with the specified characteristic in a simple random sample, you can use the normal approximation to the binomial distribution. Given a large sample size \(N = 200\) and a population proportion \(p = 0.8\), you can use the normal distribution.
The mean (\(\mu\)) of the binomial distribution is given by \(\mu = np\), and the standard deviation (\(\sigma\)) is given by \(\sigma = \sqrt{np(1-p)}\).
\[ \mu = 200 \times 0.8 = 160 \]
\[ \sigma = \sqrt{200 \times 0.8 \times 0.2} \approx 5.6569 \]
Now, you want to find the probability of obtaining \(x \geq 166\), which can be transformed into a z-score:
\[ Z = \frac{166 - \mu}{\sigma} \]
Calculate the z-score and use a standard normal distribution table or calculator to find the probability.
\[ Z = \frac{166 - 160}{5.6569} \approx 1.0607 \]
Using a standard normal distribution table, find the probability corresponding to \(Z \geq 1.0607\). This gives the probability of obtaining \(x \geq 166\).