UNIT 8
Discrete Probability Distributions
Intro Activity Summary (Ch 4 Review)
HHHH
HHTT
THHT
TTTT
HHHT
HTHT
TTTH
2x2x2x2 =
HHTH
HTTH
TTHT
16 outcomes
HTHH
THTH
THTT
THHH
TTHH
HTTT
Intro Activity Summary (Ch 3 Review)
Possible
Outcomes
0
1
2
3
4
Frequency
R.F.
C.F
C.R.F
We have already discussed that a variable is a characteristic or
attribute that can assume different values (eye color, height, weight,
etc.). Because we will now be working with variables associated with
probability, we call them random variables.
A random variable is a variable (typically represented by X)
whose values are determined by chance.
Chapter 1 review:
Discrete variable have values that can be counted.
Continuous variables are obtained from data that can be measured
rather than counted.
A probability distribution consists of the values a random
variable can assume and the corresponding probabilities of
the values.
Let’s look at flipping 3 coins simultaneously.
List the outcomes and their probabilities:
TTT
TTH
THT
HTT
1
8
1
8
1
8
1
8
HHT HTH THH HHH
1
8
1
8
1
8
1
8
Now let’s say that we are only interested in the number of heads. We
would let X be the random variable for the number of heads.
No heads
One Head
Two Heads
Three heads
TTT
TTH THT HTT
HHT HTH THH
HHH
1
8
1
8
1
8
1
8
3
8
1
8
1
8
1
8
3
8
1
8
1
8
1
8
We end up with a probability distribution that looks like this:
Number of heads X
0
1
2
3
Probability P(X)
1
8
3
8
3
8
1
8
We can also represent probability distributions graphically by
representing the values of X on the x axis and the probabilities P(X) on
the y axis.
Two requirements for a probability distribution.
1. The sum of the probabilities of all the events in the sample space must
equal 1; that is,  P( X )  1.
2. The probability of each event in the sample space must be between or
equal to 0 and 1; that is, 0  P( X )  1.
Determine whether each distribution is a probability
distribution.
X
0
5
10
15
20
P(X)
1
5
1
5
1
5
1
5
1
5
X
P(X)
X
0
5
10
15
P(X)
1
4
1
8
1
16
9
16
0
5
10
15
X
-1.0
1.5
0.3
0.2
P(X)
0
5
10
0.5
0.3
0.4
Finding the mean, variance, and standard
deviation for a discrete random variable.
The mean of a random variable with a discrete probability
distribution is
  X 1  P ( X 1 )  X 2  P ( X 2 )  ... X n  P ( X n )
The formula for the variance of a probability distribution is
2
2


    X  P( X )   
2
The standard deviation of a probability distribution is
 
2
Find the mean, variation, and standard deviation for the
number of spots that appear when a die is tossed.
X
1
2
3
4
5
6
P(X)
1
6
1
6
1
6
1
6
1
6
1
6
1 1 1 1 1 1
  1   2    3   4    5    6  
6 6 6 6 6 6
1  2 3  4 5 6
       
6 6 6 6 6 6
21
1
   3  3.5
6
2
 2  1  2  1  2  1  2  1  2  1  2  1 
2
  1    2    3    4    5    6      3.5
6
6
6
6
 6 
 6
2
 1   4   9   16   25   36  
2
                      3.5
 6   6   6   6   6   6  
2
 91
2
      3.5 2.92
6
2
So, 
2.92 1.71
The probability that 0, 1, 2, 3, or 4 people
will be placed on hold when they call is
18%, 34%, 23%, 21% and 4% respectively.
Pgs. 230-231 #19 – 24 all.
Write the probability distribution in standard form, graph it, find
the mean, and find the standard deviation.