1. Discrete Distributions
Random Variables
Discrete Distributions

2. Probability Distribution of a Random Variable -
A numerical variable whose value depends on the outcome of a chance experiment.
Probability Distribution of a Random Variable -
A probability distribution that gives all the possible values of the random variable and their probabilities.

3. Two types: Discrete – count of some random variable
Think: Shoe size Or
Number of birthdays
Discrete – count of some random variable
Continuous – measure of some random variable
Vs. Foot length Or
Age

4. Discrete and Continuous Random Variables
A random variable is discrete if its set of possible values is a collection of isolated points on the number line.
Possible values of a discrete random variable
A random variable is continuous if its set of possible values includes an entire interval on the number line.
Possible values of a continuous random variable
We will use lowercase letters, such as x and y, to represent random variables.

5. Discrete Probability Distribution
Gives the probabilities associated with each possible x value
Usually displayed in a table, but can be displayed with a histogram or formula

6. Discrete probability distributions
3)For every possible x value,
0 < P(x) < 1.
4) For all values of x,
S P(x) = 1.

7. Discrete and Continuous Random Variables
Discrete Random Variables
There are two main types of random variables: discrete and continuous. If we can find a way to list all possible outcomes for a random variable and assign probabilities to each one, we have a discrete random variable.
Discrete and Continuous Random Variables
Discrete Random Variables and Their Probability Distributions
A discrete random variable X takes a fixed set of possible values with gaps between. The probability distribution of a discrete random variable X lists the values xi and their probabilities pi:
Value: x1 x2 x3 …
Probability: p1 p2 p3 …
The probabilities pi must satisfy two requirements:
Every probability pi is a number between 0 and 1.
The sum of the probabilities is 1.
To find the probability of any event, add the probabilities pi of the particular values xi that make up the event.

8. Suppose you toss 3 coins & record the number of heads
Suppose you toss 3 coins & record the number of heads. What is the sample space?

9. The Random Variable X is defined as …
The number of heads tossed
# of heads
X=0 TTT
X=1 HTT,THT,TTH
X=2 HHT,HTH,THH
X=3 HHH

10. Create a probability distribution.
X = 0: TTT
X = 1: HTT THT TTH
X = 2: HHT HTH THH
X = 3: HHH
Possible outcomes:
Create a probability distribution.
Create a probability histogram.

11. Create a probability distribution.X
P(X)
Now we can use the probability distribution table to answer questions about the variable X.
What is the probability of getting exactly 2 heads?
P(X=2) =.375
What is the probability of getting at least 2 heads?
P(X>2) = P(X=2) + P(X=3) = .5
What is the probability of getting at least one head?
P(X>1) = P(X=1) + P(X=2) + P(X=3) = .875

12. Example: Babies’ Health at Birth
Apgar scores: rates babies on skin color, heart rate, muscle tone, breathing, and response when stimulate (on a scale)
Show that the probability distribution for X is legitimate.
Make a histogram of the probability distribution. Describe what you see.
Apgar scores of 7 or higher indicate a healthy baby. What is P(X ≥ 7)?
Value: 1 2 3 4 5 6 7 8 9 10
Probability:
0.001
0.006
0.007
0.008
0.012
0.020
0.038
0.099
0.319
0.437
0.053
(a) All probabilities are between 0 and 1 and they add up to 1. This is a legitimate probability distribution.
(c) P(X ≥ 7) = .908
We’d have a 91 % chance of randomly choosing a healthy baby.
(b) The left-skewed shape of the distribution suggests a randomly selected newborn will have an Apgar score at the high end of the scale. There is a small chance of getting a baby with a score of 5 or lower.

