Question

According the April 12, 2017 Pew Research survey, 58% of Americans approve of U. S. Missile strikes in Syria in response to reports of the use of chemical weapons by Bashar al-Assad's government (the Syrian government). A sample of 50 Americans are surveyed. Let pˆ be the sample proportion of Americans who approve the U. S. Missile strikes. What is the population proportion? Answer 0. 58 (decimal form) What is the sample size? Answer 50 Can the normal approximation be used with this distribution? Answer Yes, the sample distribution meets the "rule of thumb. " What is the mean of the sampling proportion? Answer 0. 58 What is the standard deviation sampling proportion? Answer 0. 0698 (Round to 4 decimal places. ) What is the probability that no more than 25 Americans of the 50 in the survey approve of the missile strikes? What is pˆ? Answer 0. 5 What is the z-score? Answer (Round to the nearest hundredth. ) What is the requested probability? P(pˆ Answer ≤ 0. 5)= Answer What is the probability that more than 30 of the 50 Americans in the survey approved of the missile strikes? What is pˆ? Answer 0. 6 What is the z-score? Answer 0. 29 (Round to the nearest hundredth. ) What is the requested probability? P(pˆ Answer > 0. 6)= Answer

Answer 1

The probability that more than 30 Americans of the 50 in the survey approve of the missile strikes is .3859

How to determine if the normal distribution can be applied?

The given parameters are:

Sample size, n = 50

Population proportion, p = 0.58

By central limit theorem (CLT), the normal approximation can be used with this distribution because the sample size is greater than 30

The mean of the sampling proportion?

This is the same as the population proportion.

So, the mean of the sampling proportion is 0.58

The standard deviation sampling proportion?

This is calculated using:

$\sigma = \sqrt{\frac{p(1 - p)}{n}}$

So, we have:

$\sigma = \sqrt{\frac{0.58 * (1 - 0.58)}{50}}$

σ = 0.0698

That no more than 25 Americans approve of the missile strikes?

Start by calculating the p-value

p = 25/50

p = 0.5

The z-score is:

$z = \frac{x - \mu}{\sigma}$

This gives

$z = \frac{0.5 - 0.58}{0.0698}$

This gives

z = -1.15

Using the z-table of probability, the requested probability is:

P(z ≤ -1.15) = 0.1251

The probability that more than 30 approved of the missile strikes?

Start by calculating the p-value

p = 30/50

p = 0.6

The z-score is:

$z = \frac{x - \mu}{\sigma}$

This gives

$z = \frac{0.6 - 0.58}{0.0698}$

This gives

z = 0.29

Using the z-table of probability, the requested probability is:

P(z > 0.29) = 0.3859

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