Download Section 6.1 Day 3

Section 6.1
Day 3
Find and interpret the expected value for this
lottery game. A ticket costs $1.
Winnings, x
Probability, p
$1
1/10
$2
1/14
$3
1/24
$18
1/200
$50
1/389
$150
1/20,000
$900
1/120,000
E(X) = $2.72?
Find and interpret the expected value for this
lottery game. A ticket costs $1.
Winnings, x
Probability, p
$1
1/10
$2
1/14
$3
1/24
$18
1/200
$50
1/389
$150
1/20,000
$900
1/120,000
Think again!!
Find and interpret the expected value for this
lottery game. A ticket costs $1.
Winnings, x
Probability, p
$1
1/10
$2
1/14
$3
1/24
$18
1/200
$50
1/389
$150
1/20,000
$900
1/120,000
Think again!!
Find and interpret the expected value for this lottery game.
A ticket costs $1.
Winnings, x
Probability, p
$1
1/10
$2
1/14
$3
1/24
$18
1/200
$50
1/389
$150
1/20,000
$900
1/120,000
Total
0.2207
Expected Value
Find and interpret the expected value for this lottery game.
A ticket costs $1.
Winnings, x
Probability, p
$0
$1
1/10
$2
1/14
$3
1/24
$18
1/200
$50
1/389
$150
1/20,000
$900
1/120,000
Expected Value
Find and interpret the expected value for this lottery game.
A ticket costs $1.
Winnings, x
Probability, p
$0
0.7793
$1
1/10
$2
1/14
$3
1/24
$18
1/200
$50
1/389
$150
1/20,000
$900
1/120,000
E(X) = $0.6014
How do we interpret this expected value?
E(X) = $0.6014
How do we interpret this expected value?
If we spend $1, we expect to get back
$0.6014.
Or, the state can expect to pay out $601.40
for every $1000 of tickets sold.
Linear Transformation Rule
Suppose you have a probability distribution
with random variable X, mean  x, and
standard deviation  x.
Linear Transformation Rule
Suppose you have a probability distribution
with random variable X, mean  x, and
standard deviation  x.
If you transform each value of x by
multiplying it by d and then adding c,
where c and d are constants, then
Linear Transformation Rule
Suppose you have a probability distribution
with random variable X, mean  x, and
standard deviation  x. If you transform
each value by multiplying it by d and then
adding c, where c and d are constants,
then
 c + dx = c + d  x
 c + dx = |d|●  x
Expected Value
Find and interpret the expected value for this new
lottery game. A ticket costs $1.
Winnings, x New winnings, 3x Probability, p
$0
$0
0.7793
$1
$3
1/10
$2
$6
1/14
$3
$9
1/24
$18
$54
1/200
$50
$150
1/389
$150
$450
1/20,000
$900
$2700
1/120,000
New x = 0 + 3x, so c = 0 and d = 3
Find and interpret the expected value for this new
lottery game. A ticket costs $1.
Winnings, x New winnings, 3x Probability, p
$0
$0
0.7793
$1
$3
1/10
$2
$6
1/14
$3
$9
1/24
$18
$54
1/200
$50
$150
1/389
$150
$450
1/20,000
$900
$2700
1/120,000
New x = 0 + 3x, so c = 0 and d = 3
μx = 0.6014 for original game
μ3x = 3μx = 3(0.6014) = $1.804
We expect to win $1.804 for each dollar we
spend.
Addition and Subtraction Rules
If X and Y are random variables, then
 X + Y =  X + Y
 X - Y = X -  Y
Addition and Subtraction Rules
If X and Y are random variables, then
 X + Y =  X + Y
 X - Y = X - Y
and, if X and Y are independent, then
2
2 +
2
=
 X+Y  X  Y

2
X-Y=
 2X +  2Y
For each million tickets sold, the original
New York lottery awarded one $50,000
prize, nine $5000 prizes, ninety $500
prizes, and nine hundred $50 prizes.
a. What was the expected value of a ticket?
x
0
50
500
5000
50,000
p
999,000/1,000,000
900/1,000,000
90/1,000,000
9/1,000,000
1/1,000,000
a. Expected value of a ticket is $0.185
a. Expected value of a ticket is $0.185
b. The tickets sold for $0.50 each. How
much could the state of New York expect
to earn for every million tickets sold?
a. Expected value of a ticket is $0.185
b. The tickets sold for $0.50 each. How
much could the state of New York expect
to earn for every million tickets sold?
