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AP Stats Chapter 16

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					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.
                        Spread Out
► 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
► So you own your own computer business and just made
  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
  inventory. If your client receives 2 new computers you can
  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
► To help you figure out probabilities, draw a tree
  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
         Things to think about
► 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
 trade in your old pod racer for a landspeeder.
 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
 your sale and purchase?
                       Answers:
► 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!
            Answers continued
► Show    Expected Value
     E(M)=E(43G - I)
     =43E(G)-E(I)
     =43(6940)-65,000
     =233,420 Imperial Credits
           Answers continued
► 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
    previous answer, 10,762 Imperial Credits
                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?
                     Answers
► 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.
    You were something with your big ole box and your Public
    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
     ►Discuss how you found your answers and interpret
      what they mean

				
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