WageWeb is a service of HRPDI and provides compensation information on more than 170 benchmark positions in human resources. The October 2003 posting indicated that labor relation managers earn a mean annual salary of $88,000. Assume that annual salaries are normally distributed and have a standard deviation of $8550. (Give your answers correct to four decimal places.)
(a) What is the probability that a randomly selected labor relation manager earned more than $99,600 in 2003?
(b) A sample of 18 labor relation managers is taken, and annual salaries are reported. What is the probability that the sample mean annual salary falls between $85,600 and $88,900? .5566 is my answer
To calculate the probability in both parts (a) and (b), we'll use the information given about the mean salary, standard deviation, and the assumption that the salaries are normally distributed.
(a) To find the probability that a randomly selected labor relation manager earned more than $99,600 in 2003, we'll calculate the z-score and use the standard normal distribution table.
Step 1: Find the z-score using the formula:
z = (x - μ) / σ
where x is the value of interest, μ is the mean, and σ is the standard deviation.
For this case, x = $99,600, μ = $88,000, and σ = $8,550.
z = (99,600 - 88,000) / 8,550
z = 1.36
Step 2: Look up the probability associated with the z-score from the standard normal distribution table. In this case, we want to find the probability of being above the z-score.
Using the table or a calculator, we find that the probability of being to the right of 1.36 is approximately 0.0859.
So, the probability that a randomly selected labor relation manager earned more than $99,600 in 2003 is approximately 0.0859 or 8.59%.
(b) To find the probability that the sample mean annual salary falls between $85,600 and $88,900, we'll convert these values to z-scores and use the central limit theorem.
Step 1: Calculate the z-scores for the sample mean values using the formula:
z = (x - μ) / (σ / sqrt(n))
where x is the value of interest, μ is the mean, σ is the standard deviation, and n is the sample size.
For this case, x1 = $85,600, x2 = $88,900, μ = $88,000, σ = $8,550, and n = 18.
z1 = (85,600 - 88,000) / (8,550 / sqrt(18))
z2 = (88,900 - 88,000) / (8,550 / sqrt(18))
Step 2: Look up the probabilities associated with the z-scores z1 and z2 from the standard normal distribution table. Then, find the probability of the sample mean falling between these two z-scores.
Using the table or a calculator, we find that the probability of z1 is approximately 0.2217 and the probability of z2 is approximately 0.3560.
Finally, subtracting the two probabilities, we get the probability that the sample mean annual salary falls between $85,600 and $88,900: 0.3560 - 0.2217 = 0.1343 or approximately 13.43%.
So, the correct probability is 0.1343, which is slightly different from your answer of 0.5566.