The Normal Model and Z-Scores

The normal model

• They provide a measure for how extreme a z-score is

• For every combination of mean and standard deviation, there exists a normal model

The PARAMETERS of the normal model

Parameter for “center”NOT a “statistic”

Parameter for “standard deviation”NOT a “statistic”

Together, these define the center and spread of the model

Standardizing the normal model

• When we standardize a Normally distributed variable, we write:

Standard Normal

• Its easier to standardize the data first, then we only need to model N(0,1)

• This is called the Standard normal model

• Be careful, you cant use the standard normal model on just any data set…

The Empirical Rule

Percentile Ranks:

50th 84th 97.5th 99.85th 16th 2.5th 0.15th

In other words …

Test your understanding:

• What is the purpose of a z-score?

• What does it mean when a curve is normal?

The Formal Normal

• Symmetric, single-peaked (unimodal) and bell-shaped.

• The mean is at the center of the curve, and is the same as the median.

• The normal curve describes the normal distribution N(0,1)

Why is Normal Important?

The Empirical Rule

• For a normal distribution, nearly all values lie within 3 standard deviations of the mean.

• About 68.27% of the values lie within 1 standard deviation of the mean. Similarly, about 95.45% of the values lie within 2 standard deviations of the mean. Nearly all (99.73%) of the values lie within 3 standard deviations of the mean.

Slide 6- 12

The Normal Model …

• Be careful—don’t use a Normal model for just any data set, since standardizing does not change the shape of the distribution.

• Example: If data is heavily skewed, we can’t use the Normal model for it.

Nearly normal..

• Real world data never acts like a mathematical model…

• As long as the data is Nearly Normal we can still get valuable info from the model

• We will be talking about how to determine if something is “normal” in the future.

Example

• Suppose SAT math scores are nearly normally distributed with a mean of 550 and a standard deviation of 30

ExampleSuppose you score a 600, how many people did you do better than?

You are finding the area to the left of 600

Example

• To calculate this, we will need to calculate a z-score and use the “z-table”

Example

• =550, = 30, y = 600 z=(600-550)/30 = 1.67

Area to left of 1.67 is 0.9525 thus you are in the 95th percentile

Example• You can also go the other way… meaning if you want to

know what the 75th percentile is, you :• Look up where 0.7500 is on the z-table, then find the

z-score by finding the column and row header..

Between z=0.67 and z=0.68 so lets call it z=0.675

So…

z =(y- )/ 0.675 = (y-550)/30 <--- solve for y y = (0.675*30) + 550 y = 570.25

Therefore the 75th percentile is a score of 570.25. NOTE: The area under the standard normal curve,

below a certain z-value is commonly written as:p(z<0.675) = 0.75

Slide 6- 20

Finding Normal Percentiles by Hand (cont.)

• Table A is the standard Normal table. We have to convert our data to z-scores before using the table.

• Figure 6.5 shows us how to find the area to the left when we have a z-score of 1.80:

Slide 6- 21

From Percentiles to Scores: z in Reverse

• Sometimes we start with areas and need to find the corresponding z-score or even the original data value.

• Example: What z-score represents the first quartile in a Normal model?

Slide 6- 22

From Percentiles to Scores: z in Reverse (cont.)

• Look in Table A for an area of 0.2500.• The exact area is not there, but 0.2514 is pretty

close.

• This figure is associated with z = -0.67, so the first quartile is 0.67 standard deviations below the mean.

An ExampleA forester measured 27 of the trees in a large woods that is up for sale. He found a mean diameter of 10.4 inches and a standard deviation of 4.7 inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies.

What size would you expect the central 95% of all trees to be?

About what percent of the trees should be less than an inch in diameter?

About what percent of the trees should be between 5.7 and 10.4 inches in diameter?

About what percent of the trees should be over 15 inches in diameter?