Standard Normal CalculationsStandard Normal Calculations

What you’ll learnWhat you’ll learn Properties of the standard normal distProperties of the standard normal distnn

How to transform scores into normal distHow to transform scores into normal distnn scoresscores

Determine the proportion of observations Determine the proportion of observations above, below and between two stated numbers above, below and between two stated numbers in a normal distribution.in a normal distribution.

Calculate the point for a variable with a normal Calculate the point for a variable with a normal distribution for which a stated proportion of distribution for which a stated proportion of values lie either above or below.values lie either above or below.

Comparing individuals from different Comparing individuals from different distributionsdistributions

Standard Normal DistributionStandard Normal Distribution

The Standard Normal Distribution The Standard Normal Distribution (also known as the “z-distribution”) (also known as the “z-distribution”)

N( 0, 1)N( 0, 1)

y

0.1

0.2

0.3

0.4

0.5

x-3 -2 -1 0 1 2 3

y = x normalDensity

no data Function Plot

Standardizing ScoresStandardizing Scores

We find that all normal distributions are the We find that all normal distributions are the same if we measure in units of same if we measure in units of σσ. .

We We

Using the Standard Normal Using the Standard Normal DistributionDistribution

The level of cholesterol in the blood is important The level of cholesterol in the blood is important because high cholesterol levels may increase the because high cholesterol levels may increase the risk of heart disease. We know that the risk of heart disease. We know that the distribution of blood cholesterol levels in a large distribution of blood cholesterol levels in a large population of people of the same age and sex is population of people of the same age and sex is roughly normal. For 14-year-old boys, the mean roughly normal. For 14-year-old boys, the mean is is μμ=170 mg/dl and the standard deviation, =170 mg/dl and the standard deviation, σσ=30m/dl. Levels above 240 mg/dl may require =30m/dl. Levels above 240 mg/dl may require medical attention.medical attention.

Steps to solving a “normal” distSteps to solving a “normal” distnn problem.problem.

Step 1: Step 1: – Write the question as a probability statement.Write the question as a probability statement.

Step 2:Step 2:– Calculate a z-scoreCalculate a z-score– Draw a picture and shade the regionDraw a picture and shade the region

Step 3:Step 3:– Find the appropriate region using a standard normal Find the appropriate region using a standard normal

tabletable Step 4:Step 4:

Write the answer in the context of the problemWrite the answer in the context of the problem

Question with Area Question with Area belowbelowWhat percent of 14-year-old boys have less than What percent of 14-year-old boys have less than

160 mg/dl of cholesterol?160 mg/dl of cholesterol? Step 1 (probability statement)Step 1 (probability statement)

– P(X< 160)P(X< 160)

Step 2: (z-score)Step 2: (z-score)

Since we want the percent of boys whose cholesterol is less than Since we want the percent of boys whose cholesterol is less than 160, we will find the percent of boys whose cholesterol -.33160, we will find the percent of boys whose cholesterol -.33σσ or or more more belowbelow the mean. the mean.

X

z

33.30

170160

z

y

0.1

0.2

0.3

0.4

0.5

x-3 -2 -1 0 1 2 3

y = x normalDensity = -0.33

no data Function Plot

Step 3: (Area from Table A)Step 3: (Area from Table A)We can now use Table A to find the percent of observations We can now use Table A to find the percent of observations belowbelow -0.33. (Remember that Table A always gives the area under the -0.33. (Remember that Table A always gives the area under the curve curve belowbelow a given value. a given value.

Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

-0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776

-0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121

-0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483

-0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859

-0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247

-0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

Step 3 (cont.)Step 3 (cont.) The area under the curve (the proportion of The area under the curve (the proportion of

observations) below -3.3observations) below -3.3σσ is .3707 is .3707

Step 4: (Context)Step 4: (Context)

The percent of 14-year-old boys whose The percent of 14-year-old boys whose cholesterol level is less than 160mg/dl is cholesterol level is less than 160mg/dl is approximately 37.07%approximately 37.07%

Question with Area Question with Area aboveaboveWhat percent of 14-year-od boys have more that What percent of 14-year-od boys have more that

240mg/dl of cholesterol?240mg/dl of cholesterol? Step 1 (probability statement)Step 1 (probability statement)

– P(X> 240)P(X> 240) Step 2: (z-score)Step 2: (z-score)

Since we want the percent of boys whose cholesterol is greater than 240, Since we want the percent of boys whose cholesterol is greater than 240, we will find the percent of boys whose cholesterol 2.33we will find the percent of boys whose cholesterol 2.33σσ or more or more aboveabove the the mean.mean.

X

z

33.230

170240

z

y

0.1

0.2

0.3

0.4

0.5

x-3 -2 -1 0 1 2 3

y = x normalDensity = 2.33

no data Function Plot

Step 3: (Area from Table A)Step 3: (Area from Table A)We can now use Table A to find the percent of observations below We can now use Table A to find the percent of observations below 2.33. (below because that’s what our table gives us)2.33. (below because that’s what our table gives us)

Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

Step 3: Area (continued)Step 3: Area (continued)– The value from the table is .9901. We need to The value from the table is .9901. We need to

remember that the table gives us area below a remember that the table gives us area below a value. Since the total area under the curve is 1, value. Since the total area under the curve is 1, to find the area to find the area aboveabove we can subtract the area we can subtract the area from the table from 1. So 1- .9901 = .0099from the table from 1. So 1- .9901 = .0099

Step 4: (context)Step 4: (context)– The percent of 14-year-old boys whose The percent of 14-year-old boys whose

cholesterol level is more than 240 mg/dl is cholesterol level is more than 240 mg/dl is approximately .99%.approximately .99%.

