Statistics for the Behavioral Sciences |
Lesson 6 The Normal Curve |
Roger N. Morrissette, PhD |
I. The Normal Curve (Video Lesson 6 I) (YouTube version)
A Normal Curve is a distribution of data with most of the scores clustered around the middle. The curve is bell-shaped and symmetrical. In a Normal Curve the mean, median, and mode are all equal and the frequency of the other scores gradually lessen on both sides. It has zero skew. The curve approaches but never meets the abscissa at both the high and low ends. Most data collected tends to be normally distributed or is said to have a normal curve. The normal curve reflects that most behavior is “normal.” Extremes of behavior are rare and found at either end of the curve.
Area Under the Normal Curve. If we assume that the total area under the Normal Curve equals 100% and the area below and above the mean equal 50% respectively, then the figure below gives the respective percentages within the first, second and third standard deviations:
II. Standard Scores (Video Lesson 6 II) (YouTube version)
Standard scores are used when you want to compare two distributions of data that have significantly different means and standard deviations. Standard scores allows us to compare apples and oranges, if you will. For example: Say you are a manager of a company and you are trying to decide which of your employees should get a bonus. A bonus will only be granted if an employee scores above average on two criteria. If these criteria have different means and standard deviations, then converting all of your criteria to standard scores will help you make your decision.
Employee
Customer Rating
Quarterly Sales Volume
Aparna 15 76,000 Dave 2 55,000 Latika 21 39,000 Gloria 17 100,000 Rasheed 10 20,000 X_{bar } = 13 X_{bar } = 58,000
s = 7.314
s = 31,233
A. z scores
The most commonly used standard score is the z score. Z scores have a mean of 0 and a standard deviation of 1. Once a distribution has been converted to z scores it can be compared to any other distribution that has been converted to z scores. The formula for calculating z scores is shown below:
The formula reads: z equals the raw score of a sample minus the mean of that sample divided by the standard deviation of that sample.
So now we can convert each of the two different sets of raw scores into z scores:
Employee
X_{1}
z_{1} X_{2}
z_{2} Aparna 15 0.27 76,000 0.58 Dave 2 -1.50 55,000 -0.10 Latika 21 1.09 39,000 -0.61 Gloria 17 0.55 100,000 1.34 Rasheed 10 -0.41 20,000 -1.22 X_{bar } = 13 X_{bar } = 58,000
s = 7.314
s = 31,233
With the new z scores we can do a better comparison of the two different data sets. Remember that a z score close to zero is close to the mean of the distribution. Negative z scores are below the mean and positive z scores are above the mean. For both of these categories we want values that are above the mean. Only our employees that have scores that are better than average in both categories will get a bonus.
Employee
Customer Rating Quarterly Sales Volume Aparna 0.27 0.58 Dave -1.50 -0.10 Latika 1.09 -0.61 Gloria 0.55 1.34 Rasheed -0.41 -1.22 Since Aparna and Gloria are the only two employees that had both scores above the mean, they will be the only ones to get a bonus.
B. T scores
Z scores are an excellent standard score for most people. The only real downside of z scores is that you have to deal with negative numbers. For those people that are uncomfortable with negative numbers, the T score can be used. The T score has a mean of 50 and a standard deviation of 10. To calculate a T score you must first calculate the z score as above. Once the z score has been identified for each raw score you can use the formula below to convert these z scores into T scores:
T = (z)(10) + 50
The formula reads: T equals the z score times 10 and then plus 50.
