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

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 ⋮ Calculate the standard deviation for the sampling distribution of the sample proportion using the formula $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where $p = 0.63$ is the population proportion and $n = 600$ is the sample size. ‖ ‖ step 2 ⋮ Substitute the given values into the formula to get $\sigma_{\hat{p}} = \sqrt{\frac{0.63(1-0.63)}{600}}$. ‖ ‖ step 3 ⋮ Compute the standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{0.63 \cdot 0.37}{600}} \approx 0.0202$. ‖ ‖ step 4 ⋮ Convert the range of proportions (0.61 to 0.65) into z-scores by subtracting the mean (0.63) and dividing by the standard deviation ($\sigma_{\hat{p}}$). ‖ ‖ step 5 ⋮ Calculate the z-score for 0.61: $z_{0.61} = \frac{0.61 - 0.63}{0.0202} \approx -0.99$. ‖ ‖ step 6 ⋮ Calculate the z-score for 0.65: $z_{0.65} = \frac{0.65 - 0.63}{0.0202} \approx 0.99$. ‖ ‖ step 7 ⋮ Use the standard normal distribution table to find the probabilities corresponding to the z-scores. ‖ ‖ step 8 ⋮ The probability for $z_{0.61}$ is approximately 0.1611, and for $z_{0.65}$ is approximately 0.8389. ‖ ‖ step 9 ⋮ Subtract the probability for $z_{0.61}$ from the probability for $z_{0.65}$ to find the probability that the sample proportion is between 0.61 and 0.65: $P(0.61<\hat{p}<0.65) \approx 0.8389 - 0.1611 = 0.6778$. ‖ ‖ step 10 ⋮ The closest answer choice to 0.6778 is 0.69. ‖ ***C*** ∻Key Concept∻ ⚹Normal Approximation to the Binomial Distribution⚹ ∻Explanation∻ ⚹When the sample size is large, the sampling distribution of the sample proportion can be approximated by a normal distribution with mean equal to the population proportion and standard deviation calculated using the formula $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$. The z-scores are then used to find the probabilities for a given range.⚹◊What is the formula to calculate the probability of a sample proportion falling within a certain range of values⍭ Generate me a similar question◊

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