Chapter 16 Random Variables What I will know and be able to do Recognize random variables, finding the probability

Chapter 16

Random Variables

What I will know and be able to do

Recognize random variables, finding the probability

model, mean (Expected Value) and the variance

utilizing the proper notation.

Assignment:

Read Chapter 16

Assn #4E

Expected Value: Center

•

A

random variable

assumes a value based on

the outcome of a random event.

•

_{We use a capital letter, like}

_X

_{, to denote a random}

variable.

•

A particular value of a random variable will be

denoted with a lower case letter, in this case

x

.

Expected Value: Center (cont.)

•

There are two types of random variables:

•

Discrete

random variables can take one of a finite

number of distinct outcomes.

• Example: Number of credit hours

•

_Continuous

_{random variables can take any numeric}

value within a range of values.

Expected Value: Center (cont.)

•

_A

_{probability model}

_{for a random variable consists of:}

• The collection of all possible values of a random variable, and

• the probabilities that the values occur.

•

Of particular interest is the value we expect a random

variable to take on, notated

μ

(for population mean)

or

E(X)

for expected value.

Greedy Pig

All students stand up. I will roll a die. Each student will get the number of points on the die. Students then may choose to keep their points by sitting down, or remain standing for another roll. With each roll, points accumulate for those who choose to remain standing. BUT…if I roll a 5, all those still standing lose their points, ending the round with a score of 0.

Let’s play a couple of practice rounds.

Let’s Play a Game!!

A player will pay me $5 and then draw a card from a deck. If you draw the ace of hearts, I will pay you $100. For any other ace, I will pay you $10, and for any other heart, I will pay you $5. If you draw anything else, you lose.

Who wants to play? Would you play for a top prize of $200?

Expected Value: Center (cont.)

 





E X

x P X

x













•

The

expected value

of a (discrete) random variable can

be found by summing the products of each possible

value by the probability that it occurs:

•

_{Note: Be sure that every possible outcome is included in}

Example 1: Expected Value

Given:

x 2 4 6 8

P(X=x) .3 .4 .2 .1

Example 1 continued….

We know that:

μ = E(X) = ΣxP(x)

= 2P(2) + 4P(4) + 6P(6) + 8P(8)

=2(.3) + 4(.4) + 6(.2) + 8(.1)

Example 2: Finding Probability Models

For an insurance company, x (particular value of a random variable) is $10,000 if you die that year, $5,000 if you are disabled, or $0 if

neither occurs. Suppose that the death rate in any year is 1 out of every 1000 people, and that another 2 out of 1000 suffer some kind of disability. We can display the probability model for this

insurance policy in a table like this:

Policy Holder Outcome

Payout x

Probability P(X=x) Death 10,000 1/1000

Disability 5,000 2/1000

Example 2 continued…

If we insure exactly 1000 people, find the following.

What is the expected value of total payout for the year?

What is the payout per policy?

Example 2 continued…

If we insure exactly 1000 people, find the following. What is the total payout?

We expect to pay out 10,000 + 2(5000) = $20,000

What is the expected payout per policy?(like the average)

If we charge a $50 premium for a policy, how much money would we expect to make? Charge: 1,000($50) = $50,000

Payout: $20,000 Net Profit: $30,000

Example 3: Car Repairs

Buck took his minivan in for repair recently because the air conditioner was cutting out intermittently. The mechanic

identified the problem as dirt in the control unit. He said that in about 75% of such cases, drawing down and then recharging

the coolant a couple times cleans up the problem – and costs only $60. If that fails, then the control unit must be replaced at an additional cost of $100 for parts and $40 for labor.

a. Define the random variable and construct the probability model.

Example 3 continued…

1.

2. μ=E(X)=60(.75) + 200(.25) = $95

3. Car owners with this problem will spend an average of $95 to get it fixed.

Outcome X=cost Probability Recharging

works

$60 0.75

Replace control unit

First Center, Now Spread…

• For data, we calculated the standard deviation by first

computing the deviation from the mean and squaring it. We do that with random variables as well.

• The variance for a random variable is:

• The standard deviation for a random variable is:

 

 

SD X

Var X







 









2

Var X

x

P X

x

Example 4: Repairs

Slide 16- 16 The probability model below describes the number of

repair calls that Comcast may receive during an hour.

a. How many calls should Comcast expect per hour?

b. What is the standard deviation?

Repair Calls 0 1 2 3 4

More About Means and Variances

•

_{Adding or subtracting a constant from data shifts the}

mean but doesn’t change the variance or standard

deviation:

E(X

±

c) = E(X)

±

c

Var(X

±

c) = Var(X)

More About Means and Variances (cont.)

•

In general, 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) = a

Var(X)

SD(aX) = aSD(X)

• Example: Consider everyone in a company receiving a 10% increase in salary.

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)

• If the random variables are independent, the standard deviation of their sum or difference is always the square root of the squares of the standard

deviations. 2 2

))

(

(

))

(

(

)

(

X

Y

SD

X

SD

Y

SD







Example 5: pg. 322 #25

Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of each of these variables.

• .8Y

• 2X-100

• X+2Y

• 3X-Y

• Y1+Y2

Mean SD

X 120 12

Example 6

A casino knows that people play the slot machine in hopes of hitting the

jackpot, but that most of them lose their dollar. Suppose a certain machine pays out an average of $0.92 with a standard deviation of $120.

• Why is the standard deviation so large?

• If you play 5 times, what are the mean and standard deviation of the casino’s profit?

• If gamblers play this machine 1000 times in a day, what are the mean and standard deviation of the casino’s profit?

