Transcript Bell shaped curves

Learning Objectives for Section 11.5
Normal Distributions
 The student will be able to identify what is meant by a
normal distribution.
 The student will be able to find the area under normal
curves.
 The student will be able to approximate the binomial
distribution with a normal distribution.
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Normal Distributions
We have seen that the histogram for a binomial distribution with
n = 20 trials and p = 0.50 was shaped like a bell if we join the
tops of the rectangles with a smooth curve.
Real world data, such as IQ
scores, weights of individuals,
heights, test scores have
histograms that have a
symmetric bell shape. We call
such distributions normal
distributions. They will be
the focus of this section.
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Mathematicians and the
Normal Curve
Three mathematicians contributed to the mathematical
foundation for this curve. They are Abraham De Moivre,
Pierre Laplace and Carl Friedrich Gauss.
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Abraham De Moivre
De Moivre pioneered the
development of analytic geometry
and the theory of probability. He
published The Doctrine of Chance in
1718. The definition of statistical
independence appears in this book
together with many problems with
dice and other games. He also
investigated mortality statistics and
the foundation of the theory of
annuities
De Moivre
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/De_Moivre.html
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Pierre Laplace
Laplace systematized and elaborated
probability theory in "Essai
Philosophique sur les Probabilités"
(Philosophical Essay on Probability,
1814). He was the first to publish the
value of the Gaussian integral.
We will talk about Gauss later.
Laplace
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Bell-Shaped Curves
 Many frequency distributions have a symmetric, bell
shaped histogram.
 Example 1: The frequency distribution of heights of
males is symmetric about a mean of 69.5 inches.
 Example 2: IQ scores are symmetrically distributed
about a mean of 100, with a standard deviation of 15 or
16. The frequency distribution of IQ scores is bell
shaped.
 Example 3: SAT test scores have a bell shaped,
symmetric distribution.
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Properties of Normal Curves
 Normal curves are bell-shaped and are symmetrical with
respect to the vertical line x =  (the mean).
 The curve approaches, but does not touch, the horizontal
axis as x gets very large (or x gets very small)
 The shape of a normal curve is completely determined by
its mean and standard deviation - a small standard
deviation indicates a tight clustering about the mean and
thus a tall, narrow curve; a large standard deviation
indicates a large deviation from the mean and thus a
broad, flat curve.
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Graphs of Normal Curves
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
-10
-5
0
5
10
Several normal curves
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Probability and
Area under the Normal Curve
 Key fact: For a normally distributed variable, the percentage
of observations that lie within a specified range equals the
corresponding area under its associated normal curve.
 This is approximately true for a variable that is approximately
normally distributed.
 p(a < x < b) = probability that the random variable X
is between a and b
= area under the normal curve
between x = a and x = b.
 The total area under a normal curve is 1.
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Finding Areas Under a
Normal Curve
Finding the area under a normal curve between x = a and x = b
requires calculus. We can circumvent this problem by looking up
the values in a table.
However, the shape of each normal curve is determined by the
standard deviation; the greater the standard deviation, the
“flatter” and more spread out the normal curve will be. We would
need infinitely many tables.
The solution is to standardize a normally distributed variable,
and to use the table for the standard normal curve.
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Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1. Values on the horizontal axis are called
z values. Values on the y axis are probabilities and are decimal
numbers between 0 and 1, inclusive.
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Areas under the
Standard Normal Curve
For the following examples,
1. Draw a diagram
2. Shade the appropriate area
3. Use a table or a TI 83 to
find the probability
(A) Find p(0 < z < 1.2)
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Areas under the
Standard Normal Curve
For the following examples,
1. Draw a diagram
2. Shade the appropriate area
3. Use a table or a TI 83 to
find the probability
(A) Find p(0 < z < 1.2)
Look up the z value of 1.2
Answer: 0.3849
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Areas
(continued)
(B) Find p(-1.3 < z < 0)
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Areas
(continued)
(B) Find p(-1.3 < z < 0)
We can ignore the sign of z
since the graph is
symmetrical. Look up the
z value of 1.3.
Answer: 0.432.
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Areas
(continued)
(C) Find p(-1.25 < z < 0.89)
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Areas
(continued)
(C) Find p(-1.25 < z < 0.89)
Use table to find two different
areas, and add:
area left of y axis = 0.3944,
area right of y axis = 0.3133
Answer: .7077
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Areas
(continued)
(D) Find p(z > .75)
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Areas
(continued)
(D) Find p(z > .75)
Use table to find
p(0 < z < 0.75) = 0.2744
Subtract this area from 0.5000
Answer: 0.2266
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Standardizing a
Normally Distributed Variable
To find p(a < x < b) for a normal curve with mean  and
standard deviation , we calculate
P
where
z
a