The table below shows the same data as above but as T scores:
Employee
Customer Rating Quarterly Sales Volume Aparna 52.7 55.8 Dave 35.0 49 Latika 60.9 43.9 Gloria 55.5 63.4 Rasheed 45.9 37.8
III. Using Standard Scores and the Z Table to Find Proportions under a Normal Curve: (Video Lesson 6 III) (YouTube version)
The Z Table is located in Appendix A in the back of the text book. It gives the proportions of area under the Normal Curve for all z scores. It has three very important columns:
Column 1: lists the positive z scores from 0.00 to 3.70
z | Area Between Mean and z |
Area Beyond z |
z | Area Between Mean and z |
Area Beyond z |
1 | 2 | 3 | 1 | 2 | 3 |
0.00 | .0000 | .5000 | 0.20 | .0793 | .4207 |
0.01 | .0040 | .4960 | 0.21 | .0832 | .4168 |
0.02 | .0080 | .4920 | 0.22 | .0871 | .4129 |
0.03 | .0120 | .4880 | 0.23 | .0910 | .4090 |
0.04 | .0160 | .4840 | 0.24 | .0948 | .4052 |
0.05 | .0199 | .4801 | 0.25 | .0987 | .4013 |
0.06 | .0239 | .4761 | 0.26 | .1026 | .3974 |
The table continues for 5 more pages
Since the Normal Curve is perfectly symmetrical, negative and positive z scores are treated equally. Negative z scores are the mirror opposite of the positive values. There are no negative z scores in the Z Table so simply use the positive value of the z in Column 1.
Column 2: lists the area of a Normal Curve from the z score to the mean
z | Area Between Mean and z |
Area Beyond z |
z | Area Between Mean and z |
Area Beyond z |
1 | 2 | 3 | 1 | 2 | 3 |
0.00 | .0000 | .5000 | 0.20 | .0793 | .4207 |
0.01 | .0040 | .4960 | 0.21 | .0832 | .4168 |
0.02 | .0080 | .4920 | 0.22 | .0871 | .4129 |
0.03 | .0120 | .4880 | 0.23 | .0910 | .4090 |
0.04 | .0160 | .4840 | 0.24 | .0948 | .4052 |
0.05 | .0199 | .4801 | 0.25 | .0987 | .4013 |
0.06 | .0239 | .4761 | 0.26 | .1026 | .3974 |
Column 3: lists the area of a Normal Curve from the z score to the end of the tail
z | Area Between Mean and z |
Area Beyond z |
z | Area Between Mean and z |
Area Beyond z |
1 | 2 | 3 | 1 | 2 | 3 |
0.00 | .0000 | .5000 | 0.20 | .0793 | .4207 |
0.01 | .0040 | .4960 | 0.21 | .0832 | .4168 |
0.02 | .0080 | .4920 | 0.22 | .0871 | .4129 |
0.03 | .0120 | .4880 | 0.23 | .0910 | .4090 |
0.04 | .0160 | .4840 | 0.24 | .0948 | .4052 |
0.05 | .0199 | .4801 | 0.25 | .0987 | .4013 |
0.06 | .0239 | .4761 | 0.26 | .1026 | .3974 |
A Special Note: The Z Table only gives the proportions for one half of the normal curve. If you add the proportion values from column two and column three for any z score you will find that they equal .5000 or 50%.
Once you assume that your data can be represented by a Normal Curve, you can now use the proportions of that curve to tell you information about a raw scores relative to the curve. For instance: You can calculate the proportion (or percentage) of raw scores that falls above or below a specific raw score.
Question #1: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If you received a 68 on the test, what proportion of people scored below you?
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Calculate the z score.
since the z is the raw score minus the mean divided by the standard deviation we have all the information we need
z = (68 - 71) / 14
z = - 0.21
Step 3: Identify the z score on your Normal Curve.
Step 4: Shade the area asked for in the question.
