PSY 201 Lecture Notes
The Normal Distribution
Random variable
A variable whose values are taken on randomly. A variable whose values we don’t know ahead of time.
Pretty much everything we study in statistics concerns random variables.
Probability Distribution
If we measure any characteristic across people, invariably the values of that characteristic will form a
distribution.
The formula that gives the probability of each different value is called the probability distribution for that
characteristic.
Heights of randomly selected adult males.
Circumferences of randomly selected Oak trees.
ACT scores of randomly selected students.
Every characteristic has a probability distribution.
Fortunately, thousands of physical and psychological characteristics all have distributions with the same
essential shape shown here. . .
This shape is so ubiquitous that a couple of 100 years ago, mathematicians began trying to discover a
formula that characterized it.
The formula that was discovered is
P(X) =
1
------------ *
σ SQRT(2π)
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-(X-µ)2
-----2σ2
e
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The Normal Distribution: Percentages of Scores
As mentioned above, 1000s of different characteristics – Heights, ACT scores, Conscientiousness scores,
etc – all have essentially normal distributions.
The different distributions differ only in their means, µ, and in their standard deviations, σ.
The percentages of scores between corresponding places on all normal distributions are identical.
From the text, p. 108. . .
All normal distributions have the same percentages shown in the figure.
The above figure shows that if you know the mean, M and the standard deviation, σ, of a distribution, you
can convert any score to a Z score and then you can determine the percentages of persons within specific
ranges of values. Below are some popular ranges . . .
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Finding different percentages in the normal curve.
We might not be interested in the percentage of scores between Z = -1 and Z = +1.
Instead we might want to know the percentage of scores between Z = -.5 and Z = + .5.
Fortunately, every statistics text has a table of percentages in the normal distribution between various Z
scores.
Ours is Appendix A Table 1: Area Under the Normal distribution
For a +Z score, the table gives you Start here on 9/22/15.
1. The percentage of scores in the normal distribution below that +Z.
2. The percentage of scores between 0 (the mean) and that +Z.
3. The percentage of scores above that +Z, called the tail percentage.
For a –Z score (a negative Z), the table gives you . . .
1. The percentage of scores in the normal distribution above that +Z.
2. The percentage of scores between 0 (the mean) and that +Z.
3. The percentage of scores below that +Z, called the tail percentage.
Using the Table to find percentages . . .
1. Draw a picture of the percentage that you want to find.
2. Find the Z in the table.
3. Find the column containing the required percentage.
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Example problems . . .
Corty, p. 113. What is the percentage of cases in a normal distribution that have scores above a Z score of
-2.30.
1. Picture . . .
2. Find Z in table . . .
Z pos
Z neg
3. Find the appropriate column. Match the picture you drew to the Table’s pictures
Answer: 96.93% of the Zs in a normal distribution are larger than -2.30.
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Example. Corty p. 114. “What’s the area from the mean to a Z score of +1.50?”
1. Picture . . .
2. Find Z in Table . . .
3. Match the picture in the Table to the one you drew.
Answer: 43.32% of the Zs in a normal distribution are between the mean and Z=+1.50.
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Example. Corty, p. 115. “What is the area below a Z score of -0.75?”
1. Draw a picture
2. Find Z in the Table.
3. Match the picture in the Table with the one you drew.
Answer: 22.66%.
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Example. Corty p. 116. “The mean length of pregnancy from the day of ovulation to childbirth is 266 days
with a standard deviation of 16 days. What percentage give birth at least two weeks later than average?”
Two weeks later than average would be 266 + 2*7 = 266 + 14 = 280 days or later.
0. Convert the value to a Z score.
Z = (280 – 266)/16 = +0.88.
1. Draw a picture.
2. Find the Z in the Table . . .
3. Match the picture in the Table with the one you drew.
Answer: 18.94%
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Normal Distribution Percentile Ranks, Middle percentages and Extreme Percentages
If a score is from a normal distribution, then the percentile rank of that score can be obtained by converting
the score to a Z score and using Table 1.
The following figure shows the relationship between percentile ranks and Z scores from the normal
distribution . . .
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Middle Percentages
Sometimes we want to identify a specific percentage of scores in the middle of a normal distribution.
For example, what are the Zs that separate the middle 90% of the scores in a normal distribution from the
rest.
To find the Zs corresponding to a middle percentage . . .
1. Draw a picture and put ½ the desired middle % on each side of the mean in the figure.
45%
45%
2. Find the Z corresponding to 45% from Table 1.
Search Table 1 until you find a Column B value that corresponds to 45%.
3. The answer is + 1.65 for the right side of the middle percentage and -1.65 for the left side.
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Extreme percentages.
An extreme percentage is a percentage of scores from the normal distribution that is evenly divided between
the two tails of the distribution.
Example. What are the Zs corresponding to the extreme 5% of the normal distribution.
1. Draw a picture and shade or otherwise identify ½ the extreme percentage in each tail.
2.50%
2.50%
2. Use Table 1 to find the Z scores corresponding to the tail areas.
1.96
3. Answer: The Zs that identify the extreme 5 percent of the normal distribution are -1.96 and + 1.96.
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Corty’s worked out examples, p 121. Clayton takes an IQ test which has mean = 100 and standard deviation
= 15. He does better than 95% of the population, i.e., his percentile rank is 95. What is his IQ score.
1. Draw a picture.
We can see from the picture that Clayton’s IQ is somewhere between 115 and 130.
2. Find the Z score by looking up the percentile rank in Table 1.
We’ll look in Column A, whose picture
matches the picture we drew.
3. Convert the Z score of 1.65 into an X, using Corty’s Equation 4.2 (X = µ + (Z * σ) (p105)
X = 100 + (1.65*15) = 124.75 or 125 since IQ scores are reported to the nearest whole number.
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Extreme percentages exercise on p. 123
What are the IQ scores that mark off the extreme 50% of cases.
1. Cut the percentage in half and draw a picture with the halves in the tails of the distribution.
25%
25%
2. Use Table 1 to find percentages corresponding to the tail percentages by matching a Table 1 figure to the
figure just drawn.
Note that there are TWO Zs that must be used here – the Z corresponding to the upper 25% and the Z
corresponding to the lower 25%.
The lower Z is -0.67. For it, the X value is 100 + (-0.67*15) = 89.95.
The upper Z is +0.67. For it, the X value is 100 + (+0.67*15) = 110.05.
So, 90 and 110 cut off the extreme 50% of the IQ distribution.
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