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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? choose 1 answer:

a [p(0.79< p<0.81) ≈ 0.42]
b [p(0.79< p<0.81) ≈ 0.46]
c [p(0.79< p<0.81) ≈ 0.50]
d [p(0.79< p<0.81) ≈ 0.54]

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The probability that between 79% and 81% of the 500 sampled wildlife watchers actively observed mammals in 2011 is approximately a) 0.42.

1. Central Limit Theorem: Since we are taking a random sample of 500 individuals from a large population (over 71 million), the Central Limit Theorem applies. This means that even if the population proportion is not normally distributed, the sample proportion will be approximately normally distributed, regardless of the population distribution, as long as the sample size is sufficiently large.

2. Calculating Standard Error: The standard error of the proportion (SE) represents the typical variability of sample proportions from the population proportion. We can calculate it using the formula:

SE = sqrt(p * (1 - p) / n)

where:

* p = population proportion (0.8)

* n = sample size (500)

SE = sqrt(0.8 * (1 - 0.8) / 500) ≈ 0.016

3. Calculating Z-scores: We want to find the probability of the sample proportion falling between 79% and 81%. We can convert these percentages to z-scores using the formula:

z = (sample proportion - population proportion) / SE

* Lower bound (79%): z = (0.79 - 0.8) / 0.016 ≈ -0.63

* Upper bound (81%): z = (0.81 - 0.8) / 0.016 ≈ 0.63

4. Finding the Probability: We need to find the area between -0.63 and 0.63 under the standard normal curve. Using a standard normal distribution table or calculator, we can find this area to be approximately 0.423.

Therefore, the approximate probability that between 79% and 81% of the 500 sampled wildlife watchers actively observed mammals in 2011 is 0.42, which corresponds to option a).

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