; AP Stats Chapter 16
Documents
User Generated
Resources
Learning Center
Your Federal Quarterly Tax Payments are due April 15th

# AP Stats Chapter 16

VIEWS: 231 PAGES: 50

• pg 1
```									Chapter 16
Random Variables
In stats, what is a random variable?
► Denoted    with a capital letter
► Value based on a random event
► Individual values denoted by the corresponding
lowercase letter
► If all outcomes can be listed, we denote that as a
discrete random variable
► If the amount of outcomes are infinite, we call it a
continuous random variable
► The collection of all possible values and their
probabilities are called the probability model
Where do we see this outside of the
classroom?
► Common     use is in actuary science
► Actuary science is generally associated with
insurance
► It is the science of estimating and
evaluating risk
Example 1: It’s a matter of life and
death
► You  are tasked with the job of setting prices for a
life and disability insurance plan. Currently the
payout for death is \$10,000, a payoff for a person
suffering a disability is \$5000 and should a person
have neither of the two happen the payout is \$0
(Zero). The premium for the year is \$50 a
customer. In the past year 1/1000 people died,
1/500 people suffered a disability. The insurance
company you work for wants to make a profit of
over 100% on this policy. Will they need to set a
new premium for the upcoming year?
How to solve
► First in order to solve we need to
understand what our goal is
► We need to find the average cost of the
policy per person.
► This specific type of average is called an
expected value
► Using the least common denominator we
will estimate the average cost of the policy
for 1000 people
Computation
►   Find the sum of the money being spent on 1000 policy holders using
last years statistics
►   10,000*(1)+5,000*(2)+0*(997)
►   =20,000
►   Second take that sum and divide it by the 1000 policy holders to find
the average cost per policy holder
►   20,000/1,000
►   =20
►   Now put it in context
►   If each policy holder spends \$50 per policy and the expected cost to
the insurance company is \$20 per policy, the profit for the company is
150% (50/20=1.5). So based on last years death and disability
statistics, there is no need to change the cost of the policy.
►   ***When CUSSing, the expected value is considered the center.
► If we can find a mean, we can find a standard deviation
► To find the standard deviation of a discrete variable, we
must first find the deviation of each possibility.
► Then we square each deviation to find the variance.
 Since the variance is the expected value of the squared deviations
we must multiply each squared deviation by the probability
associated with it.
► Finally to find the standard deviation, we take the square
root of the variance
► We will use our first example to help illustrate the process
AP Speak
► In this problem, the variable was discrete
► The formula for calculating the expected of
a discrete variable is as follows:
Computation and AP style
breakdown
► A)Define the random variable and construct the
probability model
 Define the variable in a complete sentence
 Put the probability model in a chart
► B)   What is the expected value of the repair
 Show all the computations that you have done and
include the formula that you use
► C)   What does this mean in context?
 Write a paragraph to make sense of what you have
done and what it means. Make sure to stay away from
the use of pronouns.
Example 2: Extended Warranty? Do I
really need it?
► So when I make a purchase I always
research the product and price, but never
the cost of an extended warranty. Too
many times an extended warranty is offered
by some pushy salesmen working on a
commission and they often feel like a rip-
off. It’s a gamble and the house almost
always wins. With that being said, in the
following problem help me decide if it is a
better deal to buy the extended warranty.
Car Problems
►A car mechanic identified two possibilities for the
cause of a broken air conditioner. Either there is
dirt in the control unit or the unit is broken down
and needs to be replaced. 75% of the time, this
problem can be fixed by simply cleaning the
control unit for \$60. In order to find out if the
control unit is broken it first must be cleaned. The
cost of replacing the unit is an additional \$100 for
the part and an additional \$40 for labor. Before
working on my air conditioner, the mechanic offers
to replace the whole unit for \$120. Am I better off
just having the unit replaced?
