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					5.1 Review and Preview
In this chapter we combine the methods of descriptive statistics presented in Chapters 2 and 3 and those of
probability presented in Chapter 4. We will be constructing probability distributions by presenting possible
outcomes along with the relative frequencies we expect. Also, we consider discrete probability distributions.



5.2 Random Variables
 A random variable    is a variable (typically represented by X) that has a single numerical value,
   determined by chance, for each outcome of a procedure.

 A Discrete probability distribution       is a description that gives the probability for each value of the
   random variable. The probabilities are determined by theory or by observation.


Probability Distribution: Probabilities    EXAMPLE 1: Create the graph of the probability distribution from
of Number of Mexican-Americans on a        the table.
  jury of 12. Assuming that jurors are
 Randomly selected from a Population
in which 80% of the eligible people are
          Mexican-Americans
   x (Mexican-
                           P(x)
    Americans)
        0                   0+
        1                   0+
        2                   0+
        3                   0+
        4                  0.001
        5                  0.003
        6                  0.016
        7                  0.053
        8                  0.133
        9                  0.236
       10                  0.283
       11                  0.206
       12                  0.069




Note: The table lists the values of x along with the corresponding probabilities. Probability values that are very
small, such as 0.000000123 are represented by 0+. We treat them as a value of 0.




Math 121 Chapter 5                                                                          Page 1
EXAMPLE 2: The probabilities of a return on an investment of $1000, $2000, and $3000 are ½, ¼, and ¼,
respectively. Construct a probability distribution for the data and draw a graph for the distribution.

                  X          P(X)
                 1000        0.50
                 2000        0.25
                 3000        0.25
                TOTAL        1.00




 A discrete random variable has either a finite number of values or a countable number of values, where
   “countable” refers to the fact that there might be infinitely many values, but they can be associated with a
   counting process.

 A continuous random variable has infinitely many values, and those values can be associated with
   measurements on a continuous scale without gaps or interruptions.

EXAMPLE 3: Identify the given random variable as being discrete or continuous.
  a) The cost of conducting a genetics experiment.

   b) The number of supermodels who at pizza yesterday.

   c) The exact life span of a kitten.

   d) The number of statistics professors who read a newspaper each day.

   e) The weight of a feather.

 Requirements for a Probability Distribution
  1)  P x   1 Where x assumes all possible values. (That is, the sum of all possibilities must be 1.)

    2) 0  P x   1 for every individual value of x. (That is, each probability value must be between 0 and 1
       inclusive.)
EXAMPLE 4: A researcher reports that when groups of four children are randomly selected from a population
of couples meeting certain criteria, the probability distribution for the number of girls is as given: Does the table
describe a probability distribution? Explain.




Math 121 Chapter 5                                                                            Page 2
EXAMPLE 5: Determine if the following distribution represents a probability distribution. Justify your reasoning
             x        P(x)             x         P(x)             x        P(x)
            0         0.502           -4         0.50             3        0.2
            1         0.465           -2         0.25             6        0.3
            2         0.098            0        -0.50             9         0
            3         0.011            2         0.25            12        0.3
            4         0.001            4         0.50            14        0.2
           Total      1.077          Total       1.00           Total      1.00



EXAMPLE 6: What value would be needed to make the following a probability distribution?


             x        P(x)
            0         0.13
            1         0.41
            2
            3         0.01
            4         0.25
           Total      1.00

EXAMPLE 7: Construct a probability distribution for a family of three children. Let X represent the number of
boys.




EXAMPLE 8: A game consists of rolling a die. If you roll an even number you win $6. If you roll an
odd number you lose. You must pay $4 to play. Find the probability distribution that reflects this
situation.




Math 121 Chapter 5                                                                        Page 3
EXAMPLE 9: Using the sample space for tossing two dice, construct a probability distribution for the
sums 2 through 12.

  x         P(x)




EXAMPLE 10: The probabilities that a customer selects 1, 2, 3, 4, or 5 items at a convenience store
are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively. Create a probability distribution for the data.

  x         P(x)




We calculate the mean, variance, and standard deviation for a distribution differently than we do for samples
(Chapter 3).

Additionally, we will discuss the measure of expectation or expected value. This is idea is used in games of
chance as well as decision theory.

 Round-Off Rule for         ,  ,  2 : Round results by carrying one more decimal place than the number of
   decimal places used for the random variable x.

 Mean for a Probability Distribution:                               x  Px 

The mean of a distribution captures the idea that if we were to repeat the experiment many, many times
(infinitely many) what on average would happen. It tells us what the “long-run” or theoretical average should
be.

