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Who Wants to be an Actuary?
An insurance company offers a “death and
disability” policy that pays $10,000 when
you die or $5,000 if you are permanently
disabled. It charges a premium of only
$50 a year for this benefit. Is the company
likely to make a profit selling such a plan?
To answer this question, the company
needs to know the probability that its
clients will die or be disabled in any year.
Expected Value: Center

The amount the company pays out on an individual policy
is called a random number because its value is based on
the outcome of a random event – use a capital letter, like
X, to denote a random variable

If we can list all the possible outcomes, we have a
discrete random variable
Example: X = payout from insurance company
All possible outcomes: x = $10,000 (death),
x = $5,000 (disabled), or x = $0 (neither)

If you can’t list all the outcomes, the random variable is
continuous
Expected Value: Center

The collection of all the possible values
and the probabilities that they occur is
called the probability model for the
random variable
Policyholder
Oucome
Payout
x
Death
10,000
Disability
5,000
Neither
0
Probability
P(X = x)
1
1000
2
1000
997
1000
Expected Value: Center

While we can’t predict what WILL
happen in any given year, we can say
what we EXPECT will happen

𝜇 = 𝐸 𝑋 = the expected value

Let’s imagine the insurance company
insures exactly 1000 people to see what
it should expect to pay out
Expected Value: Center
Policyholder
Oucome
Payout
x
Death
10,000
Disability
5,000
Neither
0
Probability
P(X = x)
1
1000
2
1000
997
1000
$10,000 1 + $5,000 2 + $0 997
= $20
1000
or
1
2
997
𝜇 = 𝐸 𝑋 = $10,000
+ $5,000
+ $0
= $20
1000
1000
1000
𝜇=𝐸 𝑋 =
The total payout is expected to be $20,000 or $20 per policy (leaving an
expected profit of $30 per policy).
Expected Value (Discrete)

Multiply each possible value by the
probability that it occurs and find the
sum
𝜇=𝐸 𝑋 =

𝑥∙𝑃 𝑥
Be sure that every possible outcome is
included in the sum!
Just Checking
One of the authors took his minivan in for
repair recently because the air conditioner
was cutting out intermittently. The mechanic
identified the problem as dirt in a control unit.
He said that in about 75% of such cases,
drawing down and then recharging the
coolant a couple of times cleans up the
problem – and only costs $60. If that fails,
then the control unit must be replaced at an
additional cost of $100 for parts and $40 for
labor.
Just Checking
a)
Define the random variable and
construct the probability model.
Outcome
X = cost
Probability
Recharging works
$60
0.75
Replace control
unit
$200
0.25
Just Checking
b)
What is the expected value of the cost
of this repair?
$60 0.75 + $200 0.25 = $95
c)
What does that mean in this context?
Car owners with this problem will spend
an average of $95 to get it fixed.
First Center, Now Spread…
The variance is the expected value of the
squared deviations from the center
Policyholder
Oucome
Payout
x
Death
10,000
Disability
5,000
Neither
0
𝑉𝑎𝑟 𝑋 = 99802
Probability
P(X = x)
1
1000
2
1000
997
1000
Deviation
(x - 𝜇)
(10000 – 20) = 9980
(5000 – 20) = 4980
(0 – 20) = –20
1
2
+ 49802
+ −20
1000
1000
2
997
= 149,600
1000
First Center, Now Spread…

𝜎2
The standard deviation is equal to the
square root of the variance
= 𝑉𝑎𝑟 𝑋 =
99802
1
2
2
+ 4980
+ −20
1000
1000
𝜎 = 𝑆𝐷 𝑋 =

