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Discrete Probability Distribution Calculations
Mean, Variance, Standard DeviationExpected Value
Remember the Structure
Required features• The left column lists the
sample space outcomes.• The right column has the
probability of each of the outcomes.
• The probabilities in the right column must sum to exactly 1.0000000000000000000.
Example of a Discrete Probability Distribution
# of children Relative Frequency
0 0.108
1 0.239
2 0.326
3 0.174
4 0.087
5 0.043
6+ 0.022
Total 1.000
The Formulas
• MEAN:
• VARIANCE:
• STANDARD DEVIATION:
TI-84 Calculations
• Put the outcomes into a TI-84 List (we’ll use L1)
• Put the corresponding probabilities into another TI-84 List (we’ll use L2)
• 1-Var Stats L1, L2
• You can type fractions into the lists, too!
•
Practice Calculations
Rolling one dieValue Probability
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
Total 1
Statistics• The mean is
• The variance is
• The standard deviation is
Practice Calculations
Statistics Total of rolling two diceValue Prob. Value Prob.
2 1/36 8 5/36
3 2/36 9 4/36
4 3/36 10 3/36
5 4/36 11 2/36
6 5/36 12 1/36
7 6/36 Total 1
• The mean is
• The variance is
• The standard deviation is
Practice Calculations
One CoinHow many heads Probability
0 1 / 2
1 1 / 2
Total 1
Statistics• The mean is
• The variance is
• The standard deviation is
Practice Calculations
Statistics Four CoinsHow many heads Probability
0 1/16
1 4/16
2 6/16
3 4/16
4 1/16
Total 1
• The mean is
• The variance is
• The standard deviation is
Expected Value Problems
The Situation• 1000 raffle tickets are sold• You pay $5 to buy a ticket• First prize is $2,000• Second prize is $1,000• Two third prizes, each $500• Three more get $100 each• The other ____ are losers.What is the “expected value” of your ticket?
The Discrete Probability Distr.Outcome Net Value Probability
Win first prize
$1,995 1/1000
Win second prize
$995 1/1000
Win third prize
$495 2/1000
Win fourth prize
$95 3/1000
Loser $ -5 993/1000
Total 1000/1000
Expected Value Problems
Statistics• The mean of this probability
is $ - 0.70, a negative value.• This is also called “Expected
Value”.
• Interpretation: “On the average, I’m going to end up losing 70 cents by investing in this raffle ticket.”
The Discrete Probability Distr.Outcome Net Value Probability
Win first prize
$1,995 1/1000
Win second prize
$995 1/1000
Win third prize
$495 2/1000
Win fourth prize
$95 3/1000
Loser $ -5 993/1000
Total 1000/1000
Expected Value Problems
Another way to do it• Use only the prize values.• The expected value is the
mean of the probability distribution which is $4.30
• Then at the end, subtract the $5 cost of a ticket, once.
• Result is the same, an expected value = $ -0.70
The Discrete Probability Distr.Outcome Net Value Probability
Win first prize
$2,000 1/1000
Win second prize
$1,000 1/1000
Win third prize
$500 2/1000
Win fourth prize
$100 3/1000
Loser $ 0 993/1000
Total 1000/1000
An Observation
• The mean of a probability distribution is really the same as the weighted mean we have seen.
• Recall that GPA is a classic instance of weighted mean– Grades are the values– Course credits are the weights
• Think about the raffle example– Prizes are the values– Probabilities of the prizes are the weights