We multiply each possible outcome by its corresponding probability and then find the sum (addition) of all
those products. It is easy to do this using an extension of the probability distribution table.




Math 121 Chapter 5                                                                        Page 4
EXAMPLE 1: Below is a probability distribution for a family of three children. Let X represent the number of
boys. Find the mean (average) number of children who will be boys.
  x         P(x)
                                            x  P  x  
                                                          
 0           1/8
 1           3/8
 2           3/8
 3           1/8
Total       8/8=1
EXAMPLE 2: A study conducted by a TV station showed the number of television per household and
the corresponding probabilities for each. Find the mean of the distribution. If you were taking a
survey on the programs watched on television, how many program diaries would you send to each
household in the survey? Why?

Number of
                              1                    2               3                        4
TVs X
Probability
                            0.32                0.51             0.12                     0.05
P(X)
x  Px                    0.32                1.02             0.36                     0.20

                   x  P  x  
                                 


 Variance for a Probability Distribution:                                  
                                                                    2    x   2  P  x  
Variance of a probability distribution gives us a measure of the spread or variability. However, we must use a
different approach to this computation than what we used in Chapter 3. (That formula incorporated the
population size, N, and in this case we are dealing with an experiment that is repeated infinitely many times.
Hence, we have no specific population size).

The formula above can be quite tedious to use in calculations so we have an equivalent “short-cut” formula that
is easier to work with. It is show below.

 Variance for a Probability Distribution:                   2  x 2  Px    2
 NOTE: When computing the variance of the distribution we must calculate the mean of the distribution
   first.




Math 121 Chapter 5                                                                                Page 5
EXAMPLE 3: A recent survey by an insurance company showed the following probabilities for the number of
bedrooms in each insured home. Find the mean and variance for the distribution.

Number of
                                  2                     3                   4                         5
Bedrooms, X
Probability
                                  0.3                   0.4                0.2                       0.1
P(X)
x  Px                          0.6                   1.2                0.8                       0.5

x 2  Px                        1.2                   3.6                3.2                       2.5


    x  P  x  
                  

 2    x2  P  x    2 
                     


EXAMPLE 4: Using the probability distribution for sums when tossing two dice, find the mean and
variance. (We created this distribution last class.)

Sums
X                  2        3            4       5        6        7        8        9        10            11       12       Total
Probability,
P(X)              1/36     2/36         3/36    4/36     5/36     6/36     5/36     4/36      3/36          2/36     1/36      1

x  Px          2/36     6/36       12/36    20/36    30/36    42/36    40/36     36/36    30/36         22/36    12/36      7

x 2  Px        4/36    18/36       48/36    100/36   180/36   294/36   320/36   324/360   300/36        242/36   144/36   1974/36




    x  P  x  
                  


 2    x2  P  x    2 
                     


 Standard Deviation for a Probability Distribution:                          x 2  Px    2

As before, we find the standard deviation by taking the square root of the variance. This means that in order to
find standard deviation we must first find the mean and the variance.

EXAMPLE 5: Use EXAMPLE 4 and find the standard deviation of the distribution.



    x2  P  x    2
                   




Math 121 Chapter 5                                                                                         Page 6
EXAMPLE 6: The number of refrigerators sold per day at a local appliance store is shown in the table, along
with the corresponding probabilities. Find the mean, variance, and standard deviation.
# of Fridges
                      0                 1              2                3               4
X
Probability
                     0.1               0.2            0.3              0.2             0.2
P(X)
x  Px                       0               0.2                 0.6            0.6                0.8

x 2  Px                     0               0.2                 1.2            1.8                3.2


          x  P  x  
                        


       2    x2  P  x    2 
                           

          x2  P  x    2 
                         



EXAMPLE 7: The probabilities that a customer selects 1, 2, 3, 4, or 5 items at a convenience store
are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively. Find the mean, variance, and standard deviation
for the probability distribution.
              x                         P(x)                         x  Px               x 2  Px 
            1                           0.32                             0.32                 0.32
            2                           0.12                             0.24                 0.48
            3                           0.23                             0.69                 2.07
            4                           0.18                             0.72                 2.88
            5                           0.15                             0.75                 3.75
           Total                        1.00                             2.72                  9.5

  x  Px   2.72                     2  x 2  Px   2  9.5  2.72 2  2.1
  x 2  Px   2  2.1  1.4
   The expected value (expectation) of a discrete random variable is denoted by E(X), and it represents
    the theoretical average value of the outcomes. It is obtained by finding
                                           E ( X )  x  Px 
Expectation is the same as finding the mean of the distribution. However, we often refer to the mean
or average as the expected value when dealing with games of chance, insurance, and decision
theory.

