### Chapter 1: Statistics

```Chapter 5: Probability
Distributions
(Discrete Variables)
P( x )
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
x
Chapter Goals
• Combine the ideas of frequency
distributions and probability to form
probability distributions.
• Investigate discrete probability distributions
and study measures of central tendency and
dispersion.
• Study the binomial random variable.
5.1: Random Variables
• Bridge between experimental outcomes and
statistical analysis.
• Each outcome in an experiment is assigned
to a number.
• This suggests the idea of a function.
Random Variable: A variable that assumes a unique
numerical value for each of the outcomes in the sample space
of a probability experiment.
Note:
1. Used to denote the outcomes of a probability experiment.
2. Each outcome in a probability experiment is assigned to a
unique value.
3. Illustration:
S
Random Variable
Outcomes










 2 1 0
1
2
Examples of random variables:
1. Let the number of computers sold per day by a local
merchant be a random variable. Integer values ranging
from zero to about 50 are possible values.
2. Let the number of pages in a mystery novel at a bookstore
be a random variable. The smallest number of pages is
125 while the largest number of pages is 547.
3. Let the time it takes an employee to get to work be a
random variable. Possible values are 15 minutes to over 2
hours.
4. Let the volume of water used by a household during a
month be a random variable. Amounts range up to several
thousand gallons.
5. Let the number of defective components in a shipment of
1000 be a random variable. Values range from 0 to 1000.
Discrete Random Variable: A quantitative random variable
that can assume a countable number of values.
Intuitively, a discrete random variable can assume values
corresponding to isolated points along a line interval. That is,
there is a gap between any two values.
Note: Usually associated with counting.
Continuous Random Variable: A quantitative random
variable that can assume an uncountable number of values.
Intuitively, a continuous random variable can assume any
value along a line interval, including every possible value
between any two values.
Note: Usually associated with a measurement.
Example: Determine whether the following random variables
are discrete or continuous.
1. The barometric pressure at 12:00 PM.
2. The length of time it takes to complete a statistics exam.
3. The number of items in the shopping cart of the person in
front of you at the checkout line.
4. The weight of a home grown zucchini.
5. The number of tickets issued by the PA State Police during
a 24 hour period.
6. The number of cans of soda pop dispensed by a machine
placed in the Mathematics building on campus.
7. The number of cavities the dentist discovers during your
next visit.
5.2 Probability Distributions of a
Discrete Random Variable
• Need a complete description of a discrete
random variable.
• This includes all the values the random
variable may assume and all of the
associated probabilities.
• This information may be presented in a
variety of ways.
Probability Distribution: A distribution of the probabilities
associated with each of the values of a random variable. The
probability distribution is a theoretical distribution; it is used
to represent populations.
Note:
1. The probability distribution tells you everything you need
to know about the random variable.
2. The probability distribution may be presented in the form
of a table, chart, function, etc.
Probability Function: A rule that assigns probabilities to the
values of the random variable.
Example: The number of people staying in a randomly
selected room at a local hotel is a random variable ranging in
value from 0 to 4. The probability distribution is known and
is given in various forms below.
x
P (x )
0
2/15
1
4/15
2
5/15
3
3/15
4
1/15
Note:
1. This chart implies the only values x takes on are 0, 1, 2, 3,
and 4.
2. P ( the random variable x equals 2)
5
 P ( 2) 
15
A line representation of the Hotel Room probability
distribution:
P( x )
0
.
4
0
.
3
0
.
2
0
.
1
0
.
0
0 1 2 3 4x
A histogram may be used to present a probability
distribution.
A histogram for the Hotel Room probability distribution:
P( x )
0
.
4
0
.
3
0
.
2
0
.
1
0
.
0
0 1 2 3 4
x
Note:
1. The histogram of a probability distribution uses the
physical area of each bar to represent its assigned
probability.
2. In the Hotel Room probability distribution: the width of
each bar is 1, so the height of each bar is equal to the
assigned probability, which is the area of each bar.
3. The idea of area representing probability is important in
the study of continuous random variables.
Reminder: Every probability function must satisfy the two
basic properties of probability.
1. The probability assigned to each value of the random
variable must be between 0 and 1, inclusive:
0  P( x)  1
2. The sum of the probabilities assigned to all the values of
the random variable must equal 1:
 P( x)  1
all x
5.3: Mean and Variance of a
Discrete Probability
Distribution
Describe the center and spread of a
population.
m, s, s2 : population parameters.
Population parameters are usually unknown
values (we would like to estimate).
Note:
1. x is the mean of the sample.
2. s2 and s are the variance and standard deviation of the
sample.
3. x , s2, and s are called sample statistics.
4. m (lowercase Greek letter “mu”) is the mean of the
population.
5. s2 (“sigma squared”) is the variance of the population.
