Transcript Section 6.1 Day 2

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
Use 5C2 = 10
3:nCr
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
1
2
3
4
0.088
0.332
0.385
0.137
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
1
2
3
4
0.088
0.332
0.385
0.137
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
(per household), x
Proportion of
Households, P(x)
0
1
2
3
4
0.088
0.332
0.385
0.137
0.058
Enter data into lists
L1
0
1
2
3
4
L2
0.088
0.332
0.385
0.137
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?