Question

Math

Posted 4 months ago

```
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) $P(0.61<\hat{p}<0.65) \approx 0.73$
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

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.

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