Section 6.1 Day 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

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.

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.

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

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 3:nCr Use 5C2 = 10

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.

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

Sampling Hockey Teams Total Possible Attendance Teams (in thousands) Devils/Islanders 19 + 16 = 35 Devils/Rangers 19 + 18 = 37 Devils/Flyers 19 + 18 = 37 Devils/Penguins 19 + 17 = 36 Islanders/Rangers 16 + 18 = 34 Islanders/Flyers 16 + 18 = 34 Islanders/Penguins 16 + 17 = 33 Rangers/Flyers 18 + 18 = 36 Rangers/Penguins 18 + 17 = 35 Flyers/Penguins 18 + 17 = 35

Sampling Hockey Teams Total Possible Attendance (in thousands), x Probability, p 33 0.1 34 0.2 35 0.3 36 0.2 37 0.2

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

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)

Sampling Hockey Teams Total Possible Attendance (in thousands), x Probability, p 33 0.1 34 0.2 35 0.3 36 0.2 37 0.2

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) = 0.2 + 0.2 P(at least 36,000) = 0.4

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)

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

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.

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

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

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

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

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) 0 0.088 1 0.332 2 0.385 3 0.137 4 0.058

Enter data into lists L1 L2 0 0.088 1 0.332 2 0.385 3 0.137 4 0.058

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

Use 1-Var Stats 1- Var Stats x = 1.745 σx = .9939693154 E(x) = μx = 1.745 σx2 = 0.987975

Estimate the expected value and standard deviation of this distribution

Calculator Notes 6A

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.

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?

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

Questions?