Car Dealer Example

Suppose a used car dealer runs autos through a two-stage process to get them ready to sell. The mechanical checkup costs $50 per hour and takes an average of 1.5 hours, with a standard deviation of .25 hours. The appearance prep (wash, polish, etc) costs $6 per hour and takes an average of 1 hour, with a standard deviation of .083 hours.

• What are the mean and standard deviation of the total time spent preparing a car?

• What are the mean and standard deviation of the total expense to prepare a car?

• What are the mean and standard deviation of the difference in costs for the two phases of the operation?

Answers

• What are the mean and standard deviation of the total time spent preparing a car?

E(T)=2.5 hours SD(T)=.263 hours

• What are the mean and standard deviation of the total expense to prepare a car?

E(T)=$81 SD(T)=$12.51

• What are the mean and standard deviation of the difference in costs for the two phases of the operation?

Continuous Random Variables

• _{Random variables that can take on any value in a range of}

values are called continuous random variables.

• _{Now, any single value won’t have a probability, but…} • Continuous random variables have means (expected

values) and variances.

• We won’t worry about how to calculate these means and variances in this course, but we can still work with models for continuous random variables when we’re given the

parameters

Continuous Random Variables (cont.)

•

Good news: nearly everything we’ve said about how

discrete random variables behave is true of

continuous random variables, as well.

•

When two independent continuous random variables

have Normal models, so does their sum or difference.

•

_{This fact will let us apply our knowledge of Normal}

Car Dealers continued…

What is the probability that it will take longer than 2.7

hours to do the appearance prep and the mechanical

check-up? Both phases of the process can be described

by a Normal model.

E(T) =E(M) + E(A) = 1.5 + 1 = 2.5 hours

SD(D) = .263 hours (WHY?)

Normal model……

Car Dealers continued…

What is the probability that it will take longer to do the

appearance prep than the mechanical check-up? Both

phases of the process can be described by a Normal

model.

**If the difference in the prep times is less than 0, it

means the appearance prep took longer.

E(D) =E(M) – E(A) = 1.5 – 1 = .5 hours

SD(D) = .263 hours (WHY?)

Example 7: Packaging Stereos

Consider a company that manufactures and ships small stereo systems that were discussed in Chapter 16 of the book. The times required to pack the stereos can be described by a

Normal model with a mean of 9 minutes and standard

deviation of 1.5 minutes. The times for the boxing stage can also be modeled as Normal, with a mean of 6 minutes and standard deviation of 1 minute.

• What is the probability that packing and boxing a system takes over 20 minutes?

• What is the probability that packing and boxing a system takes between 12 and 17 minutes?

• What percentage of the stereo systems take longer to pack

What Can Go Wrong?

•

_{Probability models are still just models.}

• Models can be useful, but they are not reality.

• Question probabilities as you would data, and think about the assumptions behind your models.

Slide 16 - 30

What Can Go Wrong? (cont.)

•

_{Don’t assume everything’s Normal.}

• You must Think about whether the Normality Assumption

is justified.

•

Watch out for variables that aren’t independent:

• You can add expected values for any two random variables, but

What Can Go Wrong? (cont.)

•

_{Don’t forget: Variances of independent random}

variables add. Standard deviations don’t.

•

Don’t forget: Variances of independent random

variables add, even when you’re looking at the

difference between them.

Slide 16 - 32

What have we learned?

• We know how to work with random variables.

• We can use a probability model for a discrete random variable to find its expected value and standard deviation.

• The mean of the sum or difference of two random variables, discrete or continuous, is just the sum or difference of their means.

• And, for independent random variables, the variance of

What have we learned? (cont.)

•

_{Normal models are once again special.}

Slide 16 - 34

Assn #4G Possible Solution problem 2

a) b)

( )

0(0.2) 1(0.4) 2(0.4) 1.2

E Y













( ) 100(0.1) 200(0.2) 300(0.5) 400(0.2)

280

E Y

Assn #4G Possible Solution problem 4

a)

b)

c) In the long run, the expected payoff of this game would be $23.61. Any amount less than this would be a reasonable

amount to pay to play.

1 5 25

(amount won) 100 50 0 $23.61

6 36 36

E



Win $100 $50 $0

P(amount won) 1

Slide 16 - 36

Assn #4G Possible Solution problem 10

a)

b)



   

   

  

( ) 0 1.2 0.2 1 1.2 0.4 2 1.2 0.4 0.56

( ) ( ) 0.75

Var Y

SD Y Var Y

            



   

   

   

  

2 2 2 2

( )

100 280 0.1 200 280 0.2 300 280 0.5 400 280 0.2

7600

( ) ( ) 87.18

Var Y

SD Y Var Y

Assn #4G Possible Solution problem 12













2 1 5 25

(Won) 100 23.61 50 23.61 0 23.61

6 36 36

1456.404

(Won) (Won) $38.16

Var

SD Var





     

   _{ }   _{ }   _{ }

     



Assn #4H Possible Solution problem 8

Assuming the two races are independent events, the probability that the horse wins both is (0.2)(0.3) = 0.06.

(profit)

$30000(0.14) $30000(0.24) $80000(0.06) $10000(0.56) $10, 600

E

  

  



Profit 1st _only

$30000

2nd _only

$30000

Both $80000

Neither -$10000

Assn #4H Possible Solution problem 18

a)

b)

c)

(profit)

100(0.9975) 9900(0.0005) 2900(0.002)

$89

E













Profit $100 -$9900 -$2900

P(profit) 0.9975 0.0005 0.002



 

 

 

 

 



2

2 2 2

(Profit)

100 89 0.9975 9900 89 0.005 2900 89 0.002 67,879

(Profit) (Profit) $260.54