Section 4.1
Probability Distributions
Larson/Farber 4th ed
Section 4.1 Objectives
• Distinguish between discrete random variables and
continuous random variables
• Construct a discrete probability distribution and its
graph
• Determine if a distribution is a probability
distribution
• Find the mean, variance, and standard deviation of a
discrete probability distribution
• Find the expected value of a discrete probability
distribution
Random Variables
Random Variable
• Represents a numerical value associated with each
outcome of a probability distribution.
• Denoted by x
• Examples
 x = The number of single occupancy vehicles on
I5 during rush hour.
 x = The amount of CO2 levels in the atmosphere.
Random Variables
Discrete Random Variable
• Has a finite or countable number of possible
outcomes that can be listed.
• Example
 x = The number of single occupancy vehicles on I5 during rush hour.
x
0
1
2
3
4
5
Random Variables
Continuous Random Variable
• Has an uncountable number of possible outcomes,
represented by an interval on the number line.
• Example
 x = The amount of CO2 levels in the atmosphere.
x
0
1
2
3
…
24
Example: Random Variables
Decide whether the random variable x is discrete or
continuous.
1. x = The amount of water the people of Seattle drink
per day (Million Gallons per day).
Solution: Continuous random
variable
x
0
1
2
3
…
32
Example: Random Variables
Decide whether the random variable x is discrete or
continuous.
2. x = The number of umbrellas sold in Seattle in
the Month of February (in thousands).
Solution: Discrete random
variable
x
0
1
2
3
…
30
Discrete Probability Distributions
Discrete probability distribution
• Lists each possible value the random variable can
assume, together with its probability.
• Must satisfy the following conditions:
In Words
In Symbols
1. The probability of each value of
the discrete random variable is
between 0 and 1, inclusive.
0  P (x)  1
2. The sum of all the probabilities is
1.
ΣP (x) = 1
Constructing a Discrete Probability
Distribution
Let x be a discrete random variable with possible
outcomes x1, x2, … , xn.
1. Make a frequency distribution for the possible
outcomes.
2. Find the sum of the frequencies.
3. Find the probability of each possible outcome by
dividing its frequency by the sum of the frequencies.
4. Check that each probability is between 0 and 1 and
that the sum is 1.
Example: Constructing a Discrete
Probability Distribution
Australia's unique wildlife apparently risks being hunted to
extinction unless the cat population is controlled. Native
fauna is ill-equipped to deal with this naturalized predator.
Three types of cat are recognized: domestic cats which are
wholly dependent on humans, unowned stray cats which
rely on humans to some extent and feral cats whose
reliance on humans is minimal. They can breed 3 or 4 times
a year, averaging 4 kittens per litter and can rapidly
establish colonies wherever there is a good food source.
Larson/Farber 4th ed
Example: Constructing a Discrete
Probability Distribution
The number of house holds with cats in Darwin Australia
from a survey of 2668 house holds.
Cats, x
Frequency, f
0
1941
1
349
2
203
3
78
4
57
5
40
∑f=
2668
Solution: Constructing a Discrete
Probability Distribution
• Divide the frequency of each score by the total
number of individuals in the study to find the
probability for each value of the random variable.
• Discrete probability distribution:
x
0
1
2
3
4
5
P(x)
0.73
0.13
0.08
0.03
0.02
0.01
Larson/Farber 4th ed
Solution: Constructing a Discrete
Probability Distribution
x
0
P(x) 0.73
1
2
3
4
5
0.13
0.08
0.03
0.02
0.01
This is a valid discrete probability distribution since
1. Each probability is between 0 and 1, inclusive,
0 ≤ P(x) ≤ 1.
2. The sum of the probabilities equals 1,
ΣP(x) = 0.73 + 0.13 + 0.08 + 0.03 + 0.02 + 0.01 = 1.
Larson/Farber 4th ed
Solution: Constructing a Discrete
Probability Distribution
• Histogram
Because the width of each bar is one, the area of each
bar is equal to the probability of a particular outcome.
Larson/Farber 4th ed
Mean
Mean of a discrete probability distribution
• μ = ΣxP(x)
• Each value of x is multiplied by its corresponding
probability and the products are added.
Larson/Farber 4th ed
Example: Finding the Mean
The probability distribution for the cats in Darwin. Find
the mean.
Solution:
x
P(x)
0
0.73
0*0.73
0
1
0.13
1*0.13
0.13
2
0.08
2*0.08
0.15
3
0.03
3*0.03
0.09
4
0.02
4*0.02
0.09
5
0.01
5*0.01
0.07
xP(x)
Variance and Standard Deviation
Variance of a discrete probability distribution
• σ2 = Σ(x – μ)2P(x)
Standard deviation of a discrete probability
distribution
•
Larson/Farber 4th ed
Example: Finding the Variance and
Standard Deviation
The probability distribution for cats per household in
Darwin. Find the variance and standard deviation. ( μ =
0.53)
0
xP(x)
P(x) x - μ (x - μ)2 (x - μ)2P(x)
0.28
0.21
00.73
0.73 -0.53
1
10.13
0.13
0.47
0.22
0.03
2
20.08
0.08
1.47
2.16
0.16
3
30.03
0.03
2.47
6.10
0.18
4
40.02
0.02
3.47
12.03
0.26
5
50.01
0.01
4.47
19.97
0.30
x
Larson/Farber 4th ed
1.14
Solution: Finding the Variance and
Standard Deviation
Variance:
Standard Deviation:
Larson/Farber 4th ed
(x - μ)2P(x)
0.21
0.03
0.16
0.18
0.26
0.30
1.14
Expected Value
Expected value of a discrete random variable
• Equal to the mean of the random variable.
• E(x) = μ = ΣxP(x)
Larson/Farber 4th ed
Example: Finding an Expected Value
#46 - A charity organization is selling $4 raffle tickets
as part of a fund-raising program. The first prize is a
boat valued at $3150 and the second prize is a camping
tent valued at $450. The remaining 15 prizes are $25
gift certificates. The number of tickets sold is 5000.
What is the expected value of your gain?
Larson/Farber 4th ed
Solution: Finding an Expected Value
• To find the gain for each prize, subtract
the price of the ticket from the prize:
 Your gain for the $3150 prize is $3150 – $4 =
$3146
 Your gain for the $450 prize is $450 – $4 = $446
 Your gain for the $25 prizes is $25 – $4 = $21
• If you do not win a prize, your gain is $0 – $4 = –$4
Larson/Farber 4th ed
Solution: Finding an Expected Value
• Probability distribution for the possible gains
(outcomes)
Gain, x
3150 - 4 = 3146
450 - 4 = 446
25 - 4 = 21
–$4
P(x)
You can expect to lose an average of $3.21 for each ticket
you buy.
Larson/Farber 4th ed
Section 4.1 Summary
• Distinguished between discrete random variables and
continuous random variables
• Constructed a discrete probability distribution and its
graph
• Determined if a distribution is a probability
distribution
• Found the mean, variance, and standard deviation of
a discrete probability distribution
• Found the expected value of a discrete probability
distribution
Larson/Farber 4th ed