D1 by garrickWilliams

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									                    B01.1305 Summer 2009 Homework 4
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1. The rate of home sales at a small real estate agency is 1.3 per day. We’ll assume that
a Poisson phenomenon can represent these home sales.
(a)     Find the probability that no homes will be sold on Monday.
(b)     Find the probability that one home will be sold on Monday.
(c)     Find the probability that two homes will be sold on Monday.
The probabilities can be found with Minitab, but they are easy enough with just a good
calculator. This “good calculator” will be needed to find e-1.3  0.272532.


2. At the Sweet Easter Company, internet orders for the very expensive Super Chocolate
Bunny ($90 each) come in at a rate of 0.9 per day, and it is believed that this phenomenon
can be described as a Poisson random variable with  = 0.9. Find
       (a)     the probability that there will be three or more orders on any day.
       (b)     the probability that, in a five-day work week, there will be at least one day
               on which there are three or more orders.


3. The probability that a “cold call” will result in selling a home termite inspection
service is 0.032. If 100 cold calls are made, find the probability that
        (a)     no sales will be made
        (b)     exactly one sale will be made
        (c)     exactly two sales will be made
        (d)     exactly three sales will be made
This problem is much easier with Minitab than with a calculator.


4. The number of customer complaints, per month, at a large brokerage firm follows a
Poisson distribution with parameter  = 3.2. Find the probability that, over the next
month,
       (a)     no customer complaints will be made
       (b)     exactly one customer complaint will be made
       (c)     exactly two customer complaints will be made
       (d)     exactly three customer complaints will be made
These can be done with a calculator, but Minitab will be easier.
       (e)     How would you compare the numbers here to the corresponding numbers
               from the previous problem?




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5. A bowl contains 18 bags of potato chips. Of these, 12 bags are standard potato chips,
and six bags are onion-flavored chips. If Lucy grabs four of these bags without looking,
what is the probability that she will get two bags of the onion-flavored chips?
        NOTE: “get two” means exactly two.


6. For each of the following situations, indicate whether the model should be binomial
or Poisson or hypergeometric.

(a)    The number of major forest fires to strike Colorado in calendar year 2006.
(b)    The number of trading days in the month of October that the stock of General
       Electric will go up in value.
(c)    The number of plays at craps, out of 50 attempted, that are winners.
(d)    The number of prize coupons, out of 800 inserted into cereal boxes, that are
       returned to collect the prizes.
(e)    The number of visitors to your web site on 25 FEB 2007.
(f)    The number of dead squirrels found on one mile of highway 93, on May 15.
       (Such schemes are actually used to estimate animal populations.)
(g)    The number of expense account claims with inadequate documentation, in a
       sample of 10 selected from a master file of 280.
(h)    The number of mattresses, out of 140 sold during the month of May, returned by
       the customers.
(i)    The number of diamonds in a single hand at hearts. (In the game of hearts, a
       single hand consists of 13 cards dealt from the deck of 52.)
(j)    The number of customers, out of 418 who made a purchase at Windham
       Supermarket, who purchased milk.


7. The multivariate hypergeometric applies in a situation in which the set of N items has
more than two types. Suppose that there are K types, and that N1 is the number of type 1,
N2 is the number of type 2, … , and NK is the number of type K. The accounting equation
N = N1 + N2 + … + NK assures that we’ve counted everything correctly. Suppose that you
take a sample of size n (without replacement, as usual) and that X1 is the random number
of these of type 1, X2 is the random number of these of type 2, …, XK is the random
number of these of type K. The accounting equation here is X1 + X2 + … + XK = n. It can
be shown without difficulty (but with a lot of notation) that

                                              N1   N 2   N 3   N K 
                                              x   x   x  ...  x 
       P[ X1 = x1 , X2 = x2 , …, XK = xK ] =  1   2   3   K 
                                                          N
                                                          n 
                                                           

As an example, suppose that you take a deck of 52 cards and select 13 cards at random.
The probability that you get exactly 5 spades, 4 hearts, 1 diamond, and 3 clubs is




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         13   13   13   13 
         5  4 1  3 
                = 1,287  715  13  286  0.0053878
                   52             635,013,559,600
                   13 
                   

                      NOTE: Some of these numbers will be useful below.

This applies the multivariate hypergeometric with N = 52, N1 = N2 = N3 = N4 = 13.
Minitab is not set up to do multivariate hypergeometric problems, so this required a
calculator.

In the game of bridge, each player receives 13 of the 52 cards.

(a)    Find the probability that a bridge player will receive exactly four spades, three
       hearts, three diamonds, and three clubs.
(b)    Argue that this is also the probability of receiving three spades, four hearts, three
       diamonds, and three clubs.
(c)    Find the probability of a “flat hand.” A “flat hand” has 4-3-3-3 suit distribution,
       meaning there are four cards of one suit and three cards of each of the other suits.


8. The ethnic makeup of American juries is a contentious issue. Suppose that a jury pool
of 117 has 62 women and 55 men. Suppose that the jury of 12 ends up as 10 men and 2
women. Is this evidence of some form of discrimination?
       NOTE: Juries are not selected randomly. Still, we make statistical calculations
       on the assumption of randomness. Here that assumption would be that the
       selection was random with respect to gender.
(a)    Let X be the random number of women selected for the jury. Find P[X = 2],
       assuming that selection is random with regard to gender. This calculation is easy
       with Minitab.
(b)    The value of P[X = 2] by itself is not indication of prejudicial selection. The
       correct comparison is P[X  2]. Find this value. Do you consider this jury
       unfairly selected with regard to gender?




