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Statistics:
Normal Curve
CSCE 115
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1
Normal distribution


The bell shaped curve
Many physical quantities are distributed in such a
way that their histograms can be approximated
by a normal curve
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Examples

Examples of normal distributions:
Height of 10 year old girls and most other body
measurements
 Lengths of rattle snakes
 Sizes of oranges
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
Distributions that are not normal:
Flipping a coin and count of number of heads
flips before getting the first tail
 Rolling 1 die

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Flipping Coins Experiment
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Experiment: Flip a coin 10 times. Count the
number of heads
Expected results: If we flip a coin 10 times, on
the average we would expect 5 heads.
Because tossing a coin is a random experiment,
the number of heads may be more or less than
expected.
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Flipping Coins Experiment



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Carry out the experiment
But the average was supposed to be 5
What can we do to improve the results?
How many times do we have to carry out the
experiment to get good results?
Faster: Use a computer simulation
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Experiment: Flipping coins
Experiment: Flipping coins compared with theory
Number of trials
Number of flips
10
20
(Use 1 to 1000)
(10, 20, or 40)
Experiment: Flip 20 coins 10 times
Experimental
4
Theoretical
3
Frequency
3
2
2
1
1
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Num ber of heads
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Experiment: Flipping coins
Experiment: Flipping coins compared with theory
Number of trials
Number of flips
100
20
(Use 1 to 1000)
(10, 20, or 40)
Experiment: Flip 20 coins 100 times
Experimental
30
Theoretical
Frequency
25
20
15
10
5
0
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Num ber of heads
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Experiment: Flipping coins
Experiment: Flipping coins compared with theory
Number of trials
Number of flips
300
20
(Use 1 to 1000)
(10, 20, or 40)
Experiment: Flip 20 coins 300 times
Experimental
70
Theoretical
60
Frequency
50
40
30
20
10
0
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Num ber of heads
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Experiment: Flipping coins
Experiment: Flipping coins compared with theory
Number of trials
Number of flips
1000
20
(Use 1 to 1000)
(10, 20, or 40)
Experiment: Flip 20 coins 1000 times
Experimental
200
Theoretical
180
160
Frequency
140
120
100
80
60
40
20
0
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
Num ber of heads
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Theory

If the number of trials of this type increase, the
histogram begins to approximate a normal curve
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Normal Curve




One can specify mean and standard deviation
The shape of the curve does not depend on
mean. The curve moves so it is always
centered on the mean
If the st. dev. is large, the curve is lower and
fatter
If the st. dev, is small, the curve is taller and
skinnier
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Normal Curve
Normal Distribution
Mean
Standard Deviation
0
1
Normal Distribution
The normal curve
with mean = 0 and
std. dev. = 1 is often
referred to as the z
distribution
0.600
0.500
0.400
0.300
0.200
0.100
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00
3.
00
3.
50
2.
50
2.
00
1.
00
1.
50
0.
50
0.
00
-4
.0
0
-3
.5
0
-3
.0
0
-2
.5
0
-2
.0
0
-1
.5
0
-1
.0
0
-0
.5
0
0.000
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Normal Curve
Normal Distribution
Mean
Standard Deviation
1.5
1
Normal Distribution
0.600
0.500
0.400
0.300
0.200
0.100
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4.
00
3.
00
3.
50
2.
50
2.
00
1.
00
1.
50
0.
50
0.
00
-4
.0
0
-3
.5
0
-3
.0
0
-2
.5
0
-2
.0
0
-1
.5
0
-1
.0
0
-0
.5
0
0.000
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Normal Curve
Normal Distribution
Mean
Standard Deviation
0
1.7
Normal Distribution
0.600
0.500
0.400
0.300
0.200
0.100
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4.
00
3.
00
3.
50
2.
50
2.
00
1.
00
1.
50
0.
50
0.
00
-4
.0
0
-3
.5
0
-3
.0
0
-2
.5
0
-2
.0
0
-1
.5
0
-1
.0
0
-0
.5
0
0.000
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Normal Curve
Normal Distribution
Mean
Standard Deviation
0
0.7
Normal Distribution
0.600
0.500
0.400
0.300
0.200
0.100
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4.
00
3.
00
3.
50
2.
50
2.
00
1.
00
1.
50
0.
50
0.
00
-4
.0
0
-3
.5
0
-3
.0
0
-2
.5
0
-2
.0
0
-1
.5
0
-1
.0
0
-0
.5
0
0.000
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Normal Curve


50% of the area is right of the mean
50% of the area is left of the mean
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Normal Curve



The total area under the curve is always 1.
68% (about 2/3) of the area is between 1 st. dev.
left of the mean to 1 st. dev. right of the mean.
95% of the area is between 2 st. dev. left of the
mean to 2 st. dev. right of the mean.
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Example: Test Scores
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Several hundred students take an exam. The
average is 70 with a standard deviation of 10.
About 2/3 of the scores are between 60 and
80
95% of the scores are between 50 and 90
About 50% of the students scored above 70
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Example: Test Scores
Normal Distribution: Test Scores
Mean
Standard Deviation
70
10
Probability
50.0%
68.2%
95.4%
0
30
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50
60
70
80
90
100
110
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Example: Test Scores
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Suppose that passing is set at 60. What percent of
the students would be expected to pass?
Solution: (60 - 70)/10 = -1. Hence passing is 1
standard deviation below the mean.
34.1% of the scores are between 60 and 70
50% of the scores are above 70
84.1% of students would expected to pass
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Example: Test Scores
Alternate solution



Suppose that passing is set at 60. What percent of
the students would be expected to pass?
Solution: According to our charts, 15.9% of the
scores will be less than 60 (less than –1 standard
deviations below the mean).
100% – 15.9% = 84.1% of the students pass
because they are right of the –1 st. dev. line.
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Grading on the Curve
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Assumes scores are normal
Grades are based on how many standard
deviations the score is above or below the mean
The grading curve is determined in advanced
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Example: Grading
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Selecting these breaks
on
the
Curve
is completely arbitrary.
One could assign other
grade break downs.
A 1 st. dev. or greater above the mean
B From the mean to one st. dev.
above the mean
C From one st. dev. below mean to
mean
D From two st. dev below mean to
one st. dev. below mean
F More than two st. dev. below mean
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Normal Distribution
Frequency
0.5000
34.1%
34.1%
13.6%
13.6%
2.2%
.13%
-4
-3
C
D
F
2.2%
B
.13%
A
0.0000
-2
-1
0
1
2
3
4
Standard deviations from mean
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Example: Grading on the Curve

Range
(in st. dev.) Expected percent
Grade from to
of scores
A
+1
50% - 34.1% = 15.9%
B
0 +1 34.1%
C
-1
0 34.1%
D
-2
-1 13.6%
F
-2 50% - 34.1 - 13.6%
= 2.3%
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Example: Grading on a Curve


Assume that the mean is 70, st. dev. is 10
Joan scores 82. What is her grade?
(82-70)/10 = 12/10 = 1.2
She scored 1.2 st. dev. above the mean.
She gets an A
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Example: Grading on a Curve


Assume that the mean is 70, st. dev. 10
Tom scores 55. What is his grade?
(55 - 70)/10 = -15/10 = -1.5
He scored 1.5 st. dev. below the mean.
He gets a D
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