Step 5: Go to the Z Table and find the proportion.
z | Area Between Mean and z |
Area Beyond z |
z | Area Between Mean and z |
Area Beyond z |
1 | 2 | 3 | 1 | 2 | 3 |
0.00 | .0000 | .5000 | 0.20 | .0793 | .4207 |
0.01 | .0040 | .4960 | 0.21 | .0832 | .4168 |
0.02 | .0080 | .4920 | 0.22 | .0871 | .4129 |
0.03 | .0120 | .4880 | 0.23 | .0910 | .4090 |
0.04 | .0160 | .4840 | 0.24 | .0948 | .4052 |
0.05 | .0199 | .4801 | 0.25 | .0987 | .4013 |
0.06 | .0239 | .4761 | 0.26 | .1026 | .3974 |
For Question #1,
We look up the z of 0.21 in Column 1 and then find the Column 3 proportion: .4168.
The answer to Question #1 is:
The proportion of the class that scored below you on the exam is .4168.
Question #2: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If you received an 89 on the test, what proportion of people scored below you?
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Calculate the z score.
z = (89 - 71) / 14
z = 1.29
Step 3: Identify the z score on your Normal Curve.
Step 4: Shade the area asked for in the question.
Step 5: Go to the Z Table and find the proportion.
Remember that the Z Table only gives you proportions for half of the Normal Curve. For this problem we want a proportion that is more than half of the Normal Curve. So in this case we go to the Z Table and find the proportion from the z to the mean and then add .5000 to that value to give us the total area in question
For Question #2,
We look up the z of 1.29 in Column 1 and then find the Column 2 proportion: .4015.
Then we add .5000 to that column 2 proportion to account for the entire shaded area.
The answer to Question #2 is:
The proportion of the class that scored below you on the exam is .9015.
Question #3: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If you received an 79 on the test, what proportion of people scored above you?
This question is sort of the mirror image of Question #1, so it is calculated the same way.
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Calculate the z score.
z = (79 - 71) / 14
z = 0.57
Step 3: Identify the z score on your Normal Curve.
Step 4: Shade the area asked for in the question.
Step 5: Go to the Z Table and find the proportion.
For Question #3,
We look up the z of 0.57 in Column 1 and then find the Column 3 proportion: .2843.
The answer to Question #3 is:
The proportion of the class that scored above you on the exam is .2843.
Question #4: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If you received an 55 on the test, what proportion of people scored above you?
This question is sort of the mirror image of Question #2, so it is calculated the same way.
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Calculate the z score.
z = (55 - 71) / 14
z = -1.14
Step 3: Identify the z score on your Normal Curve.
Step 4: Shade the area asked for in the question.
Step 5: Go to the Z Table and find the proportion.
For Question #4,
We look up the z of 1.14 in Column 1 and then find the Column 2 proportion: .3729.
Then we add .5000 to that column 2 proportion to account for the entire shaded area.
The answer to Question #4 is:
The proportion of the class that scored above you on the exam is .8729.
Question #5: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If Alicia received a 60 and Latika received a 90, what proportion of people scored between Alicia and Latika?
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Calculate each z score.
z = (60 - 71) / 14 z = (90 - 71) / 14
z = -0.79 z = 1.36
Step 3: Identify each z score on your Normal Curve.
Step 4: Shade the area asked for in the question.
Step 5: Go to the Z Table and find each proportion and add them together.
For Question #5,
We look up the z of -0.79 in Column 1 and then find the Column 2 proportion: .2852.
We look up the z of 1.36 in Column 1 and then find the Column 2 proportion: .4131.
Then we add the two proportions together to account for the entire shaded area.
The answer to Question #5 is:
The proportion of the class that scored between Alicia's score and Latika's score is .6983.
Question #6: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If Tony received a 55 and Rhonda received a 66, what proportion of people scored between Tony and Rhonda?
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Calculate each z score.
z = (55 - 71) / 14 z = (66 - 71) /b 14
z = -1.14 z = -0.36
Step 3: Identify each z score on your Normal Curve.
Step 4: Shade the area asked for in the question.
Step 5: Go to the Z Table and find each proportion and subtract the smaller from the larger.
For Question #6,
We look up the z of -1.14 in Column 1 and then find the Column 2 proportion: .3729.
We look up the z of -0.36 in Column 1 and then find the Column 2 proportion: .1406.
