### Chapter 5 - Dalton State College

```PROBABILITY &
STATISTICS FOR P-8
TEACHERS
Chapter 5
Probability Distributions
PROBABILITY DISTRIBUTIONS
We will now combine the methods of
descriptive statistics (Chapter 2 and 3) and
those of probability (Chapter 4) to describe
and analyze
probability distributions.
Probability Distributions describe what will
probably happen instead of what actually did
happen, and they are often given in the format
of a graph, table, or formula.
COMBINING DESCRIPTIVE
METHODS AND PROBABILITIES
In this chapter we will construct probability distributions
by presenting possible outcomes along with the relative
frequencies we expect.
PROBABILITY DISTRIBUTIONS
A
random variable is a variable whose
values are determined by chance.
A
discrete probability distribution
consists of the values a random variable can
assume and the corresponding probabilities of
the values.
 The
sum of the probabilities of all events in
a sample space add up to 1. Each probability
is between 0 and 1, inclusively.
RANDOM VARIABLES
Two types of random variables:
A
discrete random variable can
assume a countable number of values.

Number of steps to the top of the Eiffel Tower
A
continuous random variable can
assume any value along a given interval
of a number line.

The time a tourist stays at the top
once s/he gets there
RANDOM VARIABLES

Discrete random variables
 Number of sales
 Number of calls
 Shares of stock
 People in line
 Mistakes per page
 Continuous random
variables
 Length
 Depth
 Volume
 Time
 Weight
PROBABILITY DISTRIBUTIONS
 The
probability distribution of a
discrete random variable is a graph,
table or formula that specifies the
probability associated with each possible
outcome the random variable can
assume.
 0 ≤ P(x) ≤ 1 for all values of x
 P(x) = 1
PROBABILITY DISTRIBUTIONS
Is the following a probability
distribution?
Not a
probability
distribution
x
0
1
P(x)
2
3
4
0.22
0.10
0.30
5
-0.01
0.16
0.18
0.95
P(x) < 0
∑P(x) ≠ 1
PROBABILITY DISTRIBUTION
Construct a probability
distribution for tossing a coin
twice and recording the number of
x
P(x)
0
0.25
1
0.50
2
0.25
HH
HT
TH
TT
PROBABILITY HISTOGRAM
A probability histogram is a histogram in
which the horizontal axis corresponds to the
value of the random variable and the vertical
axis represents the probability of that value of
the random variable.
PROBABILITY HISTOGRAM
Draw a probability histogram of
the probability distribution to the
right, which represents the
number of DVDs a person rents
from a video store during a single
visit.
Probability
DVDs Rented at a Video Store
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
Number of DVDs Rented
4
5
x
P(x)
0
0.06
1
0.58
2
0.22
3
0.10
4
0.03
5
0.01
PROBABILITY DISTRIBUTION
A survey was completed regarding how many siblings
are in your family. Let X denote the number of
siblings of a randomly selected student.
a. Determine the
probability distribution
of the random variable x.
b. Construct a probability
histogram for the
random variable x.
PROBABILITY DISTRIBUTION
Solution
Probability Distribution
Probability Histogram
PROBABILITY DISTRIBUTION
Just like with any distribution, we would like
to analyze the data. The mean and standard
deviation are the most common measurements.
To find these measurements, we will treat the
probability distribution just like a frequency
distribution.
Since probability distributions represent
theoretical data, we will treat the results as a
true population.
MEAN AND STANDARD DEVIATION
Mean: µ = ∑ x P(x)
Variance: σ2 = ∑ (x – µ)2 P(x)
or
σ2 = ∑ x2 P(x) – µ2
MEAN & STANDARD DEVIATION
Find the mean and standard deviaton of the
number of spots that appear when a die is
tossed.
Outcome
x
Probability
P(x)
x P(x)
1
1/6
1/6
2
1/6
2/6
3
1/6
3/6
4
1/6
4/6
5
1/6
5/6
6
1/6
6/6
21/6
µ = ∑ x P(x)
= 21/6
= 3.5
MEAN & STANDARD DEVIATION
Find the mean and standard deviaton of the
number of spots that appear when a die is
tossed.
