HAVE YOU STARTED THE HOMEWORK
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2.
3.
Yes
33%
No
With a little help from a friend
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33%
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UPCOMING IN CLASS

Sunday Homework 6

Quiz 3 – Sept. 25th

Exam 1 – October 9th
CHAPTER 14
Random Variables
EVERYTHING
  E  X    x  P  x
  Var  X     x     P  x 
2
2
  SD  X   Var  X 
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MORE ABOUT MEANS AND VARIANCES


Adding or subtracting a constant from data
E(X ± c) = E(X) ± c Var(X ± c) = Var(X)
Multiplying each value of a random variable by a
constant multiplies the mean by that constant
and the variance by the square of the constant:
E(aX) = aE(X)
Var(aX) = a2Var(X)
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MORE ABOUT MEANS AND VARIANCES
(CONT.)

In general,


The mean of the sum of two random variables is the
sum of the means.
The mean of the difference of two random variables is
the difference of the means.
E(X ± Y) = E(X) ± E(Y)

If the random variables are independent, the
variance of their sum or difference is always the sum
of the variances.
Var(X ± Y) = Var(X) + Var(Y)
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INSURANCE POLICIES

An insurance company estimates that it
should make an annual profit of $130 on
each homeowner’s policy, with a standard
deviation of $4,000.
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WHAT IS THE MEAN AND SD IF THE
COMPANY WRITE 3 POLICIES?
1.
2.
3.
4.
Mean=3*130
SD=sqrt(3*4,000)
Mean=130+130+130
SD=sqrt(3*4,0002)
Mean=3*130
SD=3*4,000
Mean=130+130+130
SD=3*4,0002
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WHAT IS THE MEAN AND SD IF THE
COMPANY WRITE 10,000 POLICIES?
1.
2.
3.
4.
Mean=10,000*130
SD=sqrt(10,000*4,0002)
Mean=130+130+130
SD=sqrt(10,000*4,0002)
Mean=10,000*130
SD=10,000*4,000
Mean=130+130+130
SD=10,000*4,000
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WHAT ASSUMPTION DID YOU MAKE?
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2.
3.
4.
The annual profit on a policy is a continuous
random variable
Losses are dependent
Losses are independent of each other
The annual profit on a policy is a discrete
random variable.
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WILL THE COMPANY BE PROFITABLE?
1.
2.
3.
4.
No. The variance is larger than the mean.
Yes, $0 is 3.25 standard deviations below the
mean for 10,000 policies
Yes, the expected value is greater than zero.
No. Catastrophes are far too unpredictable to
expect a profit.
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APPLYING THE NORMAL MODEL

Remember the z-score involves a mean and
standard deviation,
y  y

z
s

So applying our expected value and standard
deviation of our random variable…
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DELIVERY PROBLEM
A delivery company’s trucks occasionally get
parking tickets, and based on past experience the
company plans that the trucks will average 1.2
tickets a month, with a standard deviation of 0.7
tickets.
 Suppose they have 17 trucks…

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WHAT IS THE EXPECTED NUMBER OF
TICKETS FOR 17 TRUCKS?
1.
2.
3.
4.
1.2
11.9
17
20.4
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2.
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WHAT IS THE ST. DEV. OF PARKING
TICKETS FOR 17 TRUCKS?
1.
2.
3.
4.
0.7
2.89
11.9
17
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WHAT WOULD BE AN UNUSUALLY BAD MONTH
FOR THE COMPANY? ASSUME THAT A
VARIATION OF 2 ST. DEV. IS UNUSUAL.
1.
2.
3.
4.
22
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27
30
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MORE ABOUT MEANS AND VARIANCES

In general,


The mean of the sum of two random variables is the
sum of the means.
The mean of the difference of two random variables is
the difference of the means.
E(X ± Y) = E(X) ± E(Y)

If the random variables are independent, the
variance of their sum or difference is always the sum
of the variances.
Var(X ± Y) = Var(X) + Var(Y)
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CEREAL BOWLS
The amount of cereal that can be poured into a bowl
varies…
 Large Bowl

