using a mean of 80 and a standard deviation of 12. a) what is the score that divides the distribution such that 99% of the area is below it. b) what are the scores that bound the middle 95% of the distribution

I find this webtool perfect for your kind of question

http://davidmlane.com/hyperstat/z_table.html

use the 2nd applet,
enter 80 in the mean
enter 12 in the sd
and .99 in the "shaded area" box

click on "below"

To answer these questions, we need to use the concept of the standard normal distribution, which follows a bell-shaped curve with a mean of 0 and a standard deviation of 1. However, since we are given a mean of 80 and a standard deviation of 12, we need to convert our values to match the standard normal distribution.

a) To find the score that divides the distribution such that 99% of the area is below it, we can use a z-score. The z-score represents the number of standard deviations a given value is from the mean.

To find the z-score, we can use the formula:
z = (x - μ) / σ

Where:
x = the score
μ = the mean
σ = the standard deviation

Since we want to find the score where 99% of the area is below it, we need to find the z-score that corresponds to the cumulative probability of 0.99. This can be looked up in a standard normal distribution table or calculated using statistical software.

Using the standard normal distribution table, we find that the z-score for a cumulative probability of 0.99 is approximately 2.33.

Now we can substitute the values into the formula to solve for x:
2.33 = (x - 80) / 12

Rearranging the formula to solve for x:
x - 80 = 2.33 * 12
x - 80 = 27.96
x = 27.96 + 80
x = 107.96

Therefore, the score that divides the distribution such that 99% of the area is below it is approximately 107.96.

b) To find the scores that bound the middle 95% of the distribution, we can use a similar approach. We need to find the z-scores that correspond to the cumulative probabilities of the lower and upper bounds of the middle 95%, which are (1 - 0.95) / 2 = 0.025 and 1 - (1 - 0.95) / 2 = 0.975, respectively.

Using the standard normal distribution table, we find that the z-scores for a cumulative probability of 0.025 and 0.975 are approximately -1.96 and 1.96.

Substituting these values into the formula:
-1.96 = (x - 80) / 12
1.96 = (x - 80) / 12

Solving for x in both equations:
-1.96 * 12 = x - 80
1.96 * 12 = x - 80

-23.52 = x - 80
23.52 = x - 80

x1 = -23.52 + 80
x2 = 23.52 + 80

x1 = 56.48
x2 = 103.52

Therefore, the scores that bound the middle 95% of the distribution are approximately 56.48 and 103.52.