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 cant get a part though .
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 related to the Z score.
thank you, I was using wrong formula but I got it right with your formula.
To calculate the probability in both parts of the question, we will use the z-score formula.
The z-score formula is given as follows:
z = (x - μ) / σ
where:
- x is the value you are interested in (e.g., salary),
- μ is the mean of the population (given as $88,000),
- σ is the standard deviation of the population (given as $8,550).
(a) To find the probability that a randomly selected labor relation manager earned more than $99,600 in 2003, we need to calculate the z-score for $99,600 and find the area under the normal distribution curve above that z-score.
z = (99,600 - 88,000) / 8,550
z = 1.36 (rounded to two decimal places)
Next, we need to find the area to the right of the z-score of 1.36 (which represents earning more than $99,600).
Using a standard normal distribution table or calculator, the probability corresponding to a z-score of 1.36 is approximately 0.0869.
So, the probability that a randomly selected labor relation manager earned more than $99,600 in 2003 is approximately 0.0869 or 8.69% (rounded to two decimal places).
(b) To find the probability that the sample mean annual salary falls between $85,600 and $88,900, we need to calculate the z-scores for both values and find the area under the normal distribution curve between those z-scores.
First, let's calculate the z-score for $85,600:
z1 = (85,600 - 88,000) / 8,550
z1 = -0.2815 (rounded to four decimal places)
Next, let's calculate the z-score for $88,900:
z2 = (88,900 - 88,000) / 8,550
z2 = 0.1058 (rounded to four decimal places)
Now, we need to find the area between these two z-scores.
Using a standard normal distribution table or calculator, we can find the probability associated with z1 (-0.2815) and z2 (0.1058) and subtract the smaller probability from the larger probability to get the desired result.
The probability corresponding to z1 (-0.2815) is approximately 0.3882, and the probability corresponding to z2 (0.1058) is approximately 0.5424.
So, the probability that the sample mean annual salary falls between $85,600 and $88,900 for a sample of 18 labor relation managers is approximately 0.5424 - 0.3882 = 0.1542 or 15.42% (rounded to two decimal places).