E(XL)=2.7 oz
 SD(XL)=0.2 oz


Small Bowl
E(XS)=1.7 oz
 SD(XS)=0.2 oz


You open a new box of cereal and pour one large bowl
and one small bowl.
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HOW MUCH MORE CEREAL DO YOU EXPECT TO
EAT BY USING THE LARGE BOWL?
1.
2.
3.
4.
On average 2.7oz
On average 1.7oz
On average 2.7+1.7oz
On average 2.7-1.7oz
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WHAT IS THE SD OF THE DIFFERENCE
BETWEEN THE LARGE AND SMALL BOWLS?
1.
2.
3.
4.
SD=sqrt(0.2-0.2)
SD=sqrt(0.2+0.2)
SD=sqrt(0.22-0.22)
SD=sqrt(0.22+0.22)
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ASSUMING NORMALITY, WHAT IS THE PROB. THAT
THE SMALL BOWL CONTAIN MORE CEREAL?
1.
2.
3.
4.
-3.57
-3.25
0.0002
0.9998
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2.
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4.
WHAT IS THE MEAN OF THE TOTAL
AMOUNT IN THE TWO BOWLS?
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2.
3.
4.
Mean=2.7
Mean=1.7
Mean=2.7+1.7
Mean=2.7+1.7
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WHAT IS THE SD OF THE TOTAL AMOUNT
IN THE TWO BOWLS?
1.
2.
3.
4.
SD=sqrt(0.2-0.2)
SD=sqrt(0.2+0.2)
SD=sqrt(0.22-0.22)
SD=sqrt(0.22+0.22)
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ASSUMING NORMALITY, WHAT’S THE PROB. YOU
POURED MORE THAN 4.7 OZ OF CEREAL
TOTAL?
1.
2.
3.
4.
1.0714
-1.0714
.8577
.1423
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MORE CEREAL, MORE RANDOM
VARIABLES
The amount of cereal the manufacturer puts in
the boxes is a random variable with
 Mean=16.4
 SD=0.2

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WHAT IS THE EXPECTED VALUE?
1.
2.
3.
4.
16.4-2.7
16.4-1.7
16.4+2.7+1.7
16.4-2.7-1.7
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WHAT IS THE STANDARD DEVIATION?
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2.
3.
4.
SD=sqrt(0.2-0.2-0.2)
SD=sqrt(0.2+0.2+0.2)
SD=sqrt(0.22-0.22-0.22)
SD=sqrt(0.22+0.22+0.22)
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ASSUMING NORMALITY, WHAT’S THE PROB. THE
BOX HAS MORE THAN 11 OZ OF CEREAL
TOTAL?
1.
2.
3.
4.
1.0714
-1.0714
.8577
.1423
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2.
3.
4.
APPLYING THE NORMAL MODEL

Remember the z-score involves a mean and
standard deviation,
y  y

z
s

So applying our expected value and standard
deviation of our random variable…
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COFFEE PROBLEM
At a certain coffee shop, all the customers buy a
cup of coffee and some also buy a doughnut.
 The shop owner believes that the number of cups
she sells each day is ~N(340, 23)
 And the number of doughnuts is independent of
the number of cups
 And the number of doughnuts is ~N(160, 11)


The shop is open every day but Sunday.
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WHAT’S THE PROBABILITY IT WILL SELL
OVER 2000 CUPS OF COFFEE IN A WEEK?
1.
2.
3.
4.
-0.71
0.71
0.7611
0.2389
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1.
2.
3.
4.
MORE COFFEE…

Suppose she makes a profit of $.50 on each cup of
coffee and $.40 on each doughnut
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CAN SHE EXPECT TO MAKE A PROFIT OF
OVER $300 IN ONE DAY?
1.
2.
3.
4.
Yes, her expected profit is over
25%$30025%
25%
No, her expected profit is under $300
No, $300 is more than 5 standard deviations
above the mean.
Yes, $300 is more than 5 standard deviations
below the mean.
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WHAT THE PROBABILITY THAT IN ONE DAY SHE
WILL SELL A DOUGHNUT TO MORE THAN HALF
HIS COFFEE CUSTOMERS?
1.
2.
3.
4.
0.735
0.628
0.5
0.265
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4.
UPCOMING IN CLASS

Sunday Homework 5

Quiz 3 – Sept. 25th

Exam 1 – October 9th