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8.5 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. This will
be the focus of this section.

DeMoivre
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/De_Moivre.html

Three mathematicians contributed to
the mathematical foundation for this
curve. They are Abraham De Moivre,
Pierre Laplace and Carl Frederick
Gauss

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
Laplace

Laplace also 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,
Bell shaped curves



Many frequency distributions have a symmetric, bell shaped
histogram. For example, 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 and 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.
Graph of a generic normal
distribution
0.5
0.4
0.3
Series1
0.2
0.1
0
-4
-2
0
2
4
Values on X axis represent the number of standard deviation units a
particular data value is from the mean.
Values on the y axis represent probabilities of the random variable x.
0.5
0.4
0.3
Series1
0.2
0.1
0
-4
-2
0
2
4
Area under the Normal Curve


1. Normal distribution : a smoothed out
histogram
2. P( a < x < b) = Probability that the random
variable x is between a and b is determined
by the area under the normal curve between
x = a and x = b .
Properties of Normal distributions



1. Symmetric about its mean, 
2. Approaches, but not touches, the
horizontal axis as x gets very large ( or x
gets very small)
3. Almost all observations lie within 3
standard deviations from the mean.
Area under normal curve

Example: A midwestern college has an
enrollment of 3264 female students whose
mean height is 64.4 inches and the standard
deviation is 2.4 inches. By constructing a
relative frequency distribution, with class
boundaries of 56, 57, 58, … 74, we find that
the frequency distribution resembles a bell
shaped symmetrical distribution.
Heights of Females at a College
(Relative frequency distribution with class width = 1 is smoothed out to
form a normal, bell-shaped curve) ..
Normal curve areas

Key fact: For a normally distributed variable,
the percentage of all possible observations
that lie within any specified range equals the
corresponding area under its associated
normal curve expressed as a percentage.
This holds true approximately for a variable
that is approximately normally distributed.
The area of the red portion of the graph is equal to the
prob( 66 < x < 68)
; the probability that a female student chosen at random from the
population of all students at the college has a height between 66 and
68 in.
Finding areas under a normal,
bell-shaped curve


The problem with attempting to find the area under a normal curve
between x = a and x = b ( and thus finding the probability that x is
between a and b, P( a < x < b) is that calculus is needed.
However, we can circumvent this problem by using results from
calculus. Tables have been constructed to find areas under what
is called the standard normal curve. The standard normal curve
will be discussed shortly. A normal curve is characterized by its
mean and standard deviation. The scale for the x axis will be
different for each normal curve. The shape of each normal curve
will differ since the shape is determined by the standard deviation;
the greater the standard deviation, the “flatter” and more spread
out the normal curve will be.
Standardizing a Normally
Distributed Variable

To find percentage of scores that lie within a
certain interval, we need to find the area
under the normal curve between the desired
x values. To do this, we need a table of
areas for each normal curve. The problem is
that there are infinitely many normal curves
so that we would need infinitely many tables.
Non-standard normal curves

For example, the distribution of IQ scores is normal with mean =
100 and standard deviation =16.

Ex. 2. The heights of females at a certain mid-western college is
normally distributed with a mean of 64.4 inches and a standard
deviation of 2.4 inches.
Ex. 3. The probability distribution of x, the diameter of CD’s
produced by a company, is normally distributed with a mean of 4
inches and a standard deviation of .03 inches.
Thus, for these three examples we would need three separate
tables giving the areas under the normal curve for each separate
distribution. Obviously, this poses a problem.
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
Standard normal curve
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The way out of this problem is to standardize each normal
curve which will transform individual normal distributions
into one particular standardized distribution. To find
P( a < x < b) for the non-standard normal curve, we can find
P

a

z
b

P

a
z
b

Thus P(a < x < b) =


variable z is called the standard normal variable.
The
Standard normal distribution

The standard normal distribution will have a mean of 0 and a standard
deviation of 1. Values on the horizontal axis are called z values. Z will
be defined shortly. Values on the y axis are probabilities and will be
decimal numbers between 0 and 1, inclusive.
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0.5
0.4
0.3
Series1
0.2
0.1
0
-4
-2
0
2
4
Standardized Normally
Distributed Variable

