1. Probability Distributions
§ 4.1
Probability Distributions

2. Random Variables
A random variable x represents a numerical value associated with each outcome of a probability distribution.
A random variable is discrete if it has a finite or countable number of possible outcomes that can be listed. x 2 10 6 4 8
A random variable is continuous if it has an uncountable number of possible outcomes, represented by the intervals on a number line. x 2 10 6 4 8

3. Random Variables Example:
Decide if the random variable x is discrete or continuous.
a.) The distance your car travels on a tank of gas
The distance your car travels is a continuous random variable because it is a measurement that cannot be counted. (All measurements are continuous random variables.)
b.) The number of students in a statistics class
The number of students is a discrete random variable because it can be counted.

4. Random Variables Checkpoint:
Decide if the random variable x is discrete or continuous.
a.) The number of stocks in the Dow Jones Industrial Average that have share price increases on any given day
The number of stocks whose share prices increase can be counted, so x it is a discrete variable
b.) The volume of bottled water in a 32-ounce container
The amount of water can be any volume between 0 and 32 ounces, so x is a continuous variable.

5. Discrete Probability Distributions
A discrete probability distribution lists each possible value the random variable can assume, together with its probability. A probability distribution 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

6. Verifying Probability Distributions
Verify that the distribution is a probability distribution.
Days of rain, x
Probability P(x)
0.216 1
0.432 2
0.288 3
0.064

7. Probability Distribution Checkpoint
Decide whether each distribution is a probability distribution. x 5 6 7 8
P(x)
0.28
0.21
0.43
0.15 x 1 2 3 4
P(x)
𝟏 𝟐
𝟏 𝟒
𝟓 𝟒 -1

8. Constructing a Discrete Probability Distribution
Guidelines
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.

9. Graphing a Discrete Probability Distribution Guided Practice
An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were rated on a score from 1 to 5, where 1 was extremely passive and 5 extremely aggressive. A score of 3 indicated neither trait. The results are shown in the table. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram
Score, x 1 2 3 4 5
Frequency, f 24 33 42 30 21

10. Passive-Aggressive Traitsx
P (x) 1 2 3 4 5 x 1 2 3 4 5 f 24 33 42 30 21
Σf =
Σ P(x) =

11. Graphing a Discrete Probability Distribution Checkpoint
A company tracks the number of sales new employees make each day during a 100-day probationary period. The results for one new employee are shown. Construct and graph a probability distribution.
Sales per day, x
Number of days, f 16 1 19 2 15 3 21 4 9 5 10 6 8 7

12. Number of Sales x P (x) 1 2 3 4 5 6 7 x 1 2 3 4 5 6 7 f 16 19 15 21 91 2 3 4 5 6 7 x 1 2 3 4 5 6 7 f 16 19 15 21 9 10 8
Σf =
Σ P(x) =

13. Mean The mean of a discrete random variable is given by μ = Σ x∙P(x).
Each value of x is multiplied by its corresponding probability and the products are added.
Example:
Find the mean of the probability distribution for the sum of the two spins. x
P (x) 2
0.0625 3
0.375 4
0.5625
xP (x)
2(0.0625) = 0.125
3(0.375) = 1.125
4(0.5625) = 2.25
ΣxP(x) = 3.5
The mean for the two spins is 3.5.

14. Finding the Mean of a Probability Distribution
Calculate the mean x
P (x)
x ∙ P (x) 1
0.16 2
0.22 3
0.28 4
0.20 5
0.14
Σ P(x) =
μ = Σ x ∙ P(x)=

15. Finding the Mean of a Probability Distribution
Calculate the mean x
P (x)
x ∙ P (x)
.16 1
.19 2
.15 3
.21 4
.09 5 .1 6
.08 7
.02
Σ P(x) = 1.00
μ = Σ x ∙ P(x)=

16. Variance The variance of a discrete random variable is given by
2 = Σ(x – μ)2P (x). x
P (x)
x ∙ P (x)
x - μ
(x – μ)2
P(x) (x – μ)2 1
0.16 2
0.22 3
0.28 4
0.20 5
0.14
Σ P(x) = 1
μ = Σ x ∙ P(x) = 2.94
σ2 = Σ P(x) ∙ (x – μ)2

17. Variance The variance of a discrete random variable is given by:
2 = Σ(x – μ)2P (x). x
P (x)
x ∙ P (x)
x - μ
(x – μ)2
P(x) (x – μ)2
.16 1
.19 2
.15
.30 3
.21
.63 4
.09
.36 5 .1
.50 6
.08
.48 7
.02
.14
Σ P(x) = 1
μ = Σ x ∙ P(x) = 2.6
σ2 = Σ P(x) ∙ (x – μ)2

18. Standard Deviation σ = √ (Σ P(x) ∙ (x – μ)2) =
The standard deviation of a discrete random variable is given by
σ = √ (Σ P(x) ∙ (x – μ)2) =

19. Expected Value
The expected value of a discrete random variable is equal to the mean of the random variable.
Expected Value = E (x) = μ = ΣxP(x).
Example:
At a raffle, 500 tickets are sold for $1 each for two prizes of $100 and $50. What is the expected value of your gain?
Your gain for the $100 prize is $100 – $1 = $99.
Your gain for the $50 prize is $50 – $1 = $49.
Write a probability distribution for the possible gains (or outcomes).
Continued.

20. Expected Value Example continued:
At a raffle, 500 tickets are sold for $1 each for two prizes of $100 and $50. What is the expected value of your gain?
Gain, x
P (x)
E(x) = ΣxP(x).
$99
$49
–$1
Because the expected value is negative, you can expect to lose $0.70 for each ticket you buy.
Winning no prize