SOLUTION: Suppose the number of students in a class for the Business Statistics program at a University has a mean of 23 with a standard deviation of 4.3. If 15 classes are selected randomly

Question 1151047: Suppose the number of students in a class for the Business Statistics program at a University has a mean of 23 with a standard deviation of 4.3. If 15 classes are selected randomly, find the probability that the mean number of students is between 20 and 30.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

= mu = greek letter representing the population mean
= sigma = greek letter representing the population standard deviation
n = sample size

= xbar = sample mean

In this case,

We're given two xbar values which are 20 and 30

Compute the z score for the raw score xbar = 20

Repeat for xbar = 30

We have

Use a Z table such as this one
http://www.z-table.com/
to find that,
P(Z < -2.70) = 0.0035
P(Z < 6.30) = 1.00
note: if k is larger than 3.4, then P(Z < k) will be very very close to 1.00; this is especially true of z = 6.30 as its very distant from the center z = 0. So that is how I got P(Z < 6.30) = 1.00

Subtract the values to find the area between the z scores
P(A < Z < B) = P(Z < B) - P(Z < A)
P(-2.70 < Z < 6.30) = P(Z < 6.30) - P(Z < -2.70)
P(-2.70 < Z < 6.30) = 1 - 0.0035
P(-2.70 < Z < 6.30) = 0.9965

Answer: 0.9965
This answer is approximate.

side note:
You can use a calculator like this one
http://davidmlane.com/hyperstat/z_table.html
to compute the area under the Z curve. Leave the mean and standard deviation as 0 and 1 respectively. Click the "between" radio button and type in the values -2.70 and 6.30 into the boxes. Then hit "recalculate" to have the answer come up. This can also be done on TI83 and TI84 calculators as well using the normalcdf function.

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