The Department of Tourism reported that in May of 2017, approximately $63 \%$ of the tourists to the Philippines were from Asia. They also noted that it was the first time that the Philippines had more than 500,000 tourists in the month of May. Suppose another organization had taken a simple random sample of 600 of the tourists in that population.

Assuming that the reported $63 \%$ claim is accurate, what is the approximate probability that the other organization's results were within 2 percentage points of the Department of Tourism's results?

Choose 1 answer:
(A) $P(0.61

Q: The Department of Tourism reported that in May of 2017, approximately $63 \%$ of the tourists to the Philippines were from Asia. They also noted that it was the first time that the Philippines had more than 500,000 tourists in the month of May. Suppose another organization had taken a simple random sample of 600 of the tourists in that population. Assuming that the reported $63 \%$ claim is accurate, what is the approximate probability that the other organization's results were within 2 percentage points of the Department of Tourism's results? Choose 1 answer: (A) $P(0.61<\hat{p}<0.65) \approx 0.61$ (B) $P(0.61<\hat{p}<0.65) \approx 0.65$ (C) $P(0.61<\hat{p}<0.65) \approx 0.69$ (D) $P(0.61<\hat{p}<0.65) \approx 0.73$

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the population proportion (p) and the sample size (n). ‖ ‖ step 2 ⋮ Calculate the standard error (SE) for the sample proportion using the formula $SE = \sqrt{\frac{p(1-p)}{n}}$. ‖ ‖ step 3 ⋮ Use the normal distribution to approximate the probability that the sample proportion $\hat{p}$ is within 2 percentage points of the population proportion. ‖ ‖ step 4 ⋮ Convert the percentage points into decimal form to find the range for $\hat{p}$. ‖ ‖ step 5 ⋮ Calculate the z-scores for the lower and upper bounds of the sample proportion using the formula $z = \frac{\hat{p} - p}{SE}$. ‖ ‖ step 6 ⋮ Use the standard normal distribution table or a calculator to find the probabilities corresponding to the calculated z-scores. ‖ ‖ step 7 ⋮ Subtract the probability of the lower bound from the probability of the upper bound to find the approximate probability that the sample proportion is within the given range. ‖ ∻[1] Answer∻ ⚹ The correct answer is not provided as the calculation steps are not completed. ⚹ ∻Key Concept∻ ⚹Normal Approximation to the Binomial Distribution⚹ ∻Explanation∻ ⚹The normal approximation is used when the sample size is large enough, and the question involves finding the probability of a sample proportion within a certain range around the population proportion.⚹◊What is the formula to calculate the probability that a sample proportion falls within a certain range given a reported percentage?⍭ Generate me a similar question◊

Question

The Department of Tourism reported that in May of 2017, approximately $63 \%$ of the tourists to the Philippines were from Asia. They also noted that it was the first time that the Philippines had more than 500,000 tourists in the month of May. Suppose another organization had taken a simple random sample of 600 of the tourists in that population.

Assuming that the reported $63 \%$ claim is accurate, what is the approximate probability that the other organization's results were within 2 percentage points of the Department of Tourism's results?

Choose 1 answer:
(A) $P(0.61<\hat{p}<0.65) \approx 0.61$
(B) $P(0.61<\hat{p}<0.65) \approx 0.65$
(C) $P(0.61<\hat{p}<0.65) \approx 0.69$
(D) $P(0.61<\hat{p}<0.65) \approx 0.73$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Identify the population proportion (p) and the sample size (n). ‖ ‖ step 2 ⋮ Calculate the standard error (SE) for the sample proportion using the formula $SE = \sqrt{\frac{p(1-p)}{n}}$. ‖ ‖ step 3 ⋮ Use the normal distribution to approximate the probability that the sample proportion $\hat{p}$ is within 2 percentage points of the population proportion. ‖ ‖ step 4 ⋮ Convert the percentage points into decimal form to find the range for $\hat{p}$. ‖ ‖ step 5 ⋮ Calculate the z-scores for the lower and upper bounds of the sample proportion using the formula $z = \frac{\hat{p} - p}{SE}$. ‖ ‖ step 6 ⋮ Use the standard normal distribution table or a calculator to find the probabilities corresponding to the calculated z-scores. ‖ ‖ step 7 ⋮ Subtract the probability of the lower bound from the probability of the upper bound to find the approximate probability that the sample proportion is within the given range. ‖ ∻[1] Answer∻ ⚹ The correct answer is not provided as the calculation steps are not completed. ⚹ ∻Key Concept∻ ⚹Normal Approximation to the Binomial Distribution⚹ ∻Explanation∻ ⚹The normal approximation is used when the sample size is large enough, and the question involves finding the probability of a sample proportion within a certain range around the population proportion.⚹◊What is the formula to calculate the probability that a sample proportion falls within a certain range given a reported percentage?⍭ Generate me a similar question◊

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