6. s (lowercase Greek letter “sigma”) is the standard
deviation of the population.
7. m, s2, and s are called population parameters. (A
parameter is a constant. m, s2, and s are typically unknown
values.)
Mean of a Discrete Random Variable:
The mean, m, of a discrete random variable x is found by
multiplying each possible value of x by its own probability
and then adding all the products together.
m  [ xP( x)]
Note:
1. The mean is the average value of the random variable,
what happens on average.
2. The mean is not necessarily a value of the random
variable.
Variance of a Discrete Random Variable:
Variance, s2, of a discrete random variable x is found by
multiplying each possible value of the squared deviation from
the mean, (x  m)2, by its own probability and then adding all
the products together.
s 2   [( x  m ) 2 P ( x )]
  [ x P ( x )] 
2
 [ xP( x)]
2
  [ x 2 P ( x )]  m 2
Standard Deviation of a Discrete Random Variable:
The positive square root of the variance.
s  s2
Example: The number of standby passengers who get seats on
a daily commuter flight from Boston to New York is a
random variable, x, with probability distribution given below
(in an extensions table). Find the mean, variance, and
standard deviation.
x
0
1
2
3
4
5
Totals
P( x )
0.30
0.25
0.20
0.15
0.05
0.05
1.00
 P( x )
(check)
xP( x )
0.00
0.25
0.40
0.45
0.20
0.25
1.55
[ xP( x)]
x 2 x 2 P( x )
0
0.00
1
0.25
4
0.80
9
1.35
16
0.80
25
1.25
4.45
2
[
x
 P( x)]
Solution:
Using the formulas for mean, variance, and standard
deviation:
m  [ xP( x)]  155
.
Note: 1.55 is not a value of the random variable (in this case).
It is only what happens on average.
s   [ x P( x )]   [ xP( x )]
2
2
2
 4.45  (155
. ) 2  4.45  2.4025  2.0475
s  s 2  2.0475  143
.
Example: The probability distribution for a random variable x
is given by the probability function
8 x
P( x ) 
15
for
x  3, 4, 5, 6, 7
Find the mean, variance, and standard deviation.
Solution:
Find the probability associated with each value by using the
probability function.
83 5
P(3) 

15 15
P(6) 
86 2

15
15
84 4
P(4) 

15 15
P(7) 
87 1

15
15
85 3
P(5) 

15 15
Use an extensions table to find the population parameters.
x
3
4
5
6
7
Totals
P( x )
5/15
4/15
3/15
2/15
1/15
15/15
xP( x )
15/15
16/15
15/15
12/15
7/15
65/15
x2
9
16
25
36
49
x 2 P( x )
45/15
64/15
75/15
72/15
49/15
305/15
2
[
x
 P( x)]
[ xP( x)]
65
m   [ xP( x )]   4.33
15
s 2   [ x 2 P( x )]   [ xP( x )] 
2
s  s 2  156
.  125
.
2
305  65
    156
.
15  15
5.4: The Binomial Probability
Distribution
• One of the most important discrete
distributions.
• Based on a series of repeated trials whose
outcomes can be classified in one of two
categories: success or failure.
• Distribution based on a binomial
probability experiment.
Binomial Probability Experiment:
An experiment that is made up of repeated trials that possess
the following properties:
1. There are n repeated independent trials.
2. Each trial has two possible outcomes (success, failure).
3. P(success) = p, P(failure) = q, and p + q = 1
4. The binomial random variable x is the count of the
number of successful trials that occur; x may take on any
integer value from zero to n.
Note:
1. Properties 1 and 2 are the two basic properties of any
binomial experiment.
2. Property 3 concerns the algebraic notation for each trial.
3. Property 4 concerns the algebraic notation for the complete
experiment.
4. Both x and p must be associated with “success.”
5. Independent trials mean that the result of one trial does not
affect the probability of success of any other trial in the
experiment. The probability of “success” remains constant
throughout the entire experiment.
Example: It is known that 40% of all graduating seniors on
campus have taken a statistics class. Five seniors are selected
at random and asked if they have taken a statistics class.
1. A trial is asking one student, repeated 5 times. The trials
are independent since the probability of taking a statistics
class for any one student is not affected by the results from
any other student.
2. Two outcomes on each trial:
taken a statistics class (success),
not taken a statistics class (failure)
3. p = P(taken a statistics class) = .40
q = P(not taken a statistics class) = .60
4. x = number of students who have taken a statistics class
Binomial Probability Function:
For a binomial experiment, let p represent the probability of a
“success” and q represent the probability of a “failure” on a
single trial; then P(x), the probability that there will be
exactly x successes on n trials is
 n x n  x
P( x )    ( p )(q ), for x  0, 1, 2, ... , or n
 x
Note:
1. The number of ways that exactly x successes can occur in
n trials:
 n
 
 x
2. The probability of exactly x successes: px
3. The probability that failure will occur on the remaining
(n - x) trials: qn - x
Note:
The number of ways that exactly x successes can occur in a
set of n trials is represented by the symbol
 n
 
 x
1. Must always be a positive integer.
2. Called the binomial coefficient.
3. Found by using the formula:
 n
n!