13. Why does this not start at zero?
Let x be the number of courses for which a randomly selected student at a certain university is registered.
X P(X) ?
P(x = 4) =
P(x < 4) =
What is the probability that the student is registered for at least five courses?
Why does this not start at zero?
.25
.14
P(x > 5) = .61
.39

14. Mean and Variance of Discrete Random Variables

15. Probability Distributions are also described by measures of central tendency and variability.
The MEAN of a discrete random variable X is the average of the possible outcomes of X WITH the weights (probabilities). Other names for the MEAN are the WEIGHTED AVERAGE or the EXPECTED VALUE.
Value: x1 x2 x3 …
Probability: p1 p2 p3 …
To find the mean (expected value) of X, multiply each possible value by its probability, then add all the products:

16. Winning (and losing) at Roulette
38 slots numbered 1-36, plus 0 and 00
Half of the slots 1-36 red, other half black
Both 0 and 00 are green
Suppose you place a $1 bet on red
If red, you get our original $1 back plus an addition $1
If anything else you lose

17. What is your average gain?
Money
-$1 $1
Probability
20/38
18/38
So in the long run the player loses (and the casino gains) five cents per bet!

18. Example: Apgar Scores – What’s Typical?
Consider the random variable X = Apgar Score
Compute the mean of the random variable X and interpret it in context.
Value: 1 2 3 4 5 6 7 8 9 10
Probability:
0.001
0.006
0.007
0.008
0.012
0.020
0.038
0.099
0.319
0.437
0.053
The mean Apgar score of a randomly selected newborn is This is the long-term average Agar score of many, many randomly chosen babies.
Note: The expected value does not need to be a possible value of X or an integer! It is a long-term average over many repetitions.

19. Probability Distributions are also described by measures of central tendency and variability.
The VARIANCE is an average of the squared deviation of the values of the variable X from its mean. The STANDARD DEVIATION is the square root of the variance.
Definition:
Suppose that X is a discrete random variable whose probability distribution is
Value: x1 x2 x3 …
Probability: p1 p2 p3 …
and that µX is the mean of X. The variance of X is

20. Example: Apgar Scores – How Variable Are They?
Consider the random variable X = Apgar Score
Compute the standard deviation of the random variable X and interpret it in context.
Value: 1 2 3 4 5 6 7 8 9 10
Probability:
0.001
0.006
0.007
0.008
0.012
0.020
0.038
0.099
0.319
0.437
0.053
Variance
The standard deviation of X is On average, a randomly selected baby’s Apgar score will differ from the mean by about 1.4 units.

21. Formulas for mean & variance
Found on formula card!

22. What is the mean and standard deviation of this distribution?
Let x be the number of courses for which a randomly selected student at a certain university is registered. X
P(X)
What is the mean and standard deviation of this distribution?

23. Don’t forget to change Freq back to 1!
Find the mean and standard deviation for the number of heads out of 3 tosses. X
P(X)
m = & s = .866
Don’t forget to change Freq back to 1!

24. A fair game is one where the cost to play EQUALS the expected value!
Here’s a game:
If a player rolls two dice and gets a sum of 2 or 12, he wins $20. If he gets a 7, he wins $5. The cost to roll the dice one time is $3. Is this game fair?
A fair game is one where the cost to play EQUALS the expected value!

25. NO, since m = $1.944 which is less than it cost to play ($3).
If a player rolls two dice and gets a sum of 2 or 12, he wins $20. If he gets a 7, he wins $5. The cost to roll the dice one time is $3. Is this game fair?
What is the random variable X?
Make a probability distribution table.
Outcome
Payout (X)
Probability
E(X)
NO, since m = $1.944 which is less than it cost to play ($3).

26. What is the random variable X? _____________________
Let’s play a game. A player pays $5 and draws a card from a deck. If he draws the ace of hearts, he is paid $100. For any other ace, he is paid $10, and for any other heart, he is paid $5. If he draws anything else, he gets nothing. Would you be willing to play?
What is the random variable X? _____________________
What are the values for the random variable? ____________
Outcome
Payout (X)
Probability
E(X)
Deviation

27. An insurance company offers a “death and disability” policy that pays $10,000 when you die or $5000 if you are permanently disabled. It charges a premium of on $50 a year for this benefit. Is the company likely to make a profit selling such a plan?