Question Question betweenbetween two values two valuesWhat percent of 14-year-old boys have cholesterol What percent of 14-year-old boys have cholesterol

levels between 170mg/dl and 240 mg/dllevels between 170mg/dl and 240 mg/dl Step 1 (probability statement)Step 1 (probability statement)

– P(170 < X < 240)P(170 < X < 240) Step 2: (z-scores, we need to find z-Step 2: (z-scores, we need to find z-

scores for both endpoints)scores for both endpoints)

X

z

X

z

030

170170

z

33.230

170240

z

y

0.1

0.20.3

0.40.5

x-3 -2 -1 0 1 2 3

y = x normalDensity = 2.33 = 0

no data Function Plot

Since we want the percent of boys whose cholesterol is between 170 Since we want the percent of boys whose cholesterol is between 170 mg/dl and 240mg/dl, we will find the percent of boys whose cholesterol is mg/dl and 240mg/dl, we will find the percent of boys whose cholesterol is between 0between 0σσ and 2.33 and 2.33σσ..

Step 3: (Area from Table A)Step 3: (Area from Table A)We can now use Table A to find the percent of observations below We can now use Table A to find the percent of observations below 2.33 and the area below z= 0.00 (below because that’s what our 2.33 and the area below z= 0.00 (below because that’s what our table gives us)table gives us)

Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

-0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859

-0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247

-0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

Step 3: Area (continued)Step 3: Area (continued)– The values from the table are .9901 for the z-score of The values from the table are .9901 for the z-score of

2.33 and .5000 for the z-score of 0. We need to 2.33 and .5000 for the z-score of 0. We need to remember that the table gives us area below a value. remember that the table gives us area below a value. We can take the area from 2.33 (.9901) and subtract We can take the area from 2.33 (.9901) and subtract the area from 0 (.5000) to get the area between.the area from 0 (.5000) to get the area between.

y

0.1

0.20.3

0.40.5

x-3 -2 -1 0 1 2 3

y = x normalDensity = 2.33 = 0

no data Function PlotSo:

.9901 - .5000 =

.4901

Step 4: Context---

The percent of 14-year-old boys whose cholesterol is between 170 and 240 is approximately 49.01%

Finding the value of the variable when we Finding the value of the variable when we know the percent above or belowknow the percent above or below

What cholesterol level do the top 10% of 14-What cholesterol level do the top 10% of 14-year-old boys have?year-old boys have?

y

0.1

0.2

0.3

0.4

0.5

x-3 -2 -1 0 1 2 3

y = x normalDensity

no data Function PlotStep 1: Write a probability statement

P ( X >x)= .10

This statement says: we want to find the value that separates the top 10% from the bottom 90% of the curve.

Since our table gives area below the curve, we will find a z-score that corresponds to 90% area

Step 2: Find the z-score from the table. Remember that the area is Step 2: Find the z-score from the table. Remember that the area is located on the “inside” of the table. Since the z-score that we are looking located on the “inside” of the table. Since the z-score that we are looking for is above the mean, we know the z-score will be positive. We’ll look for for is above the mean, we know the z-score will be positive. We’ll look for a value close to .9000.a value close to .9000.

Standard Normal Probability Distribution

Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

The closest value is .8997, so we will use a z-score of 1.28

Step 3: Using the z-score found, use the formula to standardize values Step 3: Using the z-score found, use the formula to standardize values substituting the three known values.substituting the three known values.

X

z30

17028.1

X

)30(30

17028.1)30(

XNow using algebra, solve the equation for X

X17028.1)30(

X40.208

Step 4: Write a statement back in contextStep 4: Write a statement back in context

A 14-year-old boys cholesterol level must be at A 14-year-old boys cholesterol level must be at least 208.40 to be in the top 10% of cholesterol least 208.40 to be in the top 10% of cholesterol levels.levels.

Comparing IndividualsComparing Individuals

One of the best reasons to standardize values One of the best reasons to standardize values (find their corresponding z-scores) is to be able to (find their corresponding z-scores) is to be able to compare individuals from different distributions.compare individuals from different distributions.

Consider again the three baseball players that we Consider again the three baseball players that we looked at earlier in the yearlooked at earlier in the yearTy Cobb Ted WilliamsTy Cobb Ted Williams George BrettGeorge Brett

.420.420 .406.406 .390.390How can we compare the batting averages of these How can we compare the batting averages of these

players when they played in different eras under players when they played in different eras under different conditions? Was Ty Cobb actually the different conditions? Was Ty Cobb actually the best hitter of these three? Let’s find out.best hitter of these three? Let’s find out.

Comparing Individuals (Cont.)Comparing Individuals (Cont.)

We know that We know that batting averages are batting averages are quite symmetric and quite symmetric and reasonably normal reasonably normal with the following with the following characteristics for characteristics for each era:each era:

DecadeDecade MeanMean Std DevStd Dev

1910s1910s .266.266 .0371.0371

1940s1940s .267.267 .0326.0326

1970s1970s .261.261 .0317.0317

Now, using that information, find the corresponding z-score for each player.

Ty CobbTy Cobb Ted Williams Ted Williams George BrettGeorge Brett

X

z

X

z

X

z

0371.

266.420. z 0326.

267.406. z 0317.

261.390. z

15.4z 26.4z 07.4z

Now that we have standardized each score onto the standard normal curve, we can compare the scores of these three individuals. Since, in this case, a larger value indicates a better batting average---it appears that Ted Williams is the best batter of these three. 4.26 > 4.15 > 4.07

Additional ResourcesAdditional Resources

Practice of Statistics, Pg 83-97Practice of Statistics, Pg 83-97