Mathematics, 21.04.2020 20:19, kennedymattkm

The average income in a certain region in 2013 was $68 comma 000 per person per year. Suppose the standard deviation is $28 comma 000 and the distribution is right-skewed. Suppose we take a random sample of 100 residents of the region. a. Is the sample size large enough to use the Central Limit Theorem for means? Explain. b. What are the mean and standard error of the sampling distribution? c. What is the probability that the sample mean will be more than $2 comma 800 away from the population mean?

Answers: 1

Show answers

Answers

answered: jonataespi5924

Using the normal distribution and the central limit theorem, we have that:

a) The sample size is of 100, which is larger than 30, hence it is large enough to use the Central Limit Theorem for means.

b) The mean is of $68,000 while the standard deviation is of $2,800.

c) There is a 0.3174 = 31.74% probability that the sample mean will be more than $2,800 away from the population mean.

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. By the Central Limit Theorem, the sampling distribution of sample means of size n has mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ . For a non-normal distribution, the Central Limit Theorem can be applied as long as the sample size is larger than 30.

In this problem:

Mean of $68,000, hence $\mu = 68000$ .Standard deviation of $28,000, hence $\sigma = 28000$ .

Item a:

The sample size is of 100, which is larger than 30, hence it is large enough to use the Central Limit Theorem for means.

Item b:

$\mu = 68000$

$s = \frac{28000}{\sqrt{100}} = 2800$

The mean is of $68,000 while the standard deviation is of $2,800.

Item c:

First, we find the z-score, which is:

$Z = \frac{2800}{s}$

$Z = \frac{2800}{2800}$

Z = 1

The probability is P(|Z| > 1), which is 2 multiplied by the p-value of Z = -1.

Looking at the z-table, Z = -1 has a p-value of 0.1587

2 x 0.1587 = 0.3174.

There is a 0.3174 = 31.74% probability that the sample mean will be more than $2,800 away from the population mean.

A similar problem is given at link

answered: pindu

(a) Yes, the sample size large enough to use the Central Limit Theorem for means.

(b) The mean and standard error of the sampling distribution are $68,000 and $2,800 respectively.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard-deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

$\mu_{\bar x}=\mu$