Lesson 7 will consist of the following topics
For lesson 7, read pages 79-87 in Practical Statistics for Educators,
Third Edition by Ruth Ravid (2005, University Press of America)
or read pages 81-84, 131-137, & 154-174 in Basic Statistics for Behavioral Science Research
2nd ed by Mary B. Harris (1998, Allyn and Bacon)
or
read pages 115-129 in Practical Statistics for
Educators, 2nd Edition by Ruth Ravid (2000, University Press of America)
or read pages 99-113 in Practical Statistics for Educators
by Ruth Ravid (1994, University Press of America).
The normal curve or the normal frequency distribution is a hypothetical distribution of scores that is widely used in statistical analysis. Since many psychological and physical measurements are normally distributed, the concept of the normal curve can be used with many scores. The characteristics of the normal curve make it useful in education and in the physical and social sciences.
Some of the important characteristics of the normal curve are:
Looking at the figure above, we can see that 34.13% of the scores lie between the mean and 1 standard deviation above the mean. An equal proportion of scores (34.13%) lie between the mean and 1 standard deviation below the mean. We can also see that for a normally distributed variable, approximately two-thirds of the scores lie within one standard deviation of the mean (34.13% + 34.13% = 68.26%).
13.59% of the scores lie between one and two standard deviations above the mean, and between one and two standard deviations below the mean. We can also see that for a normally distributed variable, approximately 95% of the scores lie within two standard deviations of the mean (13.59% + 34.13% + 34.13% + 13.59% = 95.44%).
Finally, we can see that almost all of the scores are within three standard deviations of the mean. (2.14% + 13.59% + 34.13% + 34.13% + 13.59% + 2.14% = 99.72%) We can also find the percentage of scores within three standard deviation units of the mean by subtracting .13% + .13% from 100% (100.00% - (.13% + .13%) = 99.74%).
Deviation IQ Scores, sometimes called Wechsler IQ scores, are a standard score with a mean of 100 and a standard deviation of 15. We can thus see that the average IQ for the general population would be 100. If IQ is normally distributed, we would expect that two-thirds (68.26%) of the population would have deviation IQ's between 85 and 115. That is because 85 is one standard deviation below the mean and 115 is one standard deviation above the mean. We would expect that 99.72% of the distribution would lie within three standard deviations of the mean (that is IQs between 55 and 145).
Click on this link Deviation IQs for the Normal Curve to see the normal curve with deviation IQs along the abscissa at one standard deviation intervals. After viewing the figure, click on your browser's back arrow to return to this page.
The Stanford-Binet Scale of Intelligence IQ is also a standard score with a mean of 100 but with a standard deviation of 16. Thus a Wechsler IQ of 115 (one SD above the mean) would be equivalent to a Binet IQ of 116 (also one SD above the mean). A Wechsler IQ of 130 (two SDs above the mean) would be equivalent to a Binet IQ of 132. In some definitions of mental retardation, the cut off for an IQ score indicative of mental retardation is set at two standard deviations below the mean of the general population. This would be equivalent to a Wechsler IQ of 70 but a Stanford-Binet IQ of 68. We would also expect that 2.27% (.13% + 2.14% = 2.27%) of the population would have IQs this low.
When a score is expressed in standard deviation units, it is referred to as a Z-score. A score that is one standard deviation above the mean has a Z-score of 1. A score that is one standard deviation below the mean has a Z-score of -1. A score that is at the mean would have a Z-score of 0. The normal curve with Z-scores along the abscissa looks exactly like the normal curve with standard deviation units along the abscissa.
Click on this link Z-scores for the Normal Curve to see the normal curve with Z-scores along the abscissa at one standard deviation intervals. After viewing the figure, click on your browser's back arrow to return to this page.
Another commonly used derived scores based on the normal curve, is the T-score. T-scores are derived scores with a mean of 50 and a standard deviation of 10. The average T-score for a group of scorers would be 50. We can see that a T-score of 60 would be equivalent to a Z-score of 1, and a Deviation IQ score of 115. Each of these scores would be one standard deviation above the mean and would be equal to or higher than 84.13% of the scores (50.00% + 34.13% = 84.13%)
Click on this link T-Scores for the Normal Curve to see the normal curve with T-scores along the abscissa at one standard deviation intervals. After viewing the figure, click on your browser's back arrow to return to this page.
Another useful derived score is the percentile rank. The percentile rank is the percentage or proportion of scores that score lower than a given score. If you received a percentile rank of 90 then 90% of the scores would be lower than your score and 10% of the scores would be higher. You could also say that your score is at the 90th percentile. The median for any set of scores (by definition) is at the 50th percentile. That is, 50% of the scores are lower than the median, and 50% of the scores are higher than the median. Ordinarily percentiles are reported as whole numbers so the highest percentile possible would be 99 and the lowest possible would be 1. A score that is one standard deviation below the mean would have a percentile rank of 16 (0.13 + 2.14 + 13.59 = 15.86). A score that is two standard deviations below the mean would have a percentile rank of 2 (0.13 + 2.14 = 2.27). A score that was three standard deviations below the mean would be at the 1st percentile and one that was three standard deviations above the mean would be at the 99th percentile. Some test designers have used the concept of extended percentile ranks to make finer divisions for scores at the upper half of the 99th percentile and at the lower half of the 1st percentile.
