Question

id=12876874 Chapter 13 Homew Sierra Gros > Activities and Due Dates > Chapter 13 Homework core: 11.4% Resources Give Up? Hint Check Attem 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 QCEWreported the total taxable carnings, 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 50 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 o = 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 2. Use the central limit theorem (CLT) to determine the probability P that the mean taxable wages in James' sample of 50 counties will be less than $26 million. Report your answer to four decimal places. P(* <26) > E II Fo F10 - Priser

Answer 1

solution:

From the given information

Let X be the total taxable earnings, in millions , of all wage earners in a country.

The mean total taxable earnings of all wage earners in a country across in all the countries

in james sample is $\bar x$

Mean ($\mu$) = 28.29

Standard deviation ($\sigma$) = 33.493

Sample size (n) = 50

Since n>30,

By using Central Limit Theorem

Probability that mean taxable wages in james sample countries will be less than $26 millions = P($\bar x$<26)

= $P(\frac{\bar x -\mu}{\frac{\sigma}{\sqrt n}}<\frac{26 - 28.29}{\frac{33.493}{\sqrt 50}})$

= P(Z<-0.4835)

= P(Z>0.4835)

= 1 - P(Z<=0.4835)

= 1 - 0.6856 [since, using z distribution table ]

= 0.3144

$\therefore$ Probability that mean taxable wages in james sample countries will be less than $26 millions = 0.3144