Question

According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildife watchers in the US. Of these wildlife watchers, the survey reports that 80% actively observed mammals. Suppose that one of the census workers repeated the survey with a simple random sample of only 500 wildlife watchers that same year. Assuming that the original survey's 80% claim is correct, what is the approximate probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011?

Answer 1

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

$\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018$

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

$z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55$

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

$P(0.79$