1. Section 6.1
Day 2

2. Sampling Hockey Teams Arena Seating Team (in thousands)
New Jersey Devils 19
New York Islanders 16
New York Rangers 18
Philadelphia Flyers 18
Pittsburg Penguins 17
Display 6.7

3. Sampling Hockey Teams
Suppose you pick two teams at random to play a pair of exhibition games, with one game played at each home arena.

4. Sampling Hockey Teams
Suppose you pick two teams at random to play a pair of exhibition games, with one game played at each home arena.
Construct the probability distribution of X, the total number of people who could attend the two games.

5. Sampling Hockey Teams
How many ways are there to pick two teams from the five teams?

6. Sampling Hockey Teams
How many ways are there to pick two teams from the five teams?
Since order of teams does not matter here, use combination function.
MATH PRB :nCr
Use 5C2 = 10

7. Sampling Hockey Teams
How many ways are there to pick two teams from the five teams?
Since order of teams does not matter here, use combination function.
Use 5C2 = 10
Now, list the ten pairs of teams, with the total possible attendance.

8. Sampling Hockey Teams Arena Seating Team (in thousands)
New Jersey Devils 19
New York Islanders 16
New York Rangers 18
Philadelphia Flyers 18
Pittsburg Penguins 17
Display 6.7

9. Sampling Hockey Teams Total Possible Attendance Teams (in thousands)
Devils/Islanders = 35
Devils/Rangers = 37
Devils/Flyers = 37
Devils/Penguins = 36
Islanders/Rangers = 34
Islanders/Flyers = 34
Islanders/Penguins = 33
Rangers/Flyers = 36
Rangers/Penguins = 35
Flyers/Penguins = 35

10. Sampling Hockey Teams Total Possible
Attendance (in thousands), x Probability, p

11. Sampling Hockey Teams
Now find the probability that the total possible attendance will be at least 36,000.

12. Sampling Hockey Teams
Now find the probability that the total possible attendance will be at least 36,000.
P(at least 36,000) = P(36,000) + P(37,000)

13. Sampling Hockey Teams Total Possible
Attendance (in thousands), x Probability, p

14. Sampling Hockey Teams
Now find the probability that the total possible attendance will be at least 36,000.
P(at least 36,000) = P(36,000) + P(37,000)
P(at least 36,000) =
P(at least 36,000) = 0.4

15. Expected Value
The mean of a probability distribution for the random variable X is called its expected value and is usually denoted by μx, or E(X)

16. Find the mean number of motor vehicles per household
Number of Motor Vehicles (per household), x
Proportion of Households, P(x)
0.088 1
0.332 2
0.385 3
0.137 4
0.058
Display 6.1: The number of motor vehicles per household.
Source: U.S. Census Bureau, American Community Survey, 2004, factfinder.census.gov

17. Find the mean number of motor vehicles per household
E(X) = μx = ∑xipi where pi is the probability that the random variable X takes on for the specific value xi.

18. Find the mean number of motor vehicles per household
Number of Motor Vehicles (per household), x
Proportion of Households, P(x)
0.088 1
0.332 2
0.385 3
0.137 4
0.058
Display 6.1: The number of motor vehicles per household.
Source: U.S. Census Bureau, American Community Survey, 2004, factfinder.census.gov

19. Find the mean number of motor vehicles per household
E(X) = μx = ∑xipi where pi is the probability that the random variable X takes on for the specific value xi.
μx = 0(0.088) + 1(0.332) + 2(0.385) + 3(0.137) + 4(0.058)
μx = 1.745

20. Variance and Standard Deviation
Variance: measure of spread equal to the square of the standard deviation

21. Variance and Standard Deviation
Variance: measure of spread equal to the square of the standard deviation
Var(X) = σ2x = ∑(xi – μx)2pi where pi is the probability that the random variable X takes on for the specific value xi.
Standard deviation = σ2x = σx

22. Calculator Notes 6A
Use calculator to find the expected value and variance for:
Number of Motor Vehicles Proportion of
(per household), x Households, P(x)

23. Enter data into lists
L1 L2

24. Use 1-Var Stats
1- Var Stats L1, L2

25. Use 1-Var Stats 1- Var Stats x = 1.745 σx = .9939693154
E(x) = μx = 1.745
σx2 =

26. Estimate the expected value and standard deviation of this distribution

27. Calculator Notes 6A

28. As previously stated, 90% of lung cancer cases are caused by smoking.
Suppose three lung cancer patients are randomly selected from the large population of people with that disease.
Construct the probability distribution of X, the number of patients with cancer caused by smoking.

29. Three lung cancer patients are randomly
selected.
Let S represent a patient whose lung cancer was caused by smoking and N represent a patient whose lung cancer was not caused by smoking.
How many outcomes are there?

30. Three lung cancer patients are randomly
selected.
Let S represent a patient whose lung cancer was caused by smoking and N represent a patient whose lung cancer was not caused by smoking.
How many outcomes are there? 2●2●2 = 8

33. Questions?