EXAMPLE 8: A game consists of rolling a die. If you roll an even number you win $6. If you roll an
odd number you lose. You must pay $4 to play. What are your expected winnings?
  x               P(x)
                                               E ( X )    x  P  x  
                                                                       
  2                .5
 -4                .5
Total              1


Math 121 Chapter 5                                                                                    Page 7
EXAMPLE 9: If a player rolls two dice and gets a sum of 2 or 12, she wins $20. If the person gets a
7, she win $5. The cost to play the game is $3. Find her expectation of the game. What is the
expectation for the casino?
  x         P(x)
                                    E ( X )    x  P  x  
                                                            
 17         2/36
  2         6/36
 -3        28/36
Total        1



EXAMPLE 10: A lottery offers one $1000 prize, one $500 prize, and five $100 prizes. One thousand
tickets are sold at $3 each. Find the expectation if a person buys one ticket.
  x         P(x)
                                    E ( X )    x  P  x  
                                                            
997        1/1000
497        1/1000
 97        5/1000
 -3       993/1000
Total         1



EXAMPLE 11: A 35-year-old woman purchases a $100,000 term life insurance policy for an annual
payment of $360. Based on a period life table for the US government, the probability that she will
survive the year is 0.999057. Find the expected value of the policy for the insurance company.
      x            P(x)
                             E ( X )    x  P  x  
                                                     
 -99640        0.000943
   360         0.999057




Math 121 Chapter 5                                                               Page 8
5.3 Binomial Probability Distribution
Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two
relevant categories, such as acceptable/defective or survived/died.

 A binomial probability distribution           results from a procedure that meets all the following
    requirements:
       1) The procedure has a fixed number of trials.
       2) The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in
          the other trials.)
       3) Each trial must have all outcomes classified into two categories (commonly referred to as success
          and failure).
       4) The probability of a success remains the same in all trials.

 Notation for Binomial Probability Distributions
S and F (success and failure) denote the two possible categories of all outcomes; p and q will denote the
probabilities of S and F, respectively, so
P( S )  p              denotes the probabilit y of successin one of the n trials.
P( F )  1  p  q       denotes the probabilit y of failure in one of the n trials.
n                        denotes the fixed number of trials
x                       denotes a specific # of successesin n trials, so x can be any
                        whole number between ________

P(x)                    denotes the probabilit y of getting exactly xsuccessesamong the n trials.
 Note: The word success as used here is arbitrary and does not necessarily represent something good. Be
    sure that x and p both refer to the same category being called success.

 Note:     When selecting a sample (such as a survey) for some statistical analysis, we usually sample
    without replacement, and sampling without replacement involves dependent events, which violates the
    second requirement in the above definition. However, the following rule of thumb is commonly used
    (because errors are negligible): When sampling without replacement, the events can be treated as if they
    are independent if the sample size is no more than 5% of the population size. ( n  0.05N )

EXAMPLE 1: In the case of Castaneda v. Partida it was noted that although 80% of the population in a Texas
county is Mexican-American, only 39% of those summoned for grand juries were Mexican-American. Let’s
assume that we need to select 12 jurors from a population that is 80% Mexican-American, and we want to find
the probability that among 12 randomly selected jurors, exactly 7 are Mexican-American.
a) Does this procedure result in a binomial distribution?
     1) The procedure has a fixed number of trials?

    2) The trials are independent?

    3) Each trial must have all outcomes classified into two categories?

    4) The probability of a success remains the same in all trials?



Math 121 Chapter 5                                                                           Page 9
b) If this procedure does result in a binomial distribution, identify the values of n, x, p and q.

   n = 12, x = 7, p = 0.80, q = 1 – 0.80 = 0.20




                                               P x  
                                                              n!
 Binomial Probability Formula                                         p x q n x   for   x = 0, 1, 2, … n
                                                          n  x ! x!
EXAMPLE 2: Use the binomial probability formula to find the probability of getting exactly 7 Mexican-
Americans when 12 jurors are selected at random from a population that is 80% Mexican-American.




EXAMPLE 3: Determine whether the given procedure results in a binomial distribution. For those that are not
binomial, identify at least one requirement that is not satisfied.
    a) Treating 50 smokers with Nicorette and asking them how their mouth and throat feel.



   b) Treating 50 smokers with Nicorette and recording whether there is a “yes” response when they are
      asked if they experience any mouth or throat soreness.