Then we subtract the smaller proportion from the larger to account for the entire shaded area.
The answer to Question #6 is:
The proportion of the class that scored between Tony's score and Rhonda's score is .2323.
Question #7: The mean of Dr. Morrissette's first statistics exam was 71. The standard deviation was 14. If Horatio scored better than 88% of his classmates, what score did he get?
(Note: this question is the same as asking: What score represents the 88th percentile?)
Step 1: Draw a Normal Curve with z score mean and standard deviations.
Step 2: Shade the estimated area asked for in the question.
Step 3: Convert the percentage to a proportion.
88% = .8800
Step 4: Go to the P Table and find the proportion.
The P Table is located in Appendix A in the back of the text book. It starts on page 432 and gives the proportions of area under the Normal Curve for all Percentiles. A Percentile is a score that has certain percentage of scores below it. The P Table has three very important columns:
Column 1: lists proportions for The Larger Area of the Normal Curve (Percentiles at 50% and above)
The Larger Area | z | The Smaller Area | The Larger Area | z | The Smaller Area |
1 | 2 | 3 | 1 | 2 | 3 |
.500 | 0.0000 | .500 | .625 | 0.3186 | .375 |
.505 | 0.0125 | .495 | .630 | 0.3319 | .370 |
.510 | 0.0251 | .490 | .635 | 0.3451 | .365 |
.515 | 0.0376 | .485 | .640 | 0.3585 | .360 |
.520 | 0.0502 | .480 | .645 | 0.3719 | .355 |
.525 | 0.0627 | .475 | .650 | 0.3853 | .350 |
.530 | 0.0753 | .470 | .655 | 0.3989 | .345 |
The table continues for 5 more pages
Column 2: lists the z scores relative to the proportions.
The Larger Area | z | The Smaller Area | The Larger Area | z | The Smaller Area |
1 | 2 | 3 | 1 | 2 | 3 |
.500 | 0.0000 | .500 | .625 | 0.3186 | .375 |
.505 | 0.0125 | .495 | .630 | 0.3319 | .370 |
.510 | 0.0251 | .490 | .635 | 0.3451 | .365 |
.515 | 0.0376 | .485 | .640 | 0.3585 | .360 |
.520 | 0.0502 | .480 | .645 | 0.3719 | .355 |
.525 | 0.0627 | .475 | .650 | 0.3853 | .350 |
.530 | 0.0753 | .470 | .655 | 0.3989 | .345 |
Column 3: lists proportions for The Smaller Area of the Normal Curve (Percentiles at 50% and below)
The Larger Area | z | The Smaller Area | The Larger Area | z | The Smaller Area |
1 | 2 | 3 | 1 | 2 | 3 |
.500 | 0.0000 | .500 | .625 | 0.3186 | .375 |
.505 | 0.0125 | .495 | .630 | 0.3319 | .370 |
.510 | 0.0251 | .490 | .635 | 0.3451 | .365 |
.515 | 0.0376 | .485 | .640 | 0.3585 | .360 |
.520 | 0.0502 | .480 | .645 | 0.3719 | .355 |
.525 | 0.0627 | .475 | .650 | 0.3853 | .350 |
.530 | 0.0753 | .470 | .655 | 0.3989 | .345 |
According to the P Table, the proportion .880 has a z score of 1.1750.
Step 5: Convert the z score into a raw score.
We do that with the following formula:
X = (s)(z) + X_{bar}
X = (14)(1.1750) + 71
X = 87.45
The answer to Question #7 is:
Horatio got a score of 87.45.
A Note of Caution!: If the percentile is below 50 then the z score will be negative (all z scores to the left of the mean are negative). This will NOT be shown in the P Table. You need to make sure you add the negative sign to the z score when you calculate your X value.
Additional Videos on the Concepts that might help:
Normal Distribution and z scores
z score Calculations and Percentiles in a Normal Distribution