1. QBM117 Business Statistics
Probability Distributions
Random variables and probability distributions

2. Objectives To define a random variable.
To define the probability distribution for a random variable.
To distinguish between a discrete random variable and a continuous random variable.
To introduce discrete probability distributions.
Calculate the mean, variance and standard deviation of a discrete probability distribution.

3. Random Variables
A random variable is a variable whose numerical value is determined by a the outcome of a random experiment.
It is random because the value it assumes depends on chance.

4. Examples of Random Variables
Imagine drawing a student at random from the student body.
The student’s height, weight, weekly income and grade point average are all numerical values describing properties of the randomly selected student.
They are all random variables.

5. Random Experiment
Draw a student at random from the student body
Random Variable
Height (meters) of the randomly selected student
Possible values for the random variable
Any value between about 1.5 m and 2 m

6. Random Experiment
Toss Two coins
Random Variable
The number of heads
Possible values for the random variable
0, 1 or 2

7. Random Experiment
Audit 50 tax returns
Random Variable
The number of returns containing errors
Possible values for the random variable
0, 1, 2,…,50

8. Random Experiment
Weigh a shipment of goods
Random Variable
The weight of the shipment
Possible values for the random variable
Any value greater than or equal to 0

9. Notation
We make the distinction between random variable and the values it can assume, by following the convention of using a capital letter such as X and Y to denote random variables, and using lower-case letters such as x and y to denote their values.

10. Discrete and Continuous Random Variables
There are two types of random variables
- discrete
- continuous
They are distinguished from one another by the number of possible values they can assume.

11. Discrete Random Variables
A discrete random variable has a finite number of possible values.
For example
the number of defective items in a production batch
the number of telephone calls received in a given hour
the number of customers served in a hotel reception on a given day

12. Continuous Random Variables
A continuous random variable has an infinite number of possible values.
For example
the duration of long-distance telephone calls
The lifetime of a certain brand of tyres
The total annual sales of a firm
The rate of return of a particular stock

13. Examples revisited Random Experiment
Draw a student at random from the student body
Random Variable
Height (meters) of the randomly selected student
Possible values for the random variable
Any value between about 1.5 m and 2 m
Continuous or Discrete?
Continuous

14. Random Experiment
Toss Two coins
Random Variable
The number of heads
Possible values for the random variable
0, 1 or 2
Continuous or Discrete?
Discrete

15. Random Experiment
Audit 50 tax returns
Random Variable
The number of returns containing errors
Possible values for the random variable
0, 1, 2,…,50
Continuous or Discrete?
Discrete

16. Random Experiment
Weigh a shipment of goods
Random Variable
The weight of the shipment
Possible values for the random variable
Any value greater than or equal to 0
Continuous or Discrete?
Continuous

17. Probability Distributions
A probability distribution of a random variable X tells us what the possible values of X are and the associated probabilities P(X=x) or p(x).
There are two types of probability distributions
- discrete probability distribution
- continuous distribution

18. Discrete Probability Distributions
The probability distribution of a discrete random variable is a table, formula or graph that lists all the possible values of the random variable and their associated probabilities. X x1 x2 … xn
P(X=x) p1 p2 pn

19. Requirements of Discrete Probability Distributions
If a discrete random variable X can take values
x1, x2,…, xn with probabilities p(x1), p(x2),…, p(xn) , the probabilities must satisfy two requirements:
Every probability p(xi) is a number between 0 and 1
The probabilities must add to 1

20. Example 1
Consider a study of 300 households in a town in the coast of Queensland. As a part of this study, data were collected showing the number of children in each household. The following results were obtained: 54 of the households has no children, 117 had 1 child, 72 had 2 children, 42 had 3 children, 12 had 4 children, and 3 had 5 children.
Consider the experiment of randomly selecting one of these households to participate in a follow-up study.

21. Let X = number of children in the household selected.
The possible values of X are 0, 1, 2, 3, 4, and 5.
The probability that the selected household has no children is 54/300 = 0.18.
Hence P(X=0) = 0.18
The probability that the selected household has 1 child is 117/300 = 0.39.
Hence P(X=1) = 0.39

22. The probability that the selected household has 2 children is 72/300 = 0.24
Hence P(X=2) = 0.24
The probability that the selected household has 3 children is 42/300 = 0.14
Hence P(X=3) = 0.14
The probability that the selected household has 4 children is 12/300 = 0.04
Hence P(X=4) = 0.04
The probability that the selected household has 5 children is 3/300 = 0.01
Hence P(X=5) = 0.01

23. The probability distribution of X can be presented in tabular form.
Note that each of the probabilities is between 0 and 1, and that the probabilities add to 1. X 1 2 3 4 5
P(X=x)
0.18
0.39
0.24
0.14
0.04
0.01

24. The probability distribution of X can also be presented in terms of the following formula

25. It can also be presented in the form of a graph.

26. Using a Probability Distribution
A primary advantage of defining a random variable and its probability distribution is that once the probability distribution is known, it is relatively easy to determine the probability of a variety of events that may be of interest to a decision maker.