The formula below for z can be used to standardize any normally
distributed variable x. Z is referred to as the amount of standard
deviations from the mean; A. S. D. M. = z.
represent the
mean and standard deviation of the distribution, respectively.
 ,
x


z



For example, if IQ scores are distributed normally with a mean of 100
and standard deviation of 16, the if x = IQ of an individual = 124, then
z = 124 100  1.5
16
Areas under the standard normal curve

Find the following probabilities:
A) P( 0 < z < 1.2) =

Use table or TI 83 to find area. Answer: .3849

Areas under the Standard
Normal Curve

Let z be the standard normal variable. Find the
following probabilities: Be sure to sketch a
normal curve and shade the appropriate area. If
you use a TI 83, give the appropriate commands
required to do the problem.
Examples
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
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
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Probability( -1.3 < z<0)
1. Draw diagram
2. Shade appropriate area
3. Use table or calculator to find area.
4. Answer: .4032
Examples (continued)



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Probability (-1.25 < z < .89) =
1. Draw picture
2. Shade appropriate area
3. Use table to find two different areas
4. Find the sum of the two percentages.
5. Answer: .7076
More examples:
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Probability ( z > .75)
1. Draw diagram
2. Shade appropriate area
3. Use table to find p(0<z<0.75)
4. Subtract this area from 0.5000.
Answer: 0.2266
More examples (continued)
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probability(-1.13 < z < -.79) =
1. Draw diagram
2. Shade appropriate area
3. Use table to find p(0 < z < 1.13)
4. Use table to find p( 0 < z < 0.79)
5. Subtract the smaller percentage from the
larger percentage.
6. Answer: 0.0855
Finding probabilities for nonstandard normal curves.

P( a < x < b) is the same as
a


b


P  z 










Example 1
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

IQ scores are normally distributed with a mean of 100 and a
standard deviation of 16. 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 probability: P( x > 120) , where x is IQ.
Example
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Step 2. Convert x score to a standardized z score:
Z = ( 120 – 100)/ 16 = 20/16 = 5/4 = 1.25
Probability ( x  120) = P( z > 1.25)
Step 3. Draw standard normal curve and shade appropriate
area.
Step 4. Use table or TI 83
To find area.
Answer: .1056
Areas under the Non-standard
normal curbe

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.
Solution:




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1. Draw diagram
2. Shade appropriate area
3. Use
z
x

5. Find
6. Answer: 0.3187
60  61.3 
 55  61.3
p
z

3.3 
 3.3
Non standard normal curve
areas
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
If IQ scores are normally distributed with a
mean of 100 and a standard deviation of 16,
find the probability that a randomly chosen
person will have an IQ greater than 84.
Answer: approximately .84
IQ scores example

If IQ scores are normally distributed with a
mean of 100 and a standard deviation of 16,
find the probability that a person’s IQ is
between 85 and 95.


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1. Draw diagram
2. Shade appropriate area
3. standardize variable x using
4. Find
x  
 x 
p 1
 

5. Answer: 0.2031
z
2



z
x

Areas under non-standard
normal curves

The lengths of a certain snake are normally
distributed with a mean of 73 inches and a
standard deviation of 6.5 inches. Find the
following probabilities. Let x represent the
length of a particular snake

P( 65<x<75) answer: 0 .5116
Mathematical Equation for
bell-shaped curves
Carl Frederick 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:
1
f ( x) 
e
2 
Where
data.
 ,







2
(
x


)


2
2 

are the mean and standard deviation of the
Using the Normal Curve to approximate binomial
probabilities

Example:
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.
If we wanted to find the probability
that x (number of heads) is greater
than 12, we would have to use the
binomial probability formula and
calculate P(x = 12) + P(x=13) + p(x=14)
+ … P(x=20) . The calculations would
be very tedious to say the least.

Binomial Distribution for n = 20 and p = 0.5
( A coin is tossed 20 times and the probability
of x = 0 , 1, 2, 3, …20 is calculated. Each
vertical bar represents one outcome of x. )
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.



Prob(x > 12) = P(x > 11.5) = total
area in yellow
Because the normal curve is continuous and the binomial distribution is
discrete ( x = 0 , 1 , 2, …20) 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 prob(x = 12) extends from 11.5 to 12.5 on
the horizontal axis.
Solution:


Using the procedure for finding area under a non-standard normal
curve we have the following result:
11.5  10 

p( x  11.5)  p  z 


2.24 

= 0.25