~How to work with different normal distributions that are not the Standard Normal Curve.

~Procedure: Translate the given normal distribution into a standard normal distribution by finding the appropriate z scores using the equation
  z = (x-m)/s

~Note: Making a rough sketch of the normal distribution and its corresponding standard normal curve equivalent will help in your understanding. Just remember, that the mean & standard deviation
of the normal curve in question converts to a zero z score (mean) &
a z score of +1 (standard deviation) for the standard normal curve.

~So, the area (probability) between 2 X scores for the given normal curve will convert to the area between 2 Z scores for the standard normal curve. Use the equation that ties them, given above.

~Ex: Find the probability a score is greater than 7 for a normal distribution whose mean=5 & standard deviation=2.

~Use the equation that ties the z score with the x score to get the z values associated with the standard normal curve.

( i.e., an x score of 7, gives z=(7-5)/2 = 2/2 = 1), so, want P(z>1). To get this, use your TI-83 with menu, 2nd, VARS, go to 2 (normalcdf, enter (1, 10000), enter to get .1587

~For the same distribution in the above example, find the probability that a score lies between -0.2 & 1.8.

~First , find their z scores:   z = (-0.2-5)/2 = -2.6,  z = (1.8-5)/2 = -1.6, then use the same menu to enter these,  normal cdf(-2.6,-1.6), enter, to get .0501. (make sure you use the (-) on the bottom of your TI-83)

~Note: Better yet, avoid finding z scores by entering, (-0.2, 1.8, 5, 2)

~Ex: IQ scores have a normal distribution with mean of 100 and standard deviation of 15. Find the 80th percentile score.
(i.e., the score separating the bottom 80% from the top 20%)

~Procedure:  This problem translates into finding the score for which 80% of all scores are below it. Which means, the area (probability) up to that score is .80. So, this is a reverse problem. We need to find the z score (using the standard normal curve) for that probability then convert it to an x score, using the equation which relates the two scores.

~Note: Use menu, 2nd VARS, go to 3 (invNorm), insert the probability (area) of .80, enter to get z=.8416 Now, convert this z score to an x score & you will be finished:  .8416 = (x-100)/15

First, multiply both sides by 15, to get, 12.624 = x-100, then add 100 to both sides, to get x=112.6

~Note: You can avoid most of this work by simply entering
             (.80, 100, 15) into menu 3.