SOLUTION: Solve the problem. The scores on a certain test are normally distributed with a mean score of 66 and a standard deviation of 3. What is the probability that a sample of 90 stude

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Question 1168751: Solve the problem.
The scores on a certain test are normally distributed with a mean score of 66 and a standard deviation of 3. What is the probability that a sample of 90 students will have a mean score of at least 66.3162?
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Given info:
xbar = sample mean = 66.3162
mu = population mean = 66
sigma = population standard deviation = 3
n = sample size = 90
The goal is to calculate P(xbar > 66.3162)

Calculate the standard error (SE)
SE = sigma/sqrt(n)
SE = 3/sqrt(90)
SE = 0.31622776601683

Use that to find the z score
z = (xbar - mu)/SE
z = (66.3162-66)/0.31622776601683
z = 0.9999121961453
z = 1.00

The problem of finding P(xbar > 66.3162) is approximately the same as wanting to find P(Z > 1.00)

At this point, you'll need a table or calculator.
If you go with a table, then you can use a table in the back of your book or you can use a free online resource like this
http://www.z-table.com/

That website shows a Z table
In that table, we see that P(Z < 1.00) = 0.8413
Look in the row that starts with 1.0 and the column that starts with 0 to find the value 0.8413
This represents the area under the curve to the left of the z value z = 1.00

Subtract the value from 1 to get the area under the curve to the right of the z value in question
1-0.8413 = 0.1587

The final answer is 0.1587
This means there's roughly a 15.87% chance that xbar is larger than 66.3162
Use a calculator to get better accuracy.