Normal Distribution
I. Standard Scores
Suppose that you had the averages and standard deviations from two different
distributions and you wanted to compare a data item from one distribution with a
data item from the other distribution. For instance, suppose that you had a
distribution of the heights of adult males in which H  69" and  H  3" and the
distribution of the weights of adult males in which W  173 lb and  W  20 lb, and
you wanted to compare a height of 72" to a weight of 180 lb. A useful way to
compare data items from different distributions is to compare standard scores (or
standard deviation scores, or z-scores). The standard score for data item X from a
distribution with average X and standard deviation  X can be calculated as follows:
zX 
XX
X
72  69
 1. This score is
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unitless and means that 72" is 1 standard deviation above the mean (since the zscore is positive). Similarly, the standard score for a weight of 180 lb is given by
180  173
z180
 0.35 . This means that 72" is 0.35 standard deviations above the
20
mean. The height of 72" is more “extreme” from the average height of 69" than
the weight of 180 lb is from the average weight of 173 lb.
Thus, the standard score for a height of 72" is z 72 
II. Properties of Normal Distribution
Up to this point, our histograms have consisted on rectangular blocks. Suppose that
we look at distributions in which the widths of the class intervals get very small and
the number of data items gets very large. In certain distributions, such as normal
distribution, the histogram becomes a “smooth” curve. The normal curve is shown
below:
z=-3
z=-2
z=-1
z=0
z=1
z=2
z=3
The normal curve has the following important properties:
(1) The mean (balance point), median (half of area to left, half to right), and mode
(highest point on the graph) all have the same value (with z-score of 0).
(2) The total area under the curve, as in all histograms, is 100%.
(3) The curve is symmetrical with respect to a vertical line at z = 0.
(4) The size of the z-scores is not bounded in either direction.
(5) The curve is precisely and completely defined by the mathematical formula:
y
100%
2

e
x2
2
Therefore, the area under the curve from –z to z can be determined. These
areas are given in “A Normal Table” at the end of this packet.
III. Applications of the Normal Curve
Example 1: Suppose that you are given that the weights of adult males are normally
distributed with an average of 173 lb and a standard deviation of 20 lb. What
percent of adult males weighs between 150 lb and 180 lb?
(1) Put the average weight and the other significant weights on a number line,
and shade in the region whose area is to be determined.
150
173
180
(2) Determine the z-scores of these weights. Also, locate on the number line the
the points associated with the opposite z-scores.
z150 
150  173
 1.15
20
150
-1.15
-.35
z180 
180  173
 .35
20
173
0
180
.35
2
1.15
(3) Look up the areas from –z to z for these z-scores, and show these areas above
the number line.
74.99%
27.37%
150
-1.15
-.35
173
0
180
.35
1.15
(4) Determine the area of the shaded region using these areas.
74.99%
23.81%
150
-1.15
74.99%
- 27.37%
2
47.62%
27.37%
-.35
173
0
23.81%
180
.35
1.15
Answer = 23.81% + 27.37% = 51.18%
23.81%
Thus, 51.18% of adult males weighs between 150 lb and 180 lb.
Example 2: Using the same distribution of the weights of adult males as in example
1, what percent weighs exactly 152 lb?
To answer this question, find the percent of adult males weighing between
151.5 lb and 152.5 lb.
z151 .5 
151.5  173
 1.075
20
z152 .5 
152.5  173
 1.025
20
3
 72%
 69%
1.5%
151.5 152
-1.075
152.5
-1.025
72%
- 69%
173
0
1.5%
1.025
1.075
Thus, approximately 1.5% of adult males
weighs exactly 152 lb.
2 3%
1.5%
Example 3: What is the percentile for a weight of 200 lb?
To answer this question, find the percent of adult males weighing less than
200 lb.
z 200 
200  173
 1.35
20
82.30%
8.85%
8.85%
-1.35
100.00%
- 82.30%
2 17.70%
173
0
+
200
1.35
82.30%
8.85%
91.15%
8.85%
Thus, a weight of 200 lb is the 91.15%-tile weight.
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Example 4: What is the 30%-tile weight for adult males?
(1) To answer this question, you must find a weight X such that 30% of the area
is less than X.
30%
X=?
173
(2) You could find this weight X if you had its z-score. Find the z-score that
gives 40% area between –z and z; ie. z  .53 .
40%
30%
30%
X=?
 z  .53
173
z  .53
(3) Since X is less than the average of 173 lb, X must have a negative
z-score; ie. z  .53 .
z
XX
x
 .53 
X  173
 X  173 
 20(.53)  20

20
 20 
 10.6  X  173  X  162.4 .
Thus, 162.4 lb is the 30%-tile weight.
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Practice Sheet – Normal Distribution
Suppose that IQ scores are normally distributed with an average of 100 and a standard
deviation of 15.
(1) What percent of IQ scores is between 85 and 115?
(2) What percent is greater than 85?
(3) What percent is between 79 and 133?
(4) What percent is less than 79?
(5) What percent is between 115 and 133?
(6) What percent of IQ scores is exactly 100?
(7) What percent is exactly 79?
(8) What is the percentile for an IQ score of 100?
(9) What is the percentile for an IQ score of 133?
(10) What is the 50%-tile IQ score?
(11) What is the 90%-tile IQ score?
(12) What is the 30%-tile IQ score?
(13) What is the 75%-tile IQ score?
(14) What is the 25%-tile IQ score?
(15) What is the interquartile range of IQ scores?
Solution Key for Normal Distribution
(1) 68%
(2) 84%
(3) 90.5%
(6) 2%
(7) 0.5%
(8) 50%-tile
(11) 119.5
(12) 92
(13) 110.2
(4) 8%
(5) 14.5%
(9) 98.5%-tile
(14) 89.8
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(10) 100
(15) 20.4
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