Mathematics, 07.04.2020 17:28 corban13

According to a study conducted in one city, 25% of adults in the city have credit card debts of more than $2000. A simple random sample of n equals 150 adults is obtained from the city. Describe the sampling distribution of ModifyingAbove p with caret, the sample proportion of adults who have credit card debts of more than $2000. Round to three decimal places when necessary.

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Answer from: candlewax2001

Following are the calculation to the given value:

Given:

Based on one town's study, the population percentage of adults in the city had credit card bills totaling and over $\$2000\ (p) = 0.25$

A city-wide sample of adults was recruited (n) = 150

Solution:

For step 1:

To get the sampling distribution of probability value $\hat{p}$ , we should first determine whether the sample is sufficient using the following two conditions:

$\to np \geq 10\\\\\to n(1-P) \geq 10\\\\\to np = (150 \times 0.25)= 37.5 \geq 10\\\\\to n(1-P) = 150(1 - 0.25) = 112.5\geq 10\\\\$

The two conditions are fulfilled. We understand from the "Central limit theorem" that the as being $\hat{p}$ is roughly normally distributed with the mean $\mu_{\hat{p}}=P$ and the standard deviation $\sigma_{\hat{p}} = \sqrt{\frac{(p(1-p))}{n}}$ along with $np \geq 10 \ \ and \ \ n(1-P) \geq 10$

Therefore the 2 conditions which satisfy the proportion of sample $\hat{p}$ that were normally distributed by the mean $\mu_{\hat{p}}=P$ and the standard deviation $\sigma_{\hat{p}} = \sqrt{\frac{(p(1-p))}{n}}$

For step 2:

Sample proportion mean $(\hat{p})\ \ is,\ \ \mu _{\hat{p}}=p=0.25$

For step 3:

The sample proportion $\hat{p}$ of the standard deviation:

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

$= \sqrt{\frac{(0.25(1-0.25))}{150}} \\\\= \sqrt{\frac{(0.25-0.0625)}{150}} \\\\= \sqrt{\frac{0.1875}{150}} \\\\= \sqrt{0.00125}\\\\=0.035 \\\\$

Since the distribution of the sample proportion that is $\hat{p}$ is adults which has to credit card debts of more than $\$2000$ is approximately normal; $\bold{\mu _{\hat{p}}=0.25 ,\ \sigma_{\hat{p}}=0.035}$ .

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Answer from: jwbri

The sampling distribution of $\hat p$ is N (0.25, 0.0354²).

Step-by-step explanation:

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:

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

Let p = proportion of adults in the city having credit card debts of more than $2000.

It is provided that the proportion of adults in the city having credit card debts of more than $2000 is, p = 0.25.

A random sample of size n = 150 is selected from the city.

Since n = 150 > 30 the Central limit theorem can be used to approximate the distribution of p by the Normal distribution.

The mean is:

$\mu_{\hat p}=p=0.25$

The standard deviation is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.25(1-0.25)}{150}}=0.0354$

Thus, the sampling distribution of $\hat p$ is N (0.25, 0.0354²).

Answer from: Quest

i think it is 70

step-by-step explanation:

Answer from: Quest

table 3 represents a direct variation function. : )

Which table represents a direct variation function?

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