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9. Suppose that a jury pool (meaning people available to serve on a jury) is categorized
as

       White           48
       Black           32
       Hispanic        26
       Asian           14
       TOTAL          120

It is claimed by the prosecution that the 12 jury members will be selected without regard
to ethnicity. Suppose that the jury consists of 6 whites, 2 blacks, 4 Hispanics, and no
Asians. Find the probability that this happened by chance alone. As a computational
        48                    120 
hint,   = 12,271,512 and           = 10,542,859,559,688,820.
        6                      12 


10. How would you count the homeless people in New York? Here is an approximate
version of the methodology. The city picks a target date, say April 10. Then the city
then recruits two types of people, counters and poseurs.

       The counters go through a training session to get them ready for their job. The
       job consists of the following steps.
              C1.      On April 10, approach people who seem to be homeless (in
                       subway stations, on street corners, food kitchens, and other spots).
              C2.      Identify yourself as a city survey worker.
              C3.      Ask the person whether or not he or she is homeless.
              C4.      Ask if some previous city survey worker has already interviewed
                       them.
              C5.      Report back to the city on April 11 with information in the form
                       “I interviewed 45 people, of whom 17 identified themselves as
                       homeless and 28 identified themselves as non-homeless.” This
                       could represent the information in line 1 of the table below.
                               The “17” and the “28” in this statement are both first
                               encounters. This counter perhaps questioned 29
                               non-homeless people, of whom one had been questioned
                               before. This counter perhaps also questioned 21 homeless,
                               but 4 of them had previously been questioned before.

       The poseurs will also go through a training session. The poseur job has the
       following steps.
              P1.     Dress in shabby clothes on April 10.
              P2.     Hang out in places frequented by the homeless (in subway stations,
                      on street corners, food kitchens, and other spots).




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              P3.      If questioned by a counter, the poseur is required to keep his or her
                       status a secret. (After the first meeting with a counter, the poseur
                       becomes irrelevant and can go home.)
              P4.      Report back to the city on April 11, with a simple yes-or-no result.
                       Was the poseur approached by one of the counters?

Suppose that there are H homeless people in the city. Of course H is unknown, and our
desire is to estimate H. On April 10, we will know P, the number of poseurs, and C, the
number of counters. (The value of C is irrelevant. This exercise could, in theory, be
done with one very busy counter.)

On April 11, we will know (from the poseurs) X, the number of poseurs found by the
counters.

On April 11, we will also know n, the total number of homeless people found by the
counters. This value of n will include both poseurs and genuine homeless

                    Counter      Claimed           Claimed
                                                 Non-homeless
                                                                    TOTAL
                    number       Homeless
                      1             17               28              45
                      2             11               26              37
                                                                   
                      C             20               55              75
                    TOTAL            n

Only the “Claimed Homeless” column of this table will be used.

The P poseurs will report to us information that we express as X = number of poseurs
approached by counters. This will enable us to assemble this table:

                                                          Genuine
                                          Poseurs                         TOTAL
                                                          Homeless
        Interviewed        Yes               X                              n
        by Counter?         No
                          TOTAL              P                  H         N=P+H

                                                              ˆ
Let’s suppose that we want to estimate N. Call the estimate N . Since we know P, the
                                               ˆ                   ˆ   ˆ
number of poseurs, the estimate for H (call it H ) can be found as H = N - P.

All work will be done after we know the value of n, the number of “claimed homeless”
people interviewed by the counters.

(a)    What is the distribution of the random variable X?

(b)    What is the expected value of X?



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(c)    Suppose that you set up P = 200 poseurs. Now suppose that the counters
       interview 1,000 people, of whom 80 claim to be homeless. Then suppose that we
       learn from the poseurs that 25 were interviewed. This leads to

                                                      Genuine
                                        Poseurs                       TOTAL
                                                      Homeless
        Interviewed       Yes            X = 25                        n = 80
        by Counter?        No
                         TOTAL          P = 200           H         N = 200 + H

       Use any reasonable argument to estimate N, and then give the corresponding
       estimate of H. (This is for a city smaller than New York.)


11. Suppose that Z represents a standard normal random variable. Don’t forget that
“standard” here says  = 0 and  = 1.
       (a)    Find the probability P[ Z > 1.42 ].
       (b)    Find the probability P[ -0.22 < Z < -0.13 ].
       (c)    Find the probability P[ | Z |  0.90 ].
       (d)    Find the value h for which P[ | Z | > h ] = 0.08.


12. Suppose that X is a normal random variable with mean 4,500 and with standard
deviation 1,000. Find the probability
        (a)    P[ X < 5,000 ]
        (b)    P[ X > 3,500 ]
        (c)    P[ 4,000  X  5,000 ]
        (d)    P[ | X – 4,000 | > 800 ]


13. It is maintained that, in a quiet equity market with no news, the daily number of
shares trades of EquiNimbus Corporation will be approximately normally distributed
with mean 280,000 and with standard deviation 32,000. Find the probability that the
number of shares traded tomorrow will be at most 325,000.




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