EXAMPLE 4: A test consists of multiple choice questions, each having five possible answers {a, b, c, d, e},
one of which is correct. Assume that you guess the answers to six such questions.
   a) Use the multiplication rule to find the probability that the first two guesses are wrong and the last four
       guesses are correct. That is, find P(WWCCCC), where W is wrong and C is correct.




   b) Beginning with WWCCCC, make a complete list of the different possible arrangements of two wrong
      answers and four correct answers, then find the probability for each entry in the list.




Math 121 Chapter 5                                                                                 Page 10
   c) Based on the preceding results, what is the probability of getting exactly four correct answers when six
      guesses are made?




EXAMPLE 5: Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the
binomial table to find the probability given: n = 6, x = 4, p = 0.20. Then use calculator to check results.
Compare the results you got in this problem with the answer from 4c.




EXAMPLE 6: A study was conducted to determine whether there were significant differences between medical
students admitted through special programs (such as affirmative action) and medical students admitted
through the regular admissions criteria. It was found that the graduation rate was 95% for the medical students
admitted through special programs.
    a) If 10 of the students from the special programs are randomly selected, find the probability that at least 9
       of them graduated.




   b) Would it be unusual to randomly select 10 students from the special programs and get only 7 that
      graduate? Why or why not?




Math 121 Chapter 5                                                                        Page 11
5.4: Mean, Variance, and Standard Deviation for the Binomial Distribution
 Mean for a Binomial Distribution:                                       np

 Variance for a Binomial Distribution:                                  2  npq

 Standard Deviation for a Binomial Distribution:                         npq

 Range Rule of Thumb:       Use to identify “unusual” results. We saw this same idea when we talked
   about variance and standard deviation in Chapter 3. In general, usual values lie within 2 standard
   deviations of the mean.
    Minimum “usual” value =   2

          Maximum “usual” value =   2

EXAMPLE 7: The table below describes the probability distribution for the number of Mexican-Americans
among 12 randomly selected jurors in Hidalgo County, Texas. Assuming that we repeat the process of
randomly selecting 12 jurors and counting the number of Mexican-Americans each time, find the mean number
of Mexican-Americans (among 12), the variance, and the standard deviation.

       x            P(x)               x  Px           x 2  Px 
    0               0+                     0                  0
    1               0+                     0                  0
    2               0+                     0                  0
    3               0+                     0                  0
    4              0.001                 0.004              0.016
    5              0.003                 0.015              0.075
    6              0.016                 0.096              0.576
    7              0.053                 0.371              2.597
    8              0.133                 1.064              8.512
    9              0.236                 2.124             19.116
   10              0.283                 2.830             28.300
   11              0.206                 2.266             24.926
   12              0.069                 0.828              9.936
  Total              1                   9.598             94.054


     x  Px   9.6

    2  x 2  Px    2  94 .054  9.598 2  1.9

     x 2  Px    2  1.932396  1.4




Math 121 Chapter 5                                                                  Page 12
EXAMPLE 8: Use these results and the range rule of thumb to find the maximum and minimum usual values.
Based on the results, determine whether a jury consisting of 7 Mexican-Americans among 12 jurors is usual or
unusual.




EXAMPLE 9: Grand Jury Selection: Use the jury selection problem.
Now find the mean and standard deviation for the numbers of Mexican-Americans on juries selected from this
population that is 80% Mexican-Americans using binomial distribution.




EXAMPLE 10: Grand Jury Selection: Use the jury selection problem.
  a) Assuming that groups of 870 grand jurors are randomly selected, find the mean and standard deviation
     for the numbers of Mexican-Americans.




   b) Use the range rule of thumb to find the minimum usual number and the maximum usual number of
      Mexican-Americans. Based on those numbers, can we conclude that the actual result of 339 Mexican-
      Americans is unusual? Dos this suggest that the selection process discriminated against Mexican-
      Americans?




Math 121 Chapter 5                                                                    Page 13
EXAMPLE 11: Mars, Inc. claims that 14% of its M&M plain candies are yellow, and a sample of 100 such
candies is randomly selected.
   a) Find the mean and standard deviation for the number of yellow candies in such groups of 100.




   b) A data Set consists of a random sample of 100 M&Ms in which 8 are yellow. Is this result unusual?
      Does it seem that the claimed rate of 14% is wrong?




Math 121 Chapter 5                                                                   Page 14

				
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