G According to the 2011 National X + rofile/kaid_123667084355651599004012/assignments/teacher/kaid_570905347429149155795 Q Search ... arch Khan Academy Finding probabilities with sample proportions You might need: Calculator, Z table According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildlife 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: P(0.79 < p < 0.81) - 0.42 P(0.79 < p< 0.81) 0.46 P(0.79 < p< 0.81) 0.50 P(0.79 < p< 0.81) 0.54

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G According to the 2011 National X
+
rofile/kaid 123667084355651599004012/assignments/teacher/kaid 570905347429149155795 *.
Q Search
arch
Khan Academy
Finding probabilities with sample proportions
You might need: E Calculator, Z table
According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over
71 million wildlife 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:
P(0.79 < p < 0.81) ~ 0.42
P(0.79 < p < 0.81) 0.46
P(0.79 < p < 0.81) = 0.50
P(0.79 < p < 0.81) 0.54
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Transcribed Image Text:Help G According to the 2011 National X + rofile/kaid 123667084355651599004012/assignments/teacher/kaid 570905347429149155795 *. Q Search arch Khan Academy Finding probabilities with sample proportions You might need: E Calculator, Z table According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildlife 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: P(0.79 < p < 0.81) ~ 0.42 P(0.79 < p < 0.81) 0.46 P(0.79 < p < 0.81) = 0.50 P(0.79 < p < 0.81) 0.54 Show Calculator Do 4 problems Oo o a About Math: Pre-K8th grade sto provide a free, world-class News Math: Get ready courses
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