Click on this link Percentile Ranks for the Normal Curve to see the normal curve with percentile ranks indicated along the abscissa at one standard deviation intervals. After viewing the figure, click on your browser's back arrow to return to this page.
Because of the characteristics of the normal curve we are able to find the exact proportion of scores falling below any point on the curve, above any point on the curve, or between any two points on the curve. We do this in terms of standard deviation units or Z-scores. A Z-score, you will recall, is a score expressed in standard deviation units.
Statistical tables exist that allow us to look up the exact proportion of scores between any Z-score and the mean. Such a table of values is found in your textbook as Appendix A, Z Scores and Percentage of Area under the Normal Curve Between any Given Z-Score and the Mean (Ravid, 2000) or Table I Areas Under Unit Normal Curve (Harris, 1998). In the first column of this table are listed Z scores from 0.00 on page 335 to 3.70 on page 340 (Ravid, 2000) or on pages 508-511 (Harris, 1998). In the second column is the area from the given Z-score to the mean, expressed as a proportion. To express this proportion as a percentage, multiply it by 100, i.e. move the decimal point two places to the right, and add the percent sign. This Table (and the normal curve figure) allow us to solve many problems involving scores that can be related to the normal curve.
What percentage of the general population have deviation IQs
lower than 85?
To solve this problem we have to convert the 85 IQ
to a z-score. Since 
, where X is a score,
X bar is the mean, and S is the standard deviation. So for our
problem the Z-score associated with an IQ of 85 is
So an IQ of 85 is equivalent to a z-score of -1 and our question becomes what proportion of the normal curve is lower than a z-score of -1. Appendix A only shows the proportion of scores between a z-score and the mean (we can ignore the negative sign as the table only uses positive values). So we must look at the proportion of scores between a z-score of -1 and the mean and then subtract this value from .50 (50%). So .50 - .3413 (the Appendix A value) = .1587 or 15.87% of the population has IQ scores lower than 85.
We can also solve this problem by using the normal curve figure we have seen in this lesson. Just add up the proportion of the curve below -1 standard deviation, which is 13.59 + 2.14 + 0.13 = 15.86%. The answer differs very slightly from the answer obtained from the table because of rounding errors.
What proportion of the normal curve lies between a Z-score of -2.37 and a Z-score of 1.87?
To answer this question we have to look up in Appendix A the proportion of scores between 2.37 and the mean and between 1.87 and the mean. So the answer would be .4911 + .4693 = .9604 or 96.04%.
What proportion of the scores in a normal distribution are higher than a T-score of 60?
A T-score of 60 is actually one standard deviation above the mean or equivalent to a Z-score of 1.
So we can look in Appendix A and find that a Z-score of 1 has .3413 of the distribution between it and the mean. Therefore there would be .5000 - .3413 = .1587 (15.87%) of the score above it. In other words 15.87% of the scores in a normal distribution would be higher than a T-score of 60.
What Z-score has 75% of the scores below it?
In this problem we have to enter the second column of Appendix A and find the proportion, then look at the Z-score associated with it. We need 75% of the scores, so there are 50% of the scores below the mean and then another 25% between the desired Z-score and the mean (75 - 50 = 25). So we must look in Appendix A for .2500 (25%) in column 2. The closest values are .2486 and .2517. Since .2486 is the closest to .2500 we select the Z-score as 0.67. In other words the Z-score of 0.67 has 75% of the scores below it?
What deviation IQ ranges would include 80% of the population?
This problem is a little more difficult because we have to find the Z-score that has 40% of the scores between it and the mean and then convert that Z-score to a deviation IQ score.
First looking in Appendix A we see that the Z-score associated with .4000 in the second column is 1.28, so 80% of the population lie between -1.28 and +1.28. Now the problem is to convert -1.28 and 1.28 to deviation IQs.
If you read the assignment in the text for this lesson, you may recall that the formula for converting a Z-score to a T-score is
where 10 is the standard deviation for T scores, Z is the given Z-score, and
50 is the mean for T-scores.
By analogy then the formula to convert a Z-score to a deviation IQ score would be
We can now use that formula to find the deviation IQ associated with a Z-score of -1.28
and the deviation IQ associated with a Z-score of 1.28
IQ scores, like percentile ranks, are always reported in whole numbers so our two deviation IQs become 81 and 119. So our answer becomes 80% of the population would have deviation IQs between 81 and 119.
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