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Statistics:
Normal Curve
CSCE 115
Statistics
11/7/2005
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

Statistics
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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

Distributions that are not normal:
» Flipping a coin and count of number of
heads flips before getting the first tail
» Rolling 1 die
Statistics
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Flipping Coins Experiment
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.

Statistics
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Flipping Coins Experiment
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

Statistics
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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
Statistics
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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
Statistics
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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
Statistics
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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
Statistics
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Theory

Statistics
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

Statistics
11/7/2005
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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
Statistics
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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
12
Normal Curve
Normal Distribution
Mean
Standard Deviation
1.5
1
Normal Distribution
0.600
0.500
0.400
0.300
0.200
0.100
Statistics
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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
13
Normal Curve
Normal Distribution
Mean
Standard Deviation
0
1.7
Normal Distribution
0.600
0.500
0.400
0.300
0.200
0.100
Statistics
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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
Statistics
11/7/2005
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

Statistics
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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
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

Statistics
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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
Statistics
40
50
60
70
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80
90
100
110
20
Example: Test Scores
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

Statistics
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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.

Statistics
11/7/2005
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Grading on the Curve
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

Statistics
11/7/2005
23
Example: Grading on the
Selecting
these
breaks
Curve is completely arbitrary.
One could assign other
grade break downs.
A
B
1 st. dev. or greater above the mean
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

Statistics
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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%
Statistics
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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

Statistics
11/7/2005
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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

Statistics
11/7/2005
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Example: Light bulbs

Statistics
Assume that the life span of a light bulb
is normally distributed with a mean of
1000 hours and a standard deviation of
100. The manufacturer guaranties that
the bulbs will last 800 hours. What
percent of the light bulbs will fail before
the guarantee is up?
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Example: Light bulbs
Solution: (800-1000)/100 = -2
 The percentage of successes is
13.6% + 34.1% + 50% = 97.7%
 The percentage of failures is
100% - 97.7% = 2.3%

Statistics
11/7/2005
30
Example: Mens socks

Statistics
Assume that the length of men’s feet are
normally distributed with a mean of 12
inches and a standard deviation of .5
inch. A certain brand of “one size fits all”
strechy socks will fit feet from 11 inches to
13.5 inches. What % of all men cannot
wear the socks?
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Example: Mens socks



(11-12)/.5 = -2
(13.5 – 12)/.5 = 3
Hence men from 2
standard deviations
below the mean and
3 standard
deviations above the
mean can wear the
socks
Statistics
-2 to +2 st. dev.
95.4%
+2 to +3 st. dev.
2.2%
-2 to + 3 st. dev.
97.6%
of men can wear the socks.
100% - 97.6% = 2.4%
cannot wear them.
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Normal Distributions

We might be interested in:
» The normal probability (or frequency)
» The cumulative area below the curve
Statistics
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Normal Curves
Statistics
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Normal Curves
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Excel and the Normal
Distribution



Cumulative probabilities are measured from
“- infinity”.
NORMDIST(x, mean, st dev, cumulative)
cumulative - true cumulative
false probability
NORMINV(probability, mean, st dev)
returns the x value that gives the specified
cumulative probability
Statistics
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Excel and the Normal
Distribution
Normal Distribution
Mean
Standard Deviation
3
1
Normal Distribution
1.000
NORMDIST(4, 3, 1, true)
0.750
NORMDIST(4, 3, 1, false)
0.500
0.250
0.000
-1.0
0.0
1.0
2.0
3.0
Probability
Statistics
4.0
5.0
6.0
7.0
Cumulative
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Excel and the Normal
Distribution
The following functions assume mean =
0 and st. dev. = 1
 NORMSDIST(x) returns cumulative
distribution
 NORMSINV(probability) returns the x
value that obtains the specify cumulative
probability

Statistics
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Example: Excel



Statistics
Assume mean = 12, st. dev. = .5 and the
men’s foot distribution used earlier.
What percent of the men’s feet would be
expected to be 13 inches long or shorter?
=NORMDIST(13, 12, .5, TRUE)
=.977 or 97.7%
What is the probability that a randomly chosen
foot would be exactly 13 inches long?
=NORMDIST(13, 12, .5, FALSE)
= .108
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Example: Excel

Statistics
Assume the grade distribution used earlier
with a mean of 70 and st. dev. of 10. What
percentage of the students are expected to
pass with a D or C? (That is, with scores
between 50 and 70)?
Passing with a D 70 - 2 * 10 = 50
Passing with a B- 70
We are looking for 50 <= scores < 70
= NORMDIST(70, 70, 10, TRUE)
- NORMDIST(50, 70, 10, TRUE)
= .5 - NORMDIST(50, 70, 10, TRUE)
= .477 or 47.7% 11/7/2005
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Example: Excel
Normal Distribution: Test Scores
Mean
Standard Deviation
70
10
Probability
NORMDIST(50,70,10, TRUE)
50%
C
2.3%
D
B
47.7%
A
0
30
Statistics
40
50
60
70
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80
90
100
110
42
Example: Excel

Statistics
What score does one need to be in the
top 10%?
To be in the top 10%, the student must
do better than 100% - 10% = 90% of the
students.
=NORMINV(90%, 70, 10)
= 82.8 (or 83)
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Example: Excel
Mean
Standard Deviation
70
10
Probability
NORMINV(90%,70,10)
82.8
10%
90%
C
B
D
A
0
30
Statistics
40
50
60
70
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80
90
100
110
44
Statistics
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