Mathematics, 09.11.2020 17:20 personm21

1. If 15% of adults in a certain country work from home, what is the probability that fewer than 42 out of a random sample of 350 adults will work from home? (Round your answer to 3 decimal places) 2. Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If we want to estimate this proportion to within 4% at the 95% confidence level, how many randomly selected teenagers must we survey?

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Answer from: mcfancei

1. Using the normal approximation to the binomial, it is found that there is a 0.05 = 5% probability that fewer than 42 out of a random sample of 350 adults will work from home.

2. Solving the equation for the margin of error of a confidence interval of proportions, it is found that we must survey 601 randomly selected teenagers.

Question 1:

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

15% work from home, thus p = 0.15

.Sample of 350 adults, thus n = 350

Then, for the approximation:

$\mu = np = 350(0.15) = 52.5$

$\sigma = \sqrt{np(1-p)} = \sqrt{350(0.15)(0.85)} = 6.68$

Using continuity correction, the probability is P(X < 42 - 0.5) = P(X < 41.5) , which is the p-value of Z when X = 41.5. So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41.5 - 52.5}{6.68}$

Z = -1.65

Z = -1.65 has a p-value of 0.05.

0.05 = 5% probability that fewer than 42 out of a random sample of 350 adults will work from home.

Question 2:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this problem:

Within 4%, thus M = 0.04

.We do not have an estimate for the true proportion, thus $\pi = 0.5$ .95% confidence level, thus z has a p-value of $\frac{1 + 0.95}{2} = 0.975$ , thus z = 1.96.

We have to solve for n, then:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96(0.5)$

$\sqrt{n} = \frac{1.96(0.5)}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96(0.5)}{0.04})^2$

n = 600.25

Rounding up:

We must survey 601 randomly selected teenagers.

A similar problem is given at link

Answer from: jennmcd17

(1) 0.058

(2) 601

Step-by-step explanation:

(1)

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p\\$

The standard deviation of this sampling distribution of sample proportion is:

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

As the sample size is large, i.e. n = 350 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion of adults in a certain country work from home.

Compute the probability that fewer than 42 out of a random sample of 350 adults will work from home:

Sample proportion: $\hat p=\frac{42}{350}=0.12$

$P(\hat p < 0.12)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}$

Thus, the probability that fewer than 42 out of a random sample of 350 adults will work from home is 0.058.

(2)

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Given:

MOE = 0.04

Confidence level = 95%

Assume that the sample proportion is 50%.

The critical z-value for 95% confidence level is 1.96.

Compute the required sample size as follows:

$MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$n=[\frac{z_{\alpha/2}\cdot\sqrt{\hat p(1-\hat p)}}{MOE}]^{2}\\\\=[\frac{1.96\times \sqrt{0.50(1-0.50)}}{0.04}]^{2}\\\\=600.25\\\\\approx 601$

Thus, the required sample size is 601.

Answer from: Quest

8.4 10.4 8.9 10.9 have a good day

Answer from: Quest

many reason,

step-by-step explanation:

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