27. We interpret the probabilities the same way we did last week when we were looking at probability.
Consider Example 1:
P(X=4) = 0.04 implies that the probability that a randomly selected household has 4 children is 0.04

28. We can also apply the addition rule for mutually exclusive events.
Consider Example 1:
The values of X are mutually exclusive; a household can have 0, 1, 2, 3, 4 or 5 children.
The probability that a randomly selected household has 3 or more children is

29. Example 2
Using historical records, the personnel manager of a plant has determined the probability distribution of X, the number of employees absent per day. It is
What is the probability that there are no absent employees on any given day?
What is the probability that there are no more than 2 employees absent on any given day? X 1 2 3 4 5 6 7
P(X=x)
0.005
0.025
0.310
0.340
0.220
0.080
0.019
0.001

30. What is the probability that there are no absent employees on any given day?
P(X=0) = 0.005
What is the probability that there are at most 2 absent employees on any given day?

31. Expected Value and Variance
In Topic 1 we calculated sample and population means and variances for frequency distributions.
A probability distribution is the distribution of a population.
We can calculate the population mean and variance for probability distributions.

32. Expected Value
The mean, or expected value, of a discrete random variable X is obtained by
multiplying each possible value of X by its associated probability
and then summing the resulting products.

33. Example 1 revisited X 1 2 3 4 5 P(X=x) 0.18 0.39 0.24 0.14 0.04 0.011 2 3 4 5
P(X=x)
0.18
0.39
0.24
0.14
0.04
0.01
The expected number of children per household is

34. Variance The variance of a discrete random variable X is found by
subtracting the mean from each value and squaring this difference.
multiplying squared difference by the associated probability,
and then summing the resulting products

35. A more computationally efficient method of calculating the variance of a discrete random variable is to use the following formula
This is just a rearrangement of the formula on the previous slide.

36. Example 1 revisited X 1 2 3 4 5 P(X=x) 0.18 0.39 0.24 0.14 0.04 0.011 2 3 4 5
P(X=x)
0.18
0.39
0.24
0.14
0.04
0.01
The variance of number of children per household is

37. Standard Deviation
Following on from Topic 1, the standard deviation can be found by taking the square root of the variance

38. Example 1 revisited X 1 2 3 4 5 P(X=x) 0.18 0.39 0.24 0.14 0.04 0.011 2 3 4 5
P(X=x)
0.18
0.39
0.24
0.14
0.04
0.01
The standard deviation of X, the number of children per household, is

39. Example 2 revisited
X = the number of employees absent per day
Determine the mean and standard deviation of the number of employees absent per day. X 1 2 3 4 5 6 7
P(X=x)
0.005
0.025
0.310
0.340
0.220
0.080
0.019
0.001

40. The mean number of employees absent per day is

41. The mean number of employees absent per day is

42. Example 3 (Exercise 5.19)
The owner of a small firm has just purchased a personal computer, which she expects will surge her for the next two years. The owner has just been told that she must buy a surge suppressor to provide protection for her new hardware against possible surges or variations in the electrical current. Her son David, a recent university graduate, advises that an inexpensive suppressor could be purchased that would provide protection against one surge only. He notes that the amount of damage done without a suppressor would depend on the extent of the surge. David conservatively estimates that, over the next two years, there is a 1% chance of incurring $400 damage and a 2% chance of incurring $200 damage. But the probability of incurring $100 damage is 0.1.

43. How much should the owner be willing to pay for a surge suppressor?
Determine the standard deviation of the possible amounts of damage.
To answer these questions we need to construct the probability distribution for the amount of damage incurred.

44. Let X = the amount of damage incurred.
David conservatively estimates that, over the next two years, there is a 1% chance of incurring $400 damage and a 2% chance of incurring $200 damage. But the probability of incurring $100 damage is 0.1. X
100
200
400
P(X=x)
0.87
0.10
0.02
0.01

45. To determine how much the owner should be willing to pay for a surge suppressor we need to work out the expected amount of damage to be incurred.
The expected amount of damage to be incurred is $18, therefore the owner should be willing to pay up to $18.

46. To determine the standard deviation of the possible amounts of damage we need to calculate the variance and then take the square root of the variance to obtain the standard deviation.
Hence the standard deviation of the possible amounts of damage is $55.46.

47. Reading for next lecture
Chapter 5 Section 5.4
Exercises
5.1
5.5
5.11
5.22 a and b only