z
b


x

The variable z is called the standard normal variable.
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Example
Assume IQ scores are distributed normally with a mean of
100 and standard deviation of 16.
(A) If the IQ of an individual is x = 124, what z value
corresponds to this?
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Example
Assume IQ scores are distributed normally with a mean of
100 and standard deviation of 16.
(A) If the IQ of an individual is x = 124, what z value
corresponds to this?
z=
124 100  1.5
16
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Example
(continued)
(B) Find the probability that a randomly chosen person has an IQ
greater than 120.
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Example
(continued)
(B) Find the probability that a randomly chosen person has an IQ
greater than 120.
Step 1. Draw a normal curve and shade appropriate area.
State the probability:
p(x > 120)
where x is IQ.
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Example
(continued)
Step 2. Convert x score to a standardized z score:
z = (120 – 100) / 16 = 20/16 = 5/4 = 1.25
p(x > 120) = p(z > 1.25)
Step 3. Use table or TI 83 to find area.
Answer: 1 - 0.6021 = 0.1056
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Example 2
A traffic study at one point on an interstate
highway shows that vehicle speeds are
normally distributed with a mean of 61.3
mph and a standard deviation of 3.3 miles
per hour. If a vehicle is randomly
checked, find the probability that its speed
is between 55 and 60 miles per hour.
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Example 2
(continued)
Step 1. Draw a normal curve
and shade appropriate area.
State the probability:
p(55 < x < 60)
where x is speed.
Step 2. Convert x score to a standardized z score:
60  61.3 
 55  61.3
p
z
 = p(-1.91 < z < -0.39)
3.3 
 3.3
Step 3. Use table or TI 83 to find area.
Answer: 0.4719 - 0.1517 = 0.3202
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Mathematical Equation for
Bell-Shaped Curves
Carl Friedrich Gauss, a mathematician, was probably the first to
realize that certain data had bell-shaped distributions. He
determined that the following equation could be used to describe
these distributions:


 ( x )2 


2

2 

1
f ( x) 
2 
e


where ,  are the mean and standard deviation of the data.
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Using the Normal Curve to
Approximate Binomial Probabilities
Example: Find the probability of getting 12 or more Heads when
you toss a coin 20 times.
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Using the Normal Curve to
Approximate Binomial Probabilities
Example: Find the probability of getting 12 or more Heads when
you toss a coin 20 times.
Solution: We have seen that the histogram for a binomial
distribution with n = 20 trials and p = 0.50 was shaped like a bell
if we join the tops of the rectangles with a smooth curve.
To find the probability that x
(number of heads) is greater than
or equal to 12, we would have to
calculate p(x=12) + p(x=13) +
p(x=14) + … p(x=20) . The
calculations would be very
tedious, to say the least.
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Using the Normal Curve to
Approximate Binomial Probabilities
 We could, instead, treat the binomial distribution as a normal
curve, since its shape is pretty close to being a bell-shaped
curve, and then find the probability that x is greater than 12
using the procedure for finding areas under a normal curve.
 = np = 10
= sqrt(np(1-p)) = sqrt(5) = 2.24
Probability that x ≥ 12
= total area in yellow
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Approximating Binomial by Normal
(continued)
Because the normal curve is continuous and the binomial
distribution is discrete, we have to make what is called a
correction for continuity.
Since we want p(x ≥ 12), we must include the rectangular
area corresponding to x = 12. The base of this rectangle
starts at 11.5 and ends at 12.5.
Therefore, we must find p(x > 11.5). The rectangle
representing the p(x = 12) extends from 11.5 to 12.5 on the
horizontal axis.
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Approximating Binomial by Normal
(continued)
Using the procedure for finding area under a non-standard
normal curve we get
11.5  10 

p( x  11.5)  p  z 


2.24 

= p(z > 0.770) = 0.5 - 0.2794 = 0.2206
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Rule of Thumb Test
 Use a normal distribution to approximate a binomial
distribution only if the interval
[   3 ,   3 ]
lies entirely in the interval from 0 to n. (n is sample size.)
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