1,000,000(0.50 – 0.185) = $315,000
Page 377, P7
Page 377, P7
Claire
Charlotte
Max
Alisa
Shaun
List all possible random samples of size 3
from this group of 5 students.
Page 377, P7
Claire
Charlotte
Max
Alisa
Shaun
List all possible random samples of size 3
from this group of 5 students.
5C3 = 10
Page 377, P7
Claire, Charlotte, Max
Claire, Charlotte, Alisa
Claire, Charlotte, Shaun
Claire, Max, Alisa;
Charlotte, Alisa, Shaun
Claire, Max, Shaun;
Max, Alisa, Shaun
Claire, Alisa, Shaun
Charlotte, Max, Alisa
Charlotte, Max, Shaun
Page 377, P7
Page 377, P7
Claire, Charlotte, Max
Claire, Charlotte, Alisa
Claire, Charlotte, Shaun
Claire, Max, Alisa;
Charlotte, Alisa, Shaun
Claire, Max, Shaun;
Max, Alisa, Shaun
Claire, Alisa, Shaun
Charlotte, Max, Alisa
Charlotte, Max, Shaun
Page 377, P7
Page 377, P10
Page 377, P10
How many possible samples of size 2?
Page 377, P10
How many possible samples of size 2?
6C 2
= 15
Page 377, P10
(a) The 15 possible samples of size 2 are:
1 and 2; 1 and 3; 1 and 4; 1 and 5; 1 and 6
2 and 3; 2 and 4; 2 and 5; 2 and 6
3 and 4; 3 and 5; 3 and 6
4 and 5; 4 and 6
5 and 6
Page 377, P10
Assume computers 1, 2, and 3 are the
defective monitors. (The probabilities
would be the same no matter which 3
were assigned as the defective monitors).
Page 377, P10
Assume computers 1, 2, and 3 are the
defective monitors. (The probabilities
would be the same no matter which 3
were assigned as the defective monitors).
1 and 2; 1 and 3; 1 and 4; 1 and 5; 1 and 6
2 and 3; 2 and 4; 2 and 5; 2 and 6
3 and 4; 3 and 5; 3 and 6
4 and 5; 4 and 6
5 and 6
Page 377, P10
Page 377, P10
Page 377, P10
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Questions?
Fathom Activity 6.1a
Activity 6.1a
x
Die 1
Die 2
Sum
Difference
 2x
Activity 6.1a
Die 1
Die 2
Sum
Difference
x
 2x
3.5
2.917
Activity 6.1a
x
 2x
Die 1
3.5
2.917
Die 2
3.5
2.917
Sum
Difference
Fathom Activity 6.1a
Rolls
Die1
S1 = mean
S2 = variance
3.42
2.99459
Fathom Activity 6.1a
Rolls
Die2
S1 = mean
S2 = variance
3.516
2.94469
Activity 6.1a
x
 2x
Die 1
3.5
2.917
Die 2
3.5
2.917
Sum
7
5.834
Difference
Activity 6.1a
x
 2x
Die 1
3.5
2.917
Die 2
3.5
2.917
Sum
7
5.834
Difference
0
5.834
Activity 6.1a, 500 rolls
Variance
5.276
5.895
Fathom Activity 6.1a
Rolls
Sum
S1 = mean
S2 = variance
6.936
5.76567
Fathom Activity 6.1a
Rolls
Difference
S1 = mean
S2 = variance
-0.096
6.1129