SOLUTION: Scores on a biology final exam are normally distributed with a mean of 220 and a standard deviation of 16 Determine the percentage of samples of size 4 that will have mean scores w

Question 1182261: Scores on a biology final exam are normally distributed with a mean of 220 and a standard deviation of 16 Determine the percentage of samples of size 4 that will have mean scores within 12 points of the population mean score of 220.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

mu = population mean = 220
sigma = population standard deviation = 16
n = 4 = sample size

Compute the z score for x = 220-12 = 208
z = (x - mu)/(sigma/sqrt(n))
z = (208 - 220)/(16/sqrt(4))
z = -1.5

Repeat for x = 220+12 = 232
z = (x - mu)/(sigma/sqrt(n))
z = (232 - 220)/(16/sqrt(4))
z = 1.5

Computing P(208 < x < 232) is the same as P(-1.5 < z < 1.5)

Use a z table such as this one here
https://www.ztable.net/
(or similar found in the back of your textbook) to determine the following
P(z < -1.5) = 0.06681
P(z < 1.5) = 0.93319

We can now use this formula
P(a < z < b) = P(z < b) - P(z < a)
to get...

P(a < z < b) = P(z < b) - P(z < a)
P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5)
P(-1.5 < z < 1.5) = 0.93319 - 0.06681
P(-1.5 < z < 1.5) = 0.86638
You can use a calculator to get more accuracy.

Roughly 86.6% of the samples (of size 4) will have mean scores within 12 points of the population mean 220.

--------------------------------------------

Here's a thought experiment (to either try out or just think about).

Let's say you were to randomly select a group of 4 people, and compute the sample mean, then that's one xbar value. Repeat this process as many times as you like to get various other xbar values. You most likely won't select the same four people each time.

This produces the xbar distrbution. It's normal since the parent distribution is normal. This xbar distribution has mean of mu = 220 and standard deviation sigma/sqrt(n) = 16/sqrt(4) = 8.

Now after you computed many many xbar values, the question is: how many of those xbar values are within 12 points of 220? In other words, how many are between 220-12 = 208 and 220+12 = 232? You could simply look at all the data you collected, and do a comparison. Add 1 to the tally for any value that is in the range specified.

If you have say 1000 xbar values, then you should get around 886 values that fit the interval 208 < xbar < 232. Of course, due to the random nature of these types of problems, it likely won't be exact. But it should be close enough. The 886 is from 86.6% of 1000.

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