Question 1204966: Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 169000 dollars. Assume the standard deviation is 36000 dollars. Suppose you take a simple random sample of 91 graduates. Assume the population is normally distributed.
Note: You should carefully round any intermediate values you calculate to 4 decimal places to match wamap's approach and calculations.
Find the probability that a single randomly selected salary exceeds 166000 dollars.
P(X > 166000) = 0.2023
Enter your answer as a number accurate to 4 decimal places.
Find the probability that a sample of size 91 is randomly selected with a mean that exceeds 166000 dollars.
P(x > 166000) =
Enter your answer as a number accurate to 4 decimal places.
I did the first part but can't find P(xbar > 166000) with sample of 91
part but can not figure out the bottom because when I do it I get a number that does not / can not exist for this problem. please help
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part 1
mu = 169000 = mean
sigma = 36000 = population standard deviation.
x = 166000
z = (x - mu)/sigma
z = (166000 - 169000)/36000
z = -0.08333333333333 approximately
z = -0.08
Then use Z table to find that P(z < -0.08) = 0.46812 approximately.
Such tables should be found at the back of your stats textbook.
For exams, the teacher will likely hand out the table.
Then,
P(Z > -0.08) = 1-P(Z < -0.08)
P(Z > -0.08) = 1-0.46812
P(Z > -0.08) = 0.53188
This leads back to P(x > 166000) = 0.53188 approximately
Rounding to four decimal places gets us 0.5319
I'm not familiar with wamap, so I don't have access to that particular stats calculator.
However, there are many other available calculators out there as alternatives.
For example, using the TI83 or TI84 gives
normalCDF(-0.08, 99, 0, 1) = 0.531 881 440 4
I put a space between each group of 3 decimal digits to make the number a bit more readable.
That value rounds to 0.5319 when rounding to four decimal places.
Or another TI83 calculation we could do is
normalCDF(-0.0833, 99, 0, 1) = 0.533 205 493 9
That value rounds to 0.5332 when rounding to four decimal places.
Or another TI83 calculation we could do is
normalCDF(-0.083333, 99, 0, 1) = 0.533 206 686 5
That value rounds to 0.5332 when rounding to four decimal places.
It appears that the results steadily approach the value 0.5332 as the z score gets closer to -0.0833333 where the '3's go on forever.
I'm not sure how you came up with 0.2023
Please let me know your scratch work and thought process. Thank you.
Here are some alternative calculators (for future students reading this question who do not have a TI83/TI84 and also don't have access to wamap)- A very user friendly calculator by professor David M Lane. The calculator also displays the shaded diagram which is a nice bonus.
- Use WolframAlpha. Type in P(Z > -0.0833) for instance. Make sure the "referring to statistics" option is selected.
- Use GeoGebra. It can be accessed through either the normal input bar, the CAS mode, or the probability distribution mode. The reference page can be found here
- Use the normDist function on a spreadsheet. See this reference page for more info.
There are many other alternatives that I haven't listed. You can do an internet search to find your favorite one.
Of all of the choices, the 1st option (the David M Lane one) is probably the best for new students.
The only drawback is there doesn't appear to be a way to change the precision.
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Part 2
mu = 169000 = mean
sigma = 36000 = population standard deviation.
xbar = 166000 = sample mean
n = 91 = sample size
xbar means we have x with a horizontal line over top
When working in the xbar distribution, aka sample mean distribution, we have this z score conversion formula:
z = (xbar - mu)/( sigma/sqrt(n) )
So,
z = (xbar - mu)/( sigma/sqrt(n) )
z = (166000 - 169000)/( 36000/sqrt(91) )
z = -0.79494933451412
z = -0.79
The result is approximate.
Use a table to find that
P(Z < -0.79) = 0.21476
which leads to
P(Z > -0.79) = 1-P(Z < -0.79)
P(Z > -0.79) = 1-0.21476
P(Z > -0.79) = 0.78524
P(Z > -0.79) = 0.7852
and ultimately leads back to
P(xbar > 166000) = 0.7852
If using a TI83, then
normalCDF(-0.7949, 99, 0, 1) = 0.786 664 228 9
and
normalCDF(-0.7949493, 99, 0, 1) = 0.786 678 568 6
Play around with the precision and you'll see the more accurate the input, the results tend to approach 0.7867
It's not too far from the table result 0.7852
Similar calculator results should happen with other calculators.
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