Question 1191672: Question 10 options:
The average teacher's salary in a certain state is $37,764. Assume a normal distribution with σ = $5100.
a) What is the probability that a randomly selected teacher's salary is greater than $46,800?
For a sample of 75 teachers, what is the probability that the sample mean is greater than $38,900?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the population mean = 37,764.
the population standard deviation = 5,100.
the z-score formula is z = (x - m) / s
z is the z-score
x is the sample score.
m is the mean
s is the standard deviation if you are looking at a sample element.
s is the standard error if you are looking at the mean of a sample of a number of sample elements.
your questions are answered below.
i assigned b) to the second question, since you assigned a) to the first question.
a) What is the probability that a randomly selected teacher's salary is greater than $46,800?
here, you are dealing with a sample element, so the standard deviation is used.
the z-score formula becomes:
z = (46,800 - 37,764) / 5,100.
solve for z to get z = 1.771764706.
using the ti-84 plus calculator, i get the area to the right of that z-score equals .0382167637.
if you are dealing with a z-score rounded to 2 decimal places and an area to the right of that z-score rounded to 5 decimal places, then your z-score would be equal to 1.77 and you would get the area to the right of that z-score equal to .03836.
here's the table that i used.
it gives you the area to the left.
the area to the right is equal to 1 minus that.
the area to the left was .96164 and the area to the right was 1 minus that = .03836.
https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf
b) For a sample of 75 teachers, what is the probability that the sample mean is greater than $38,900?
since you are dealing with the mean of a sample of 75 teachers, you would use the standard error, rather than the standard deviation.
in that case, s = standard deviation divided by square root of sample size.
that would be 5,100 / sqrt(75) = 588.8972746.
the z-score would be equal to (38,900 - 37,764) / 588.8972746 = 1.929029135.
using the ti-84 plus calculator, the area to the right of that would be equal to .0268635538.
if you were using the z-score rounded to 2 decimal places, and you were using the z-score table referenced above, then you would get the area to the left of a z-score of 1.93 that is equal to .97320 which would lead to an area to the right of that z-score equal to 1 - .97320 = .02680.
i confirmed with the ti-84 plus that the z-score table results were accurate.
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