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Probability Distributions and Expected Value
We plan to look at models for distributions that show the probabilities of all possible
outcomes.
Recall:
Discrete variables: values that are separate from each other (unique); often integer
values; number of possibilities is sometimes small
Continuous variables: have an infinite number of possible values in a continuous
interval; real numbers (i.e. decimals and fractions)
In this unit we will focus on distributions involving discrete random variables.
Classify each of the following random variables as discrete or continuous?
a) Number of phone calls made by a salesperson
b) Length of time the salesperson spent on the phone
c) Company’s annual sales
d) The number of widgets sold by a company
e) The distance from the Earth to the sun.
Uniform Probability Distribution:
Ex: rolling one die – the frequency of each number
Outcome - Random variable (x)
1
2
3
4
5
6
Probability P(x)
1/6 = 16.7%
1/6 = 16.7%
1/6 = 16.7%
1/6 = 16.7%
1/6 = 16.7%
1/6 = 16.7%
The sum of all probabilities is 1 (or 100%).
As we discussed earlier in the course, all probability distributions must sum to 1 since
they include all possible outcomes.
For a discrete uniform probability distribution P(x) =
possible outcomes.
1
where n is the number of
n
Mound Shaped Probability Distribution:
Ex: rolling two dice – the frequency of each sum
Outcome - Random variable (x)
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
P(x)
In conclusion, the probability distribution of a discrete random variable, x, is a function
that provides the probability of each possible value of x. This function can be presented
as a table of values, a graph or a mathematical expression.
Expected Value:
E(x) is the predicted average of all possible outcomes of a probability experiment.
Ex: After many times of rolling 2 dice you would predict that the sum of six would show
up 5/36 of the time. So what would you EXPECT your sum to be if you rolled two dice?
The expected value is equal to the sum of the products of each outcome (random variable
= x with its probability P(x). RECALL WEIGHTED MEANS!
E(sum) = 2P(sum=2) + 3P(sum=3) + 4P(sum=4) + … + 11P(sum=11) + 12P(sum=12)
= 2(1/36) + 3(2/36) + 4(3/36) + … + 11(2/36) + 12(1/36)
= 252/36
=7
Therefore, you would expect to roll a sum of 7.
Ex: A summer camp has seven 4.6m canoes, ten 5.0m canoes, four 5.2m canoes and four
6.1m canoes. Canoes are assigned randomly to campers going on a canoe trip.
a) Show the probability distribution for the length of an assigned canoe?
b) What is the expected length of an assigned canoe?
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