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Business Statistics 4

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  • 7/29/2019 Business Statistics 4

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    Discrete Probability Distributions

    Chapter-4

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    Probability Distribution

    Probabilities tell you???

    how likely certain events.

    What probability does not tell.

    overall impact of these events

    and, what it means to you?

    In this chapter we will see how to use

    probability to predict long-term outcomes and

    also measure the certainty of these predictions

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    Probability Distribution

    Random Variables

    Discrete Probability Distributions

    Expected value and Variance Binomial Distribution

    Poisson Distribution

    Hypergeometric Distribution

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    Random Variables

    A random variable is a numerical description of theoutcome of an experiment.

    A discrete random variable may assume either a

    finite number of values or an infinite sequence ofvalues.

    A continuous random variable may assume any

    numerical value in an interval or collection ofintervals. Eg:- Time, weight, distance, and temperaturecan be described by continuous random variables.

    Which is a variable that

    can take a set of values,

    where each value is

    associated with a

    specific probability

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    Example: XYZ MartNumber of customers

    visiting the store

    No. of days this level was

    Observed

    Probability that Random

    Variable will take on this

    value

    101 5 0.05

    102 7 0.07

    1038 0.08

    104 9 0.09

    105 10 0.1

    106 12 0.12

    107 12 0.12

    108 11 0.11

    109 10 0.1

    110 9 0.09

    111 7 0.08

    Total 100 1.00

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    Example: XYZ Mart

    0

    0.02

    0.04

    0.06

    0.08

    0.1

    0.12

    0.14

    101 102 103 104 105 106 107 108 109 110 111

    Probability

    Probability

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    Random Variables

    Question Random Variable x Type

    Family

    size

    x = Number of dependents

    reported on tax return

    Discrete

    Distance from

    home to store

    x = Distance in miles from

    home to the store site

    Continuous

    Own dogor cat

    x = 1 if own no pet;= 2 if own dog(s) only;

    = 3 if own cat(s) only;

    = 4 if own dog(s) and cat(s)

    Discrete

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    Discrete Probability Distribution

    The probability distribution for a random variabledescribes how probabilities are distributed overthe values of the random variable.

    We can describe a discrete probability distributionwith a table, graph, or equation.

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    Discrete Probability Distribution

    The probability distribution is defined by aprobability function, denoted byf(x), which providesthe probability for each value of the random variable.

    The required conditions for a discrete probabilityfunction are:

    f(x) > 0

    f(x) = 1

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    Discrete Probability Distribution

    A tabular representation of the probability distribution for TV sale

    Number

    Units Sold of Days0 80

    1 50

    2 40

    3 104 20

    200

    x f(x)0 .40

    1 .25

    2 .20

    3 .054 .10

    1.00

    80/200

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    Discrete Probability Distribution

    .10

    .20

    .30

    .40

    .50

    0 1 2 3 4Values of Random Variable x (TV sales)

    Probabi

    lity

    Graphical Representation of Probability Distribution

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    Discrete Uniform Probability

    Distribution

    The discrete uniform probability distribution is thesimplest example of a discrete probabilitydistribution given by a formula.

    The discrete uniform probability function is

    f(x) = 1/n

    where:n = the number of values the random

    variable may assume

    the values of therandom variable

    are equally likely

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    Expected Value

    The expected value, or mean, of a random variableis a measure of its central location.

    The variance summarizes the variability in thevalues of a random variable.

    The standard deviation, , is defined as the positivesquare root of the variance.

    Var(x) = 2 = (x - )2f(x)

    E(x) = = xf(x)

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    Expected Value

    expected number ofTVs sold in a day

    x f(x) xf(x)

    0 .40 .00

    1 .25 .25

    2 .20 .403 .05 .15

    4 .10 .40

    E(x) = 1.20

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    Expected Value

    Variance and Standard Deviation

    0

    12

    3

    4

    -1.2

    -0.20.8

    1.8

    2.8

    1.44

    0.040.64

    3.24

    7.84

    .40

    .25

    .20

    .05

    .10

    .576

    .010

    .128

    .162

    .784

    x - (x - )2 f(x) (x - )2f(x)

    Variance of daily sales = 2 = 1.660

    x

    Standard deviation of daily sales = 1.2884 TVs

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    Binomial Distribution

    It is associated with a multi-step experiment

    Four Properties of a Binomial Experiment

    3. The probability of a success, denoted byp, doesnot change from trial to trial.

    4. The trials are independent.

    2. Two outcomes, success and failure, are possibleon each trial.

    1. The experiment consists of a sequence of nidentical trials.

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    Binomial Distribution

    where:

    f(x) = the probability of x successes in n trials

    n = the number of trialsp = the probability of success on any one trial

    ( )!( ) (1 )!( )!

    x n xnf x p px n x

    Binomial Probability Function

    Could x be a Continuous Random Variable?

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    Binomial Distribution

    Example:-

    Evans is concerned about a low retention rate foremployees. In recent years, management has seen aturnover of 10% of the hourly employees annually.Thus, for any hourly employee chosen at random,

    management estimates a probability of 0.1 that theperson will not be with the company next year.

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    Binomial Distribution

    Using the Binomial Probability Function

    Choosing 3 hourly employees at random, what isthe probability that 1 of them will leave the companythis year?

    f xn

    x n xp px n x( )

    !

    !( )!( )

    ( )

    1

    1 23!(1) (0.1) (0.9) 3(.1)(.81) .2431!(3 1)!

    f

    Let: p = .10, n = 3, x = 1

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    DecisionTree

    Binomial Distribution

    1st Worker 2nd Worker 3rd Worker x Prob.

    Leaves

    (.1)

    Stays(.9)

    3

    2

    0

    2

    2

    Leaves (.1)

    Leaves (.1)

    S (.9)

    Stays (.9)

    Stays (.9)

    S (.9)

    S (.9)

    S (.9)

    L (.1)

    L (.1)

    L (.1)

    L (.1) .0010

    .0090

    .0090

    .7290

    .0090

    1

    1

    .0810

    .0810

    .0810

    1

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    Binomial Distribution

    (1 )np p

    E(x) = = np

    Var(x) = 2 = np(1 p)

    Expected Value

    Variance

    Standard Deviation

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    Binomial Distribution

    3(.1)(.9) .52 employees

    E(x) = = 3(.1) = .3 employees out of 3

    Var(x) = 2 = 3(.1)(.9) = .27

    Expected Value

    Variance

    Standard Deviation

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    Binomial DistributionHarish is in charge of the electronics section of a large

    departmental store. He has noticed that the probability

    that a customer who is just browsing will buy somethingis 0.3. Suppose that 15 customers browse in the

    electronics section each Hour, then

    1. What is the Prob. that at least one browsing customerwill buy something during a specific hour.

    2. What is the Prob. that at least four browsing customer

    will buy something during a specific hour.

    3. What is the Prob. that no browsing customer will buy

    anything during a specific hour.

    4. What is the Prob. that no more than four browsing

    customers will buy something during a specific hour.


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