## Question 559:

1

So we need to find the area in between two points in a normal distribution. To do so we need to convert this normal distribution in to the standard normal distribution so we can make some inferences about the probability of events.  First thing we need to do is find the z-score associated with 8% of the area above and below a point in this distribution.

We can use the percentile to z-score calculator to find our z-score. The tricky thing about this question is we have to think about what we're looking up. We want the middle area of the bell-curve so there is 8% of the area above and 8% of the area below our area of interest. That means a total area of 8+8=16% above and below total and 100-16=84% of the area in the middle (told you it was tricky). So we need to enter .84 in the percentile to z-score calculator and select 2-sided area. This choice will distribute the area above and below. We get a 2-sided z-score of 1.4053. So now we need to setup a simple equation to solve for the unknown value which gets us the z-score of 1.4053. This process is basically equating a normal distribution with mean of 185.1 and standard deviation of 37.8 to a mean of 0 and standard deviation of 1.

1. Ok, so using the formula for a z-score, we subtract the mean from a value and divide that results by the standard deviation= (x-185.1)/37.8 = 1.4053.
2. Now we continue with a little algebra. x-185.1= 53.12034
3. x = 238.2203
4. Since we need two values, which are symmetrically opposite on the curve, we also need to find the value for the z-score of -1.4053 (notice the negative sign).
5. (x-185.1)/37.8 = -1.4053.
6. x-185.1= -53.12034
7. x = 131.9797

The level that separates the bottom 8% is then 131.97 and the level that separates the top 8% is 238.2203 mg/100ml.    We can confirm our results visually by using the interactive graph of the standard normal curve and entering the mean of 185.1 and 37.8 in the 1st graph. Then hover your mouse over around where the x-value says 238. You should see "White area: 16% which would be the area in both tails of the curve (8 above and below) thus confirming our answers.