# Normal Distribution

1.) A normal distribution has a mean=60 and a standard deviation=6. Find the following probabilities:

p(X>56)

p(X<70)

p(52<X<68)

p(61<X<75)

2.) A normal distribution has a mean=35 and standard deviation=5. Find the scores associated with the following regions:

a.) the score needed to be in the 43rd percentile.

b.) The score needed to be in the top 15% of the distribution.

c.) the scores that rank off the middle 50% of the distribution.

#### Solution Preview

1.) A normal distribution has a mean=60 and a standard deviation=6. Find the following probabilities:

We're going to use z-scores to figure this out. The z-score for a given number x is defined as:

z = (x - mean)/standard deviation

For each of the numbers below, we're going to calculate z, then look at a z distribution table to see what percentage of the area of the z-distribution is below or above the z-score. That area will be the probability that we're looking for.

p(X>56)

z = (56 - 60)/6 = -0.667

Look at the z-distribution table here: http://www.statsoft.com/textbook/sttable.html#z. (If you want to use a distribution table from your textbook, go ahead, just follow the same steps below.) Find the p-value corresponding to z = 0.67. It is 0.2486. That means that the area under the curve between z = 0 and z = 0.67 is 0.2486. It also means that the area under the curve between z = 0 and z = -0.67 is 0.2486.

We are interested in the area under the curve above z = -0.67. We can find that by adding

the area under the curve between z = 0 and z = -0.67 (0.2486) to the area ...

#### Solution Summary

The solution includes answers and step-by-step explanations of how to solve the problems, including how to calculate a z-score and how to find a probability using a z distribution table.