The Normal Distribution. The Distribution The Standard Normal Distribution We simply transform all X values to have a mean = 0 and a standard deviation.

The Normal Distribution

The DistributionThe Distribution

X

3.503.00

2.502.00

1.501.00

.500.00

-.50-1.00

-1.50-2.00

-2.50-3.00

-3.50-4.00

1200

1000

800

600

400

200

0

Std. Dev = 1.00

Mean = -.01

N = 10000.00

The Standard Normal The Standard Normal DistributionDistribution

• We simply transform all We simply transform all XX values to values to have a mean = 0 and a standard have a mean = 0 and a standard deviation = 1deviation = 1

• Call these new values Call these new values zz

• Define the area under the curve to Define the area under the curve to be 1.0be 1.0

zz Scores Scores

• Calculation of Calculation of zz

where where is the mean of the population and is the mean of the population and

is its standard deviation is its standard deviation

This is a simple This is a simple linear transformationlinear transformation of of XX..

σμX

z

Tables of Tables of zz

• We use tables to find areas under We use tables to find areas under the distributionthe distribution

• A sample table is on the next slideA sample table is on the next slide

• The following slide illustrates areas The following slide illustrates areas under the distributionunder the distribution

z

3.503.00

2.502.00

1.501.00

.500.00

-.50-1.00

-1.50-2.00

-2.50-3.00

-3.50-4.00

Normal Distribution

Cutoff at +1.6451200

1000

800

600

400

200

0

z = 1.6454

545

Area = .05 .05

Using the TablesUsing the Tables

• Define “larger” versus “smaller” Define “larger” versus “smaller” portionportion

• Distribution is symmetrical, so we Distribution is symmetrical, so we don’t need negative values of don’t need negative values of zz

• Areas between Areas between zz = +1.5 and = +1.5 and zz = -1.0 = -1.0 See next slideSee next slide

Calculating areasCalculating areas

• Area between mean and +1.5 = 0.4332Area between mean and +1.5 = 0.4332

• Area between mean and -1.0 = Area between mean and -1.0 = 0.34130.3413

• Sum equals 0.7745Sum equals 0.7745

• Therefore about 77% of the Therefore about 77% of the observations would be expected to fall observations would be expected to fall between between zz = = -1.0 and -1.0 and zz = +1.5 = +1.5

Converting Back to Converting Back to XX• Assume Assume = 30 and = 30 and = 5 = 5

• 77% of the distribution is expected 77% of the distribution is expected to lie between 25 and 37.5to lie between 25 and 37.5

5.3755.130

2550.130

Therefore

X

X

zX

Xz

Probable LimitsProbable Limits

• XX = = + + zz

• Our last example has Our last example has = 30 and = 30 and = 5= 5

• We want to cut off 2.5% in each tail, We want to cut off 2.5% in each tail, soso zz = = ++ 1.96 1.96

2.20596.1308.39596.130

XX

zX

Cont.

Probable LimitsProbable Limits--cont.--cont.

• We have just shown that 95% of the We have just shown that 95% of the normal distribution lies between normal distribution lies between 20.2 and 39.820.2 and 39.8

• Therefore the probability is .95 that Therefore the probability is .95 that an observation drawn at random will an observation drawn at random will lie between those two valueslie between those two values

Measures Related to Measures Related to zz

• Standard scoreStandard score Another name for a Another name for a zz score score

• Percentile scorePercentile score The point below which a specified percentage The point below which a specified percentage

of the observations fallof the observations fall

• TT scores scores Scores with a mean of 50 and a standard Scores with a mean of 50 and a standard

deviation of 10deviation of 10

Cont.