Answered

The United States Bureau of Labor Statistics (BLS) conducts the Quarterly Census of Employment and Wages (QCEW) and reports a variety of information on each county in America. In the third quarter of 2016, the QCEW reported the total taxable earnings, in millions, of all wage earners in all 3222 counties in America.

Suppose that James is an economist who collects a simple random sample of the total taxable earnings of workers in 56 American counties during the third quarter of 2016. According to the QCEW, the true population mean and standard deviation of taxable earnings, in millions of dollars, by county are ?=28.29 and ?=33.493, respectively.

Let X be the total taxable earnings, in millions, of all wage earners in a county. The mean total taxable earnings of all wage earners in a county across all the counties in James' sample is x??.

Use the central limit theorem (CLT) to determine the probability P that the mean taxable wages in James' sample of 56 counties will be less than $33 million. Report your answer to four decimal places.

P(x??<33)=

Use the CLT again to determine the probability that the mean taxable wages in James' sample of 56 counties will be greater than $30 million. Report your answer to four decimal places.

P(x??>30)=

Answer :

Answer:

The mean total taxable earnings of all wage earners in a county is same as the population mean, $28.29.

The probability that the mean taxable wages in James' sample of 56 counties will be less than $33 million is 0.8508.

The probability that the mean taxable wages in James' sample of 56 counties will be greater than $30 million is 0.3520.

Step-by-step explanation:

Let X = total taxable earnings of workers.

The expected value of the random variable X is:

E (X) = μ = $28.29

The standard deviation of the random variable X is:

SD (X) = σ = $33.493.

The data was collected from 56 American counties.

The sample mean of a random variable is the an unbiased estimator of the population mean.

If repeated samples are collected from a population and the mean for each sample is computed then the expected value of the sample means is same as the population mean.

So the mean total taxable earnings of all wage earners in a county is same as the population mean, $28.29.

According to the Central limit theorem if we have a population with mean μ and standard deviation σ and take appropriately huge random-samples (n ≥ 30) from the population with replacement, then the distribution of the sample-means will be approximately normally distributed.

Then, the mean of the sample means is given by,  [tex]\mu_{\bar x}=\mu[/tex].

And the standard deviation of sample means is, [tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

Compute the value of P (\bar X < 33) as follows:

[tex]P(\bar X<33)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}<\frac{33-28.29}{33.943/\sqrt{56}})\\=P(Z<1.04)\\=0.8508[/tex]

Thus, the probability that the mean taxable wages in James' sample of 56 counties will be less than $33 million is 0.8508.

Compute the value of P (\bar X > 30) as follows:

[tex]P(\bar X>30)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{30-28.29}{33.943/\sqrt{56}})\\=P(Z>0.38)\\=1-P(Z<0.38)\\=1-0.64803\\=0.35197\\\approx0.3520[/tex]

Thus, the probability that the mean taxable wages in James' sample of 56 counties will be greater than $30 million is 0.3520.

Parrain

Based on the Central Limit Theorem, the probability that the mean taxable wages will be less than $33 million is 0.8537 and the probability that mean taxable income will be more than $30 million is 0.3512.

What is the probability that mean taxable wages are less than $33 million?

= P (x < 33)

= P ( x - u) / (σ / √n)

= P (Z < (33 - 28.29) / (33.493 / √56))

= P (Z < 1.0524)

= 0.8537

What is the probability that mean taxable wages are more than $33 million?

= P (Z > (30 - 28.29) / (33.493 / √56))

= P (Z > 0.3821)

= 0.3512

Find out more on  Central Limit Theorem at https://brainly.com/question/898534.

Other Questions