Outcome
x
Probability
P(x)
x2 P(x)
1
1/6
1/6
σ2 = ∑ x2 P(x) – µ2
2
1/6
4/6
3
1/6
9/6
4
1/6
16/6
5
1/6
25/6
6
1/6
36/6
σ = √ 105/36
91/6
= 1.708
= 91/6 – (21/6)2
= 105/36
MEAN & STANDARD DEVIATION
The probability distribution shown represents
the number of trips of five nights or more that
American adults take per year. (That is, 6% do
not take any trips lasting five nights or more,
70% take one trip lasting five nights or more
per year, etc.) Find the mean.
MEAN & STANDARD DEVIATION
# Trips
x
Probability
P(x)
x P(x)
x2 P(x)
0
0.06
0.00
0.00
1
0.70
0.70
0.70
2
0.20
0.40
0.80
3
0.03
0.09
0.27
4
0.01
0.04
0.16
1.23
1.93
µ = ∑ x P(x) = 1.23
σ2 = ∑ x2 P(x) – µ2 = 1.93 – (1.23)2 = 0.4171
σ = √0.4171 = 0.6458
EXPECTATION
 The
expected value, or expectation, of a
discrete random variable of a probability
distribution is the theoretical average of the
variable.
 The
expected value is, by definition, the mean
of the probability distribution.
E(x) = µ = ∑ x P(x)
WINNING TICKETS
One thousand tickets are sold at \$1 each for four prizes
of \$100, \$50, \$25, and \$10. After each prize drawing,
the winning ticket is then returned to the pool of tickets.
What is the expected value if you purchase a ticket?
Winnings = prize amount – \$1 ticket price
Winnings
x
Probability
P(x)
x P(x)
99
1/1000
99/1000
49
1/1000
49/1000
24
1/1000
24/1000
9
1/1000
9/1000
-1
996/1000
-996/1000
-815/1000
E(x) = ∑ x P(x)
= –0.815
On average, you
will lose 82 cents
for every dollar
you spend
WINNING TICKETS (ALTERNATE APPROACH)
One thousand tickets are sold at \$1 each for four prizes
of \$100, \$50, \$25, and \$10. After each prize drawing,
the winning ticket is then returned to the pool of tickets.
What is the expected value if you purchase a ticket?
Winnings
x
Probability
P(x)
x P(x)
100
1/1000
100/1000
50
1/1000
50/1000
25
1/1000
25/1000
10
1/1000
10/1000
0
996/1000
0/1000
185/1000
E(x)
= ∑ x P(x) –
initial cost
= 0.185 – \$1
= –0.815
THE BINOMIAL DISTRIBUTION
 Many
types of probability problems have
only two possible outcomes or they can
be reduced to two outcomes.
 Examples
include: when a coin is tossed
it can land on heads or tails, when a baby
is born it is either a boy or girl, etc.
THE BINOMIAL DISTRIBUTION
The binomial experiment is a probability
experiment that satisfies these requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be
independent of each other.
4. The probability of success must remain the
same for each trial.
NOTATION FOR THE BINOMIAL
DISTRIBUTION
p
The numerical probability of success
q
The numerical probability of failure
n
The number of trials
x
The number of successes
Note that x = 0, 1, 2, 3,...,n
THE BINOMIAL DISTRIBUTION
In a binomial experiment, the probability of
exactly X successes in n trials is
n!
X
n X
P X  
 p q
 n - X ! X !
or
P X  
n Cx
number of possible
desired outcomes
 p X  q n X
probability of a
desired outcome
THE BINOMIAL DISTRIBUTION
 The

Binomial Probability Distribution
the probability of exactly x successes in n trials is
P(x) = nCx
x
p
n-x
q
THE BINOMIAL DISTRIBUTION

If 40% of the class is
female, what is the
probability that 1 of
the next 2 students
walking in will be
female?
P(x) = nCx px qn-x
P(1) = 2C1 p1 q2-1
= (2) (.4)1 (.6)1
= 0.48
n = 2 (# trials)
x = 1 (# successes)
p = .4 (probability of success)
q = .6 (probability of failure)
THE BINOMIAL DISTRIBUTION
A survey found that one out of five Americans
say he or she has visited a doctor in any given
month. If 10 people are selected at random, find
the probability that exactly 3 will have visited a
doctor last month.
n = 10
x=3
p = 1/5
q = 4/5
P(x) = nCx px qn-x
P(3) = 10C3 p3 q10-3
3
10!  1 
P  3 
 
7!3!  5 
7
4
    0.201
5
THE BINOMIAL DISTRIBUTION
Individual baseball cards, chosen at random from a
set of 20, are given away inside cereal boxes. Stan
needs one more card to complete his set so he buys
five boxes of cereal. What is the probability that he
will complete his set?