28. Suppose that the death rate in any year is 1 out of every 1000 people, and that another 2 out of 1000 suffer some kind of disability.
Policyholder
Outcome
Payout X
Probability
P(x)
Death
$10,000
Disability
$5000
Neither $0

29. E (X) = Expected value = μ = = $10,000( ) + $5000( ) + $0( )
Policyholder
Outcome
Payout X
Probability
P(x)
Death
$10,000
Disability
$5000
Neither $0
E (X) = Expected value = μ =
= $10,000( ) + $5000( ) + $0( )
= $10 + $10 + $0
= $20

30. Policyholder
Outcome
Payout X
Probability
P(x)
Death
$10,000
Disability
$5000
Neither $0
Deviation
(x – μ)
(10,000 – 20) = 9980
(5000 – 20) = 4980
(0 – 20) = - 20
Var(x) = 99802( ) ( ) + (-20)2 ( ) = 149,600
Standard deviation (X) = √149,600 = $386.78
The insurance company can expect an average payout of $20 per policy, with a standard deviation of $380.78

31. Discrete and Continuous Random Variables
Discrete random variables commonly arise from situations that involve counting something. Situations that involve measuring something often result in a continuous random variable.
Discrete and Continuous Random Variables
Definition:
A continuous random variable X takes on all values in an interval of numbers. The probability distribution of X is described by a density curve. The probability of any event is the area under the density curve and above the values of X that make up the event.
The probability model of a discrete random variable X assigns a probability between 0 and 1 to each possible value of X.
A continuous random variable Y has infinitely many possible values. All continuous probability models assign probability 0 to every individual outcome. Only intervals of values have positive probability.

32. Example: Young Women’s Heights
Read the example on page 351. Define Y as the height of a randomly chosen young woman. Y is a continuous random variable whose probability distribution is N(64, 2.7).
What is the probability that a randomly chosen young woman has height between 68 and 70 inches?
P(68 ≤ Y ≤ 70) = ???
P(1.48 ≤ Z ≤ 2.22) = P(Z ≤ 2.22) – P(Z ≤ 1.48)
= – =
There is about a 5.6% chance that a randomly chosen young woman has a height between 68 and 70 inches.

33. What is the probability that a randomly selected young woman is taller than 68.5 inches?
.0478
What is the probability that a randomly selected young woman is shorter than 63.5 inches?
.4265
40% of young women are what height or shorter?
63.3 inches

34. Linear function of a random variable
The mean is changed by addition & multiplication!
If x is a random variable and a and b are numerical constants, then the random variable y is defined by
and
The standard deviation is ONLY changed by multiplication!

35. Let x be the number of gallons required to fill a propane tank
Let x be the number of gallons required to fill a propane tank. Suppose that the mean and standard deviation is 318 gal. and 42 gal., respectively. The company is considering the pricing model of a service charge of $50 plus $1.80 per gallon. Let y be the random variable of the amount billed. What is the mean and standard deviation for the amount billed?
m = $ & s = $75.60

36. Linear combinations Just add or subtract the means!
If independent, always add the variances!

37. The mean of the sum of two random variables is the sum of the means.
E(X + Y) = E(X) + E(Y)
The mean of the difference of two random variables is the difference of the means.
E(X - Y) = E(X) - E(Y)
If the random variable are independent, the variance of their sums OR difference is always the sum of the variances.
Var(X + Y) = Var(X) + Var(Y)

38. What is the mean and standard deviation of:SD X 10 2 Y 20 5
What is the mean and standard deviation of:
a) 3X
Mean = 30; standard deviation = 6
b) Y + 6
Mean = 26; standard deviation = 5
c) X + Y
Mean = 30; standard deviation = √ = 5.39
d) X - Y
Mean = -10; standard deviation = √ = 5.39
e) X1 + X1
Mean = 20; standard deviation = √ = 2.83

39. A nationwide standardized exam consists of a multiple choice section and a free response section. For each section, the mean and standard deviation are reported to be
mean SD MC FR
If the test score is computed by adding the multiple choice and free response, then what is the mean and standard deviation of the test?
m = & s =

40. Example
Suppose x is the number of sales staff needed on a given day. If the cost of doing business on a day involves fixed costs of $255 and the cost per sales person per day is $110, find the mean cost (the mean of x or mx) of doing business on a given day where the distribution of x is given below.

41. We need to find the mean of y = 255 + 110x
Example continued
We need to find the mean of y = x

42. We need to find the variance and standard deviation of y = 255 + 110x
Example continued
We need to find the variance and standard deviation of y = x