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Chapter 6
The Normal Probability
Distribution
CONTINUOUS RANDOM VARIABLES


Continuous random variables can assume the
infinitely many values corresponding to points on a
line interval. This means that the range of the variable
contains an interval of real numbers.
Examples:

Heights, weights

length of life of a particular product

experimental laboratory error
Recall that the probability distribution p(x) of a
discrete random variable is just the histogram of
its values. We had the following examples in
chapter 5:
 1. Toss a fair coin three times and let x be the
number of heads. Then x is binomial and

x
0
1
2
3
p(x)
1/8
3/8
3/8
1/8

We can create what we call the probability
polygon h(x) by starting from the midpoint of the
left side of the first bar and connecting the
midpoints of the upper sides of the subsequent
bars by straight lines and finishing at the
midpoint of the right side of the last bar.


2. A marksman hits a target 80% of the time. He
fires five shots at the target. What is the probability
that exactly 3 shots hit the target? If x is the number
of shots that hit the target, then x is a binomial and
k
p=
.80
0
.000
1
.007
2
.058
3
.263
4
.672
5
1.000
Again we can draw the probability polygon h(x) in
the same fashion.

Note that the total area under the probability
histogram is 1 and that the area under the
probability polygon is very close to 1

Now let x be a continuous random variable and
consider a set of n measurements of x. we can
construct the relative frequency histogram of
these values and the corresponding probability
polygon h(x) . As mentioned before, the area
under h(x) is very close to 1. We can use this
polygon approximate
as the area
under h(x) over the interval
.
rel freq
10
8
6
rel freq
4
2
0
0.835 0.865 0.895 0.925 0.955 0.985 1.015 1.045 1.075 More

As
(the number of measurements
increases without bound) the relative frequency
polygon h(x) approaches the graph of a smooth
function
whose area under the graph is one.
This function is called the probability density
function, pdf, of the random variable x.
rel freq
10
8
h(x)
6
rel freq
4
2
0
0.835 0.865 0.895 0.925 0.955 0.985 1.015 1.045 1.075 More

The pdf
of x can be used, among other
things, to find the exact value of
which is the area under the graph of
over
the interval
which is, from calculus,
a
b
 The
probability density function
for the random
variable x describes the probability distribution of the
continuous random variable x. This function is
sometime referred to as the depth of probability.
•In many situations the density function is given
by a mathematical formula where as in many
many situations finding a nice formula may not
be possible.
PROPERTIES OF CONTINUOUSPROBABILITY
DISTRIBUTIONS
1.
2. The total area under the curve =
3.
=area under the curve between a and b
Exercise: Show that there is no probability attached to any single
value of x. That is, P(x = a) = 0.
There are many different types of continuous random
variables. Here are some that occur often:
1. Uniform Random variable

2.
Exponential Random Variable
Some other random variable include The Gamma, Chi
Square, and Bata random variables. The most important
random variable is the Normal Random variable.
To choose a model for real life situation,
we try to pick a model that

Fits the data well

Allows us to make the best possible
inferences using the data.
As mentioned, one important continuous
random variable is the normal
random variable and we will discuss
that in more detail.
THE NORMAL DISTRIBUTION
• The formula that generates the
normal probability distribution is:
1  x 
 

2  
2
1
f ( x) 
e
for   x 
 2
e  2.7183
  3.1416
 and  are the population mean and standard deviation.
• The shape and location of the normal
curve changes as the mean and standard
deviation change.
THE STANDARD NORMAL
DISTRIBUTION
 To
find P(a < x < b), we need to find the
area under the appropriate normal curve.
 To simplify the tabulation of these areas,
we standardize each value of x by
expressing it as a z-score, the number of
standard deviations  it lies from the mean
.
z
x

THE STANDARD
NORMAL (Z)
DISTRIBUTION
 Mean
= 0; Standard deviation = 1
 When x = , z = 0
 Symmetric about z = 0
 Values of z to the left of center are negative
 Values of z to the right of center are positive
 Total area under the curve is 1.
USING TABLE 3
The four digit probability in a particular row
and column of Table 3 gives the area under
the z curve to the left that particular value of
z.
Area for z = 1.36
EXAMPLE
Use Table 3 to calculate these probabilities:
P(z 1.36) = .9131
P(z >1.36)
= 1 - .9131 = .0869
P(-1.20  z  1.36) =
.9131 - .1151 = .7980
mpirical Rule:
% of the
within 2
ns of the mean.
USING TABLE 3
To find an area to the left of a z-value, find the area
directly from the table.
To find an area to the right of a z-value, find the area
in Table 3 and subtract
from
Remember
the 1.
Empirical Rule:
99.7%
of the of z, find the two
To find the areaApproximately
between two
values
within 3
areas in Table 3, measurements
and subtractlieone
from the other.
standard deviations of the mean.
P(-3  z  z3) 1.96)
P(-1.96
= .9987
.9750 - .0250 =
.9500
.0013=.9974
WORKING BACKWARDS
Find the value of z that has area .25 to its left.
1. Look for the four digit area
closest to .2500 in Table 3.
2. What row and column does
this value correspond to?
3. z = -.67
4. What percentile
does this value
represent?
25th percentile,
or 1st quartile (Q1)
Find the value of z that has area .05 to its right.
1. The area to its left will be 1 - .05
= .95
2. Look for the four digit area
closest to .9500 in Table 3.
3. Since the value .9500 is
halfway between .9495 and
.9505, we choose z halfway
between 1.64 and 1.65.
4. z = 1.645
FINDING PROBABILITIES FOR THE
GENERAL NORMAL RANDOM
VARIABLE
To find an area for a normal random variable x
with mean  and standard deviation , standardize
or rescale the interval in terms of z.
Find the appropriate area using Table 3.
Example: x has a normal
distribution with  = 5 and  =
2. Find P(x > 7).
75
P ( x  7)  P ( z 
)
2
 P( z  1)  1  .8413 .1587
1
z
EXAMPLE
The weights of packages of ground beef are
normally distributed with mean 1 pound and
standard deviation .10. What is the probability
that a randomly selected package weighs
between 0.80 and 0.85 pounds?
P(.80  x  .85) 
P(2  z  1.5) 
.0668 .0228 .0440
What is the weight of a package
such that only 1% of all packages
exceed this weight?
P( x  ?)  .01
? 1
P( z 
)  .01
.1
? 1
From Table 3,
 2.33
.1
?  2.33(.1)  1  1.233