1. Probability Distributions
Section 4.1
Probability Distributions
Larson/Farber 4th ed

2. 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

3. 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.

4. 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 I- 5 during rush hour. x 1 5 3 2 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 1 24 3 2 …

6. Example: Random Variables
Decide whether the random variable x is discrete or continuous.
x = The amount of water the people of Seattle drink per day (Million Gallons per day).
Solution: Continuous random variable x 1 32 3 2 …

7. Example: Random Variables
Decide whether the random variable x is discrete or continuous.
x = The number of umbrellas sold in Seattle in the Month of February (in thousands).
Solution: Discrete random variable x 1 30 3 2 …

8. Discrete Probability Distributions
Lists each possible value the random variable can assume, together with its probability.
Must satisfy the following conditions:
In Words In Symbols
The probability of each value of the discrete random variable is between 0 and 1, inclusive.
The sum of all the probabilities is 1.
0 ≤ P (x) ≤ 1
ΣP (x) = 1

9. Constructing a Discrete Probability Distribution
Let x be a discrete random variable with possible outcomes x1, x2, … , xn.
Make a frequency distribution for the possible outcomes.
Find the sum of the frequencies.
Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies.
Check that each probability is between 0 and 1 and that the sum is 1.

10. 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

11. 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
1941 1
349 2
203 3 78 4 57 5 40
∑ f =
2668

12. 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 1 2 3 4 5
P(x)
0.73
0.13
0.08
0.03
0.02
0.01
Larson/Farber 4th ed

13. Solution: Constructing a Discrete Probability Distributionx 1 2 3 4 5
P(x)
0.73
0.13
0.08
0.03
0.02
0.01
This is a valid discrete probability distribution since
Each probability is between 0 and 1, inclusive, 0 ≤ P(x) ≤ 1.
The sum of the probabilities equals 1, ΣP(x) = = 1.
Larson/Farber 4th ed

14. 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

15. 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

16. Example: Finding the Mean
The probability distribution for the cats in Darwin. Find the mean. x
P(x)
0.73 1
0.13 2
0.08 3
0.03 4
0.02 5
0.01
xP(x)
0*0.73
1*0.13
0.13
2*0.08
0.15
3*0.03
0.09
4*0.02
5*0.01
0.07
Solution:

17. 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

18. 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) x
P(x)
0.73 1
0.13 2
0.08 3
0.03 4
0.02 5
0.01 x
P(x)
x - μ
(x - μ)2
(x - μ)2P(x)
0.73
-0.53
0.28
0.21 1
0.13
0.47
0.22
0.03 2
0.08
1.47
2.16
0.16 3
2.47
6.10
0.18 4
0.02
3.47
12.03
0.26 5
0.01
4.47
19.97
0.30
1.14
Larson/Farber 4th ed

19. Solution: Finding the Variance and Standard Deviation
(x - μ)2P(x)
0.21
0.03
0.16
0.18
0.26
0.30
1.14
Variance:
Standard Deviation:
Larson/Farber 4th ed

20. 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

21. 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

22. 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

23. Solution: Finding an Expected Value
Probability distribution for the possible gains (outcomes)
Gain, x
= 3146
= 446
= 21
–$4
P(x)
You can expect to lose an average of $3.21 for each ticket you buy.
Larson/Farber 4th ed

24. 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