𝑉𝑎𝑟 𝑋 =
2
997
= 149,600
1000
149,600 = $386.78
General Formulas:
𝜎 2 = 𝑉𝑎𝑟 𝑋 =
𝜎 = 𝑆𝐷 𝑋 =
𝑥−𝜇
𝑉𝑎𝑟 𝑋
2
P 𝑋
Another Example
As the head of inventory for the Knowway computer company,
you were thrilled that you had managed to ship 2 computers
to your biggest client the day the order arrived. You are
horrified, though, to find out that someone had restocked
refurbished computers in with the new computers in your
storeroom. The shipped computers were selected randomly
from the 15 computers in the stock, but 4 of those were
actually refurbished.
If your client gets 2 new computers, things are fine. If the
client gets one refurbished computer, it will be sent back at
your expense - $100 – and you can replace it. However, if
both computers are refurbished, the client will cancel the
order this month and you lose $1000. What is the expected
value and the standard deviation of your loss?
Refurbished Computers
RR 0.057
3
14
4
15
11
14
4
14
11
15
10
14
RN 0.2095
NR 0.2095
NN 0.524
To find the probabilities of each event, use a tree diagram
Refurbished Computers
Outcome
x
P(X = x)
Two refurbs
$1000
P(RR) = 0.057
One refurb
$100
P(NR ∪ RN) = 0.2095 + 0.2095 =
0.419
Two new
0
P(NN) = 0.524
E(X) = 0(0.524) + 100(0.419) + 1000(0.057) = $98.90
I expect this mistake to cost the firm $98.90.
Refurbished Computers
Outcome
x
P(X = x)
Squared Deviation
(x - 𝜇)2
Two refurbs
$1000
0.057
1000 − 98.90
One refurb
$100
0.419
100 − 98.90
Two new
0
0.524
0 − 98.90
2
2
2
Var(X) = 1000 − 98.90 2 (0.057) + 100 − 98.90 2 (0.419) + 0 − 98.90 2 (0.524)
= 51,408.79
SD(X) = 51408.79 = $226.74
I expect a standard deviation of $226.74. The large standard deviation
reflects the fact that there’s a pretty large range of possible losses.
More About Means & Variances

Recall that adding/subtracting a constant shifts
the mean but does not change the variance or
standard deviation
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 = 𝑎2 𝑉𝑎𝑟 𝑋
More About Means & Variances

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
More About Means & Variances

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

Does X + X + X = 3X?
 Answer depends on context and
independence – don’t rely on what you know
from Algebra
A Few Good Examples
Mr. Ecks and Ms. Wye each have a policy
through the insurance company in the previous
slides. (𝜇 = 𝐸 𝑥 = $20, 𝜎 2 = 149,600)
1.
What is the expected variance the sum of Mr. Ecks’
and Ms. Wye’s policies?
Var(X + Y) = Var(X) + Var(Y) = 149,600 + 149, 600 = 299,200
2.
What is the expected standard deviation of the sum
of the policies?
𝜎=
3.
𝑉𝑎𝑟(𝑋 + 𝑌) =
299,200 = $546.00
What if Mr. Ecks’ policy is doubled. What is the
new expected variance of his policy?
𝑉𝑎𝑟 2𝑋 = 22 𝑉𝑎𝑟 𝑋 = 4 × 149,600 = 598,400
Just Checking
Suppose the time it takes a customer to
get and pay for seats at the ticket window
of a baseball park is a random variable
with a mean of 100 seconds and a
standard deviation of 50 seconds. When
you get there, you find only two people in
line in front of you.
Just Checking
A.
How long do you expect to wait for your
turn to get tickets?
𝐸 𝑋1 + 𝑋2 = 100 + 100 = 200 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
B.
What’s the standard deviation of your wait
time?
𝜎=
C.
𝑉𝑎𝑟 𝑋1 + 𝑋2 =
502 + 502 = 70.7 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
What assumption did you make about the
two customers in finding the standard
deviation?
The times for the two customers are independent.
Continuous Random Variables

Continuous random variables follow the
same “rules” as discrete random
variables – difference is the use of a
Normal model

When two independent continuous
random variables have Normal models,
so do their sum or difference
A Continuous Example
A company manufactures small stereo systems. At
the end of the production line, the stereos are
packaged and prepared for shipping. Stage 1 of this
process is called “packing,” where workers collect all
system components (a main unit, speakers, a power
cord, an antenna, and some wires) and wrap
everything inside a styrofoam form. Stage 2, called
“boxing,” requires workers to place the styrofoam form
and directions into a cardboard box, close it, seal it,
and label the box for shipping. The company claims
the time required for packing can be described by a
Normal model with a mean of 9 minutes and a
standard deviation of 1.5 minutes and the time
required for boxing can also be described by a Normal
model with a mean of 6 minutes and a standard
deviation of 1 minute.
A Continuous Example
A.
What is the probability that packing two
consecutive systems takes over 20
minutes?
B.
What percentage of the stereo systems
take longer to pack than to box?
What Can Go Wrong?

Probability models are still just models

If the model is wrong, so is everything
else

Don’t assume everything’s Normal

Watch out for variables that are not
independent
Don’t Forget…

Variances of independent random variables
add, standard deviations don’t (think –
Pythagorean Theorem)

Variances of independent random variables
add, even when you’re looking at the
differences between them

Don’t write independent instances of a
random variable with notation that looks
like they are the same variables