Salaries for various positions can vary significantly depending on whether or not the company is in the public or private sector. The U.S. department of labor posted the 2007 mean salary for human resource managers employed by the federal government, as &76, 503. Assume the annual salaries for this type of job are normally distributed and have a standard deviation of $8850.

What is the probability that a randomly selected human resource manager received over $100,000 in 2007?

A sample of 20 human resource managers is taken and annual salaries are reported. What is the probability that the sample mean annual salary falls between $70,000 and $80,000

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for that Z score.

For the sample mean,

Z = (score-mean)/SEm

SEm = SD/√n

Use same table for the Z scores.

Thank you! I was able to complete the first question. Is SE Standard Error? and m=mean? Thank you again

Yes and Yes.

What do you mean by score?

To calculate the probability in these scenarios, we will use the z-score formula and the standard normal distribution table.

1. Probability of a human resource manager receiving over $100,000:

First, we need to standardize the value of $100,000 using the z-score formula:
z = (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.

Given:
x = $100,000
μ = $76,503
σ = $8850

z = ($100,000 - $76,503) / $8850
z = 2.78

Using the standard normal distribution table or a calculator, we can find that the area to the left of z = 2.78 is approximately 0.9973.
To find the probability of receiving over $100,000, we subtract this value from 1:
P(X > $100,000) = 1 - 0.9973 = 0.0027

Therefore, the probability that a randomly selected human resource manager received over $100,000 in 2007 is approximately 0.0027.

2. Probability that the sample mean annual salary falls between $70,000 and $80,000:

To calculate this probability, we need to calculate the z-scores for both values using the sample mean and the standard deviation of the population.

Given:
Sample size (n) = 20
Sample mean (x̄) = ($70,000 + $80,000) / 2 = $75,000
Standard deviation (σ) = $8850 (same as before)

To calculate the standard error (SE), which measures the variability of the sample mean:
SE = σ / √n
SE = $8850 / √20 ≈ $1977.05

Now, we need to find the z-scores for $70,000 and $80,000 using the sample mean and the standard error:
z1 = ($70,000 - $75,000) / $1977.05 ≈ -2.52
z2 = ($80,000 - $75,000) / $1977.05 ≈ 2.52

Using the standard normal distribution table or a calculator, we can find the area to the left of z = -2.52 and z = 2.52. Let's assume it is approximately 0.0062 for z = -2.52 and 0.9938 for z = 2.52.

To find the probability that the sample mean annual salary falls between $70,000 and $80,000, we subtract the areas:
P($70,000 < X < $80,000) = 0.9938 - 0.0062 = 0.9876

Therefore, the probability that the sample mean annual salary falls between $70,000 and $80,000 is approximately 0.9876.