Computation
► Deviations
 Death: 10,000-20=9980
 Disability: 5,000-20=4980
 Neither: 0-20=-20
► Variance
 1           2   2           2  997 
Var ( X )  9980 2         4980         (20 )          149 ,600
 1000          1000             1000 

► Standard Deviation
SD( X )  149,600  \$386.78
AP Speak
► The   formulas you need to know
 Variance

 2  Var( X )   ( x   )2 P( X )
 Standard Deviation

  SD( X )  Var( X )
Example 3: Computer Glitch
your first big sale. You were so excited that you shipped
the 2 computers your client bought that same day.
Unfortunately though, your regional manager got into the
stock room again and mixed up the refurbished computers
with the new computers. The 2 that you shipped were
randomly selected and are well on their way. You had 15
new computers and 4 refurbished computers in your
breathe easy. If he gets one refurbished it will cost you
\$100 to ship and replace. If both are refurbished, your
client will do business elsewhere costing you \$1000.
► What is the expected value of your loss?
► What is the standard deviation of your loss?
AP Practice
► Writeout a plan to ensure the reader you
understand the question.
 I am going to find my company’s expected loss for
incorrect shipping.
► To create an equation first define the random
variable.
 X=amount loss
 R=refurbished
 N=New
diagram. (You can not create your equation until
you do this)
Creating the model and
computations
► Using   the tree create the model
Outcome          X (Cost)      P(X=x)

2 Refurbished    1000          P(RR)=.057

1 Refurbished    100           P(NR U
RN)=.2095+.20
95=.419
0 Refurbished    0             P(NN)=.524
Computations
► To find the expected value multiply the cost
of the error times the probability for each
possibility. Then take the sum
 E(X) = 0(.524)+100(.419)+1000(.057)
 =\$98.9
Variance and Standard Deviation
► Variance Var(X)=
(0  98.90) 2 (.524)  (100  98.90) 2 (.419)  (0  98.90) 2 (.057)

 51408.79
► Standard Deviation             =

51408 .78  \$226 .735
So what does it all mean?
►I expect the mistake to cost \$98.90, with a
standard deviation of \$226.74. There is
such a large standard deviation because of
the large gap in cost between 1 and 2
refurbished computers. I will be demoting
my regional manager to assistant
regional…no assistant to the regional
manager.
Day 1 Homework
► Page   381 – 385
 4, 7, 8, 16, 18
Day 2 Problem of the Day
► Using  today’s handout go through the process of
think-show-tell for the following problem: A couple
plans to have children until they get a girl, but
they agree that they will not have any more than
three children even if all 3 are boys. Assuming
that gender probabilities are equivalent.
► A) Create a probability model for the number of
children they will have
► B) Find the expected number of children
► C) Find the expected number of boys
Day 2
►   When we first introduced
mean and standard
deviation we said that
when we add a constant to
E ( X  c)  E ( X )  c
the mean, or shift the
mean, the variance and
Var ( X  c)  Var ( X )
standard deviation stayed
the same. The same is
true when we have an
expected value
What about if we use multiplication?
► When we multiplied
the data points by a
constant, the mean      E (aX )  aE( X )
was multiplied by the
constant and the        Var(aX )  a Var( X )
2

variance by the
constant squared. The
same holds true with
expected value
► Yesterday  we simplified the process of insurance
for the sake of showing you how to calculate the
expected value, now lets investigate a little
deeper.
► Does it cost the same to insure one person for at
payouts of \$20,000 and \$10,000 as it does to
insure two people at rates of \$10,000 and \$5,000?
► Or if you would rather think about it this way, is
the risk the same?
► Does either sentence phrasing make a difference?
► What if that number changed to 1000 people?
► Let’s discuss…What I want to do here is get an
idea of what would seem logical to you.
Hopefully we came to a consensus that
risks and payouts are not the same
► Main reason it is different is that the odds of
one person dying are greater than 2 people
both dying.
► The risk gets spread out among more
people.
► As the risk spreads out the variance
becomes more predictable.