P(x) = nCx px qn-x
n=5
x=1
p = 1/20
q = 19/20
P(1) = 5C1 p1 q5-1
= (5) (.05)1 (.95)4
= 0.204
The probability of Stan completing
his set is 20%.
THE BINOMIAL DISTRIBUTION
A test consists of 10 multiple choice questions, each
with four possible answers. To pass the test, one
must answer at least nine questions correctly. Find
the probability of passing, if one were to guess the
P(x successes) = nCx px qn - x
P(x ≥ 9 successes) = P(9 successes) + P(10 successes)
n = 10
9
1
x = 9, 10
 1    3  
P(x ≥ 9) 10 C9           
4
4
p = 1/4
q = 3/4
= 0.000028610 +
= 0.000029564
10
0
1
 
 3  
10 C10          
4
4
0.000000954
The probability of
passing is 0.003%.
THE BINOMIAL DISTRIBUTION
A survey from Teenage Research Unlimited
(Northbrook, Illinois) found that 30% of teenage
consumers receive their spending money from part-time
jobs. If 5 teenagers are selected at random, find the
probability that at least 3 of them will have part-time
jobs.
n=5
p = 0.3
q = 0.7
x = 3, 4, 5
5!
3
2
P  3 
  0.30    0.70   0.132
2!3!
5!
4
1
P  4 
  0.30    0.70   0.028
1!4!
5!
5
0
P  5 
  0.30    0.70   0.002
0!5!
P(x ≥ 3) = 0.132 + 0.028 + 0.002 = 0.162
THE BINOMIAL DISTRIBUTION
A family has nine children. What is the
probability that there is at least one girl?
This can be best solved using the compliment,
that is, the probability of zero girls:
n=9
p = 0.5
q = 0.5
x=0
P(x successes) = nCx px qn - x
P (0) = 9C0 p0 q9 - 0
= (1) (.5)0 (.5)9
= 0.001953
P(x ≥ 1) = 1 – P(0)
= 1 – 0.001953
= 0.998
BINOMIAL DISTRIBUTION
Create a probability distribution table for
tossing a coin 3 times
P(x) = nCx px qn-x
x
P(x)
0
0.125
1
0.375
2
0.375
3
0.125
P(0)
P(1)
P(2)
P(3)
= 3C0 p0 q3-0
= (1) (.5)0 (.5)3
= (3) (.5)1 (.5)2
= (3) (.5)2 (.5)1
= (1) (.5)3 (.5)0
= 0.125
= 0.375
= 0.375
= 0.125
TOSSING COINS
A coin is tossed 3 times. Find the probability of
getting exactly two heads, using Table B.
n  3, p  12  0.5, X  2  P  2   0.375
BINOMIAL DISTRIBUTION
(a) Construct a binomial probability histogram
with n = 8 and p = 0.15.
(b) Construct a binomial probability histogram
with n = 8 and p = 0. 5.
(c) Construct a binomial probability histogram
with n = 8 and p = 0.85.
For each histogram, comment on the shape of the
distribution.
THE BINOMIAL DISTRIBUTION
The mean, variance, and standard deviation
of a variable that has the binomial distribution
can be found by using the following formulas.
Mean:   np
Variance:   npq
2
Standard Deviation:   npq
THE BINOMIAL DISTRIBUTION
Insurance Co. reported that 2% of all American births
result in twins. If a random sample of 8000 births is
taken, find the mean, variance, and standard deviation of
the number of births that would result in twins.
  np  8000  0.02   160
  npq  8000  0.02 0.98  156.8  157
2
  npq  8000  0.02  0.98   12.5  13
THE BINOMIAL DISTRIBUTION
According to the Experian Automotive, 35% of all carowning households have three or more cars. Find the
mean and standard deviation for a random sample of
400 homes.
µ = np
σ = √npq
= (400)(0.35)
= √(400)(0.35)(0.65)
= 140
= 9.54
```