Question

Salaries for various positions can vary significantly depending on whether the company is in the public or private sector. The US Department of Labor posted the 2007 average salary for human resource managers employed for the Federal goverenment at 76,503. Assume that annual salaries for this job type are normally distributed and have a standard deviation of 8,850 dollars. a. What is the probality that a randomly selected human resource manager received over a 100,000 in 2007? B. A sample of 20 labor relations managers is taken and annual salaries reported. What is the probality that the sample mean annual salary falls between 70,000 and 80,000 dollars?

Answer 1

Answer:

a) 0.004

b) 0.9609                                                

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 76503

Standard Deviation, σ = 8850

We are given that the distribution of annual salaries is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(salary greater than 100,000)

P(x > 100000)

$P( x > 100000) = P( z > \displaystyle\frac{100000 - 76503}{8850}) = P(z > 2.655)$

$= 1 - P(z \leq 2.655)$

Calculation the value from standard normal z table, we have,  

$P(x > 100000) = 1 - 0.996 = 0.004  = 0.4\%$

b) P(sample mean annual salary falls between 70,000 and 80,000 dollars)

Sample size, n = 20

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{8850}{\sqrt{20}} = 1978.92$

$P(70000 \leq x \leq 80000) = P(\displaystyle\frac{70000 - 76503}{1978.92} \leq z \leq \displaystyle\frac{80000-76503}{1978.92})\\\\ = P(-3.286 \leq z \leq 1.767)\\\\= P(z \leq 1.767) - P(z < -3.286)\\=0.9614 - 0.0005= 0.9609 = 96.09\%$

$P(70000 \leq x \leq 80000) = 96.09\%$