 
 x x !(n  x )!
n! is an abbreviation for n factorial.
n!  n(n  1)(n  2)(3)(2)(1)
6!  6  5  4  3  2  1  720
Example: According to a recent study, 65% of all homes in a
certain county have high levels of radon gas leaking into their
basements. Four homes are selected at random and tested for
radon. The random variable x is the number of homes with
high levels of radon (out of the four).
Properties:
1. There are 4 repeated trials: n = 4. The trials are
independent.
2. Each test for radon is a trial, and each test has two
3. p = P(radon) = .65, q = P(no radon) = .35
p+q=1
4. x is the number of homes with high levels of radon,
possible values: 0, 1, 2, 3, 4
The binomial probability function:
 4
P( x )    (.65) x (.35) 4  x , for x  0, 1, 2, 3, 4
 x
 4
P(0)    (.65) 0 (.35) 4  (1)(1)(0.0150)  0.0150
 0
 4
P(1)    (.65)1 (.35) 3  (4)(0.65)(0.0429)  01115
.
 1
 4
P(2)    (.65) 2 (.35) 2  (6)(0.4225)(01225
.
)  0.3105
 2
 4
P(3)    (.65) 3 (.35)1  (4)(0.2746)(0.35)  0.3845
 3
 4
P(4)    (.65) 4 (.35) 0  (1)(0.1785)(1)  0.1785
 4
Example: In a certain automobile dealership, 70% of all
customers purchase an extended warranty with their new car.
For 15 customers selected at random:
1. Find the probability that exactly 12 will purchase an
extended warranty.
2. Find the probability at most 13 will purchase an extended
warranty.
Solution:
Let x be the number of customers who purchase an extended
warranty. x is a binomial random variable.
The probability function associated with this experiment:
 15
P( x )    (.7) x (.3)15 x , for x  0, 1, 2, ... ,15
 x
Probability exactly 12 purchase an extended warranty:
 15
P(12)    (.7)12 (.3) 3 .1700
 12
Probability at most 13 purchase an extended warranty:
P( x  13)  P(0)  P(1)    P(13)
 1  P(14)  P(15)
15  14 1 15  15
0
 1   (.7) (.3)   (.7) (.3) 
15 
14 

 1  [.0305  .0047]
 1  .0352  .9648
Note:
1. The value of many binomial probabilities (n < 15 and
common values of p) are found in Table 2, Appendix B.
2. Minitab has special commands for computing binomial
probabilities or cumulative probabilities.
PDF 12;
Binomial 15 .7.
Probability Density Function
Binomial with n = 15 and p = 0.700000
x
P( X = x)
12.00
0.1700
3. Many graphing calculators also have built-in functions for
computing binomial probabilities and cumulative
probabilities.
Notation:
If x is a binomial random variable with n trials and probability
of a success p, this is often denoted: x ~ B(n, p).
Example: Suppose x is a binomial random variable with n =
18 and p = .35. A convenient notation to identify this random
variable is: x ~ B(18, .35).
5.5: Mean and Standard
Deviation of the Binomial
Distribution
• Population parameters of the binomial
distribution help to describe the distribution.
• Mean and standard deviation indicate where
the distribution is centered and the spread of
the distribution.
The mean and standard deviation of a theoretical binomial
distribution can be found by using the following two
formulas:
m  np
s  npq
Note:
1. Mean is intuitive: number of trials multiplied by the
probability of a success.
2. The variance of a binomial probability distribution is:
s   npq   npq
2
2
Example: Find the mean and standard deviation of the
binomial distribution when n = 18 and p = .75.
Solution:
n = 18, p = .75,
q = 1  .75 = .25
m  np  (18)(.75)  135
.
s  npq  (18)(.75)(.25)  3.375  18371
.
The probability function is
 18
P( x )    (.75) x (.25)18 x for x  0, 1, 2, ... , 18
 x
Table of values and probabilities:
x
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
15.00
16.00
17.00
18.00
P( X = x)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0010
0.0042
0.0139
0.0376
0.0820
0.1436
0.1988
0.2130
0.1704
0.0958
0.0338
0.0056
Histogram:
m
s
P( x )
0
.
2
2
0
.
2
0
0
.
1
8
0
.
1
6
0
.
1
4
0
.
1
2
0
.
1
0
0
.
0
8
0
.
0
6
0
.
0
4
0
.
0
2
0
.
0
0
x
0
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
```