Day 3
► Nothing New
► Nothing Due
Take 2 People as an example. We will use our data
from yesterday’s insurance question.
► Married   Mac              ► Single   Mac
 Mac (X) CJ (Y)              Payouts are doubled
► Are our risk factors           since he is only one
person and he is
independent…generally          insured for twice as
speaking?                      much.
 Since they are then we      Use E(2X) and Var(2X)
will use E(X+Y) and
 Var(2X)=4(149600)
Var(X+Y)
 =598,400
► Using the data from           Standard deviation =
yesteday                       \$773.56 compared to
 Var(X)=149,600               \$546.99 for Married
 Var(Y)=149,60                Mac
 Var(X+Y)=299,200
Example 5: Intergalactic Currency
Exchange
► You decide to take an intergalactic trip to Tatooine
to explore the deserts. In order to do so you must
Currently your pod is worth 6,940 Galactic credits
with a standard deviation of 250 Galactic credits.
A land speeder is going for 65,000 Imperial Credits
with a standard deviation of 500 Imperial Credits.
1 Galactic credit is worth 43 Imperial credits. How
much money can you expect to have in your
pocket to spend on the Cantina in Mos Eisley after
► Think:
 I want to estimate my spending money at Mos Eisley.
► My   variables are
 G=sale of my pod racer
 I=price of a landspeeder
 M=Profits for Mos Eisley
► Equation
 M=43G – I
► Independence?
 From the empire no, for my purchases YES!
► Show    Expected Value
   E(M)=E(43G - I)
   =43E(G)-E(I)
   =43(6940)-65,000
   =233,420 Imperial Credits
► Show   Variance and Standard Deviation
 Var(M)=Var(43G – I)***This is the proper
notation since we are subtracting the two
values however….
 =Var(43G)+Var(I) ***We always add variances
 =43*43Var(G)+Var(I)
 =1849(250*250)+(500*500)
 Var(M) = 115,812,500
 Standard deviation is the square root of the
Conclusion
►I  can expect to have 233,420 Imperial
credits with a standard deviation of 10,762
Imperial Credits
► This is a profit of 5428 Galactic Credits with
a standard deviation of 250 Galactic credits
Continuous Random Variable
►A   variable that can take on any value.
 Time is a continuous random variable
► What   type of models can occur
 Normal, skewed, symmetric, bimodal.
depth will we go into for the AP
► What
exam?
 Our main concern is when they are normal. We
will use what we know about normal
distributions to make calculations.
What’s the difference between Continuous
and Discrete Random variables?
► Values   of discrete variables are all known and
have a probability associated with them
► Continuous variables must fit in some kind of
model as listed above but can be anywhere on
the spectrum. (i.e. the possible outcomes are
infinite.)
► When a continuous variable is normal, the
process of calculating the expected value,
variance, and standard deviation are the same
as in a discrete variable.
Example 4: Baseball’s been very very
good to me
► While working for the flyers, it took an
average of 100 seconds to serve a customer
with a standard deviation of 50 seconds.
(What can I say occasionally a fried carrot
worked its way into the hot dog bun). If
you came to my stand with two people in
front of you how long would you expect to
wait? What is the standard deviation of
your wait time? What assumptions must
you make about those in front of you?
► How   long do you expect to wait?
 E(customer 1 + customer 2)=100+100
 200 seconds or 3 minutes and 20 seconds
► What   is the standard deviation of your wait time?
 Variance is the standard deviation squared. Since we
are adding the two, we must first convert them back to
variances to solve.
 Variance =50 squared + 50 squared = 5000
 The square root of that is 70.7 seconds
► What   assumptions must you make?
 That each customer ahead of you in line is
independent from the other.
Example 6: Gone are the days of the
boom box
►   Back in my day it was cool to carry around a boom box.
