If x is an observation from a roughly symmetric distribution that has mean and standard deviation , then the standard value of x is :

z = x -

Note: A standardized score is often called a z-score.

Example : Men’s IQ’s have a roughly symmetric distribution with amean of 72 and a standard deviation of 8.

What is the standard score for a man with an IQ of 66 ?

z =66 - 72

8=

-6

8= - .75

The Standard Normal Distribution

Q: What percentage of people have a score below 66 ?

The Standard Normal TableTable A is a table of areas under the standard normal curve. Thetable entry for each z value is the area under the curve to the left of z

.1357

The Standard Normal TableExample : Imagine we have done an experiment, and we want to findwhat percentage of people fell under a score, namely x.

We then proceed to find that the z-score for the value x is -1.10.

The Standard Normal TableExample : The Graduate Record Examinations (GRE) are widelyused to help predict the performance of applicants to graduate schools.The range of possible sores on a GRE is 200 to 900. The psychologydepartment at a university finds the scores of its applicants on thequantitative GRE are approximately normal with mean = 544 andstandard deviation = 103. Answer the following :

1) Find the percentage of people who scored 700 or higher on the test.

2) Find the percentage of people who scored below 500 on the test.

3) Find the percentage of people who scored between 500 and 800 on the test.

1) Find the percentage of people who scored 700 or higher on the test.

Find the percentage to the right of the 700 marker.

1) Find the percentage of people who scored 700 or higher on the test.

Find the z-score : z =700 - 544

103=

156

103= 1.51

Answer: 1 - .9345 = .0655

.9345

.0655

2) Find the percentage of people who scored below 500 on the test.

Find the percentage to the left of 500

Find the z-score : z =500 - 544

103=

- 44

103= - 0.43

2) Find the percentage of people who scored below 500 on the test.

0.3336

Answer : 0.3336

3) Find the percentage of people who scored between 500 and 800 on the test.

Find the percentage between 500 and 800

3) Find the percentage of people who scored between 500 and 800 on the test.

Find the first z-score : z =500 - 544

103=

- 44

103= - 0.43

Find the second z-score : z =800 - 544

103=

256

103= 2.49

0.3336

0.9936Area =

.9936 - .3336 =

0.66

How To Tell If The Data Is “Normal”

1) Line up the data from least to greatest.

2) Determine the mean and standard deviation.

3) If the 68 - 95 - 99.7 rule is close, then the data can be considered normal.

How To Tell If The Data Is “Normal”Example : 25 people were chosen at random and were asked their age.Is this data set roughly normal?

39 41 47 58 65 37 37 49 56 59 62 36 4852 64 29 44 47 49 52 53 54 72 50 50

• The mean is 50 and the standard deviation is 10.• If this is normal, then 68 % will be within one standard deviation. That means 68 % of the ages will be between 40 and 60.

• The number of ages between 40 and 60 is : 16 ( This is 64 % )

• If this is normal, then 95 % will be within two standard deviations. That means 95 % of the ages will be between 30 and 70.

• The number of ages between 30 and 70 is : 23 ( This is 92 % )

• If this is normal, then 99.7 % will be within three standard deviations. That means 99.7 % of the ages will be between 20 and 80.• The number of ages between 20 and 80 is : 25 ( This is 100 % )

Homework92, 94, 98, 100, 104, 107