Enemy. I decide I want to bring them back into style and
hire you all to work in the packing department. There are
two parts to the packing process, preparing the boom
boxes for shipping and boxing them. As a class, the time it
takes you to prepare the packages (Stage 1) is normally
distributed with a mean of 9 minutes standard deviation of
1.5 minutes. The time it takes to put the boom boxes into
their shipping box (Stage 2) is also normally distributed
with a mean of 6 minutes and a standard deviation of 1
minute
Questions
► What  is the probability that packing two
consecutive systems takes over 20 minutes?
► What percentage of stereos take longer to
pack than to box?
Where do we begin?
► Describe      what you are doing
 Rephrase the question
►Iwant to estimate the probability of packing two consecutive
boom boxes takes over 20 minutes
 Define the variables
► P1 = Packing first boom box
► P2= Packing Second Boom Box
► T = Total packing time
 Create an equation
►T   = P1 + P2
► Do I have to do this? How many points will you
take off if I don’t?
 YES! And Most to all points for this part will be lost if
you chose not to
Showing the Answer & Conclusion Part 1

► E(P1+P2)     = 18 minutes
► Var (T) = 4.5
► SD(T) = 2.12
► Show a Normal Curve See board
► Calculate a z score for 20 minutes
► Z score = .94
► Probability associated with that score on the
right tail is 17.36%
► We can say that there is a 17.36% chance that
the it will take us more than 20 minutes to pack
2 boom boxes.
Part 2 Thinking out the plan
► Describe   what you are doing
 Rephrase the question
►I  would like to estimate how often the boom boxes will take
longer to pack than to box.
 Define the variables
►P  = Packing the boom box
► B = boxing the Boom Box
► D = difference in times to pack and box a system
 Create an equation
►D   =P–B
 Ultimately what we are looking for is for P – B to be
greater than zero, that would mean that packing would
take longer than boxing.
Day 4
January 22, 2010
Pick up Practice Quiz C
Due today Problems 28, 29
Day 4 Problem of the day
► Find     the expected value, and the standard
deviation of the following data. The top row
is x and the bottom row is P(X)
100           200         300         400

.1            .2          .5          .2
Are the conditions met?
► Is   this a normal distribution?
 The text of the problem stated that each distribution
was normal.
► Is   there independence?
 The packing of the boom box and the boxing of the
boom box should not effect each other. ***Yes, I
know that the quality of the packing would have an
effect on the boxing, but we are assuming that the
boom box does not leave stage 1 unless it is in
perfect condition.
Show and Tell
► First   let’s find the expected value
 E(D) = E(P – B)
 =E(P) – E(P)
 9 – 6 = 3 minutes
► Find    the variance and standard deviation
   Var(D) = Var(P – B)
   Var(P) + Var(B)
   1.5*1.5 + 1*1
   Var(D) = 3.25
   SD (D) = the square root of 3.25 which is
approximately 1.8
Show a picture
► Findthe z score associate with the
difference
 (0-3)/1.8 = -1.67
► Seethe board for the picture
► P(D>0)=P(z>-1.67)=.9525
► Conclusion
 About 95% of boom boxes will take more time
in stage 1 than in stage 2
Day 5 Announcements
► Homework  check Quiz C
► Review Today
 Problem Solving Recap
 Expected Value Game
► Quiz   Tomorrow on Chapter 16
Step By Step Recap
► Think
 Plan: Actually state what you are going to try and solve
► In order to decide if I need to change the cost of the life
insurance policy I must…
 Variable: Define the discrete variable in words
►X    is the cost of the insurance policy on the insurance company
 Plot
► Make   a picture of the probabilities. Use a tree diagram to
illustrate this
 Model
► Createa table of all possible values and outcomes of the
random variable
Step By Step Recap Continued
► Show
 Find the expected value
►Show   formulas and computations
 Find the Variance
►Show   formulas and computations
 Find the Standard Deviation
►Show   formulas and computations
Step By Step Recap Continued
► Tell
 Conclusion
►Write out a paragraph in complete sentences
►Use the context of the problem and avoid pronouns