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    DOM105 Session 3

    Uploaded by

    Tanish Gandhi
    This document provides information on various probability distributions including discrete, binomial, uniform, hypergeometric, Poisson, and normal distributions. It defines key terms like probability, expected value, variance, and standard deviation. Examples are provided to illustrate concepts like calculating the probability of certain outcomes for each distribution and using normal distributions to find probabilities based on z-scores. Formulas for each distribution are also outlined.

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    DOM105 Session 3

    Uploaded by

    Tanish Gandhi
    12 views17 pages
    This document provides information on various probability distributions including discrete, binomial, uniform, hypergeometric, Poisson, and normal distributions. It defines key terms like probability, expected value, variance, and standard deviation. Examples are provided to illustrate concepts like calculating the probability of certain outcomes for each distribution and using normal distributions to find probabilities based on z-scores. Formulas for each distribution are also outlined.

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    This document provides information on various probability distributions including discrete, binomial, uniform, hypergeometric, Poisson, and normal distributions. It defines key terms like probability, expected value, variance, and standard deviation. Examples are provided to illustrate concepts like calculating the probability of certain outcomes for each distribution and using normal distributions to find probabilities based on z-scores. Formulas for each distribution are also outlined.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    12 views17 pages

    DOM105 Session 3

    Uploaded by

    Tanish Gandhi
    This document provides information on various probability distributions including discrete, binomial, uniform, hypergeometric, Poisson, and normal distributions. It defines key terms like probability, expected value, variance, and standard deviation. Examples are provided to illustrate concepts like calculating the probability of certain outcomes for each distribution and using normal distributions to find probabilities based on z-scores. Formulas for each distribution are also outlined.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 17

    DOM105 2019

    Session 3
    Reference: SfM Ch. 5
    Probability distribution of a discrete variable
    A discrete variable is a variable that takes only discrete values. These
    values may not be integer, but they do not form a continuous function.
    It is a mutually exclusive list of all possible numerical outcomes along
    with the probability of each outcome occurring.
    Eg: The number of possible absentees in an office:
    No. of absentees (x) Probability
    0 0.15
    1 0.35
    2 0.2
    3 0.15
    4 0.1
    5 0.05
    Expected value of a discrete variable
     The expected value of a discrete variable is the weighted average of all the
    outcomes, the weights being the probability scores.

    µ

    In the previous example, µ =


    0.15(0)+0.35(1)+0.2(2)+0.15(3)+0.1(4)+0.05(5) = 1.85
    Variance and standard deviation of discrete variable
     The variance of the discrete variable is the sum of the squared difference
    between outcome and expected value, multiplied by the probability of that
    outcome.

    Variance =

    The standard deviation is the square root of the variance.

    σ==
    The Uniform Distribution
     Also called rectangular dist. Has the same chance of occurrence anywhere
    in its range.
    P(X<=x)=(x-a)/(b-a), where b is max(x) and a is min(x)
    Mean:
    Variance:
    Standard Deviation:
    The Binomial Distribution
     A discrete random variable distribution created by a Bernoulli Process,
    which has the following properties –
    It is a series of trials, each trial has only two outcomes, with probabilities
    p and 1-p
    The value of p stays fixed over the course of the process
    The trials are statistically independent
    If there are n trials, the chances of obtaining exactly r successes (r<=n) is
    given by the binomial formula (let q = 1-p):
    Central Tendency of the Binomial Distribution
     

    Mean: np
    Variance: npq
    Standard deviation:

    Final note: To apply the binomial distribution, we must first


    ensure that the process meets the conditions for a Bernoulli
    Process.
    Hypergeometric Distribution
     Where the binomial distribution the sample data are selected with
    replacement from a finite pool (or without from an infinite pool) the
    hypergeometric distribution is found when the samples are taken from a
    finite pool without replacement.
    If n samples are taken from population N, and out of the population A
    members are of interest, then the probability of exactly x successes out of
    n samples is:

    Mean = Std. dev.


    Out of 20 people 12 have a Masters degree. If you select 5 at random without
    replacement, what is the chance of 3 people having masters?
    (12C3*8C2)/20C5

    P of there being at least one masters?


    P(x=1)+P(x=2)+…+P(x=5) = 1 - P(x=0) = 1 – (12C0*8C5)/20C5
    Poisson Distribution
    Characteristics of the Poisson Process:
    The process is applied to a discrete random variable that takes integer
    values
    The average value of the random variable over the given time period is
    already known or can be calculated given past data
    At any one second, the possibility of a positive outcome is very small, and
    a fixed value.
    At any one second, the possibility of two or more positive outcomes is so
    small we can assign it a value of zero.
    The probability of a positive outcome at any given second is not only
    fixed, but independent of the actual time as well as the result in any other
    second.
    The Poisson Formula
     Let λ be the mean number of occurrences in the interval of time under
    study.
    e is the base of the natural logarithm system, approx. 2.71828
    Poisson probability of x number of incidents occurring

    If a binomial process has a large number of trials (n>20) and a small
    probability of success (p<0.05), we can use the Poisson formula after
    substituting the binomial mean np.
    A store receives 5 customers an hour on average. If the hourly customer
    arrival follows a poisson distribution, what is the probability of receiving 10
    or more customers in an hour?
    The normal distribution
    It is a continuous probability distribution. Also called Gaussian
    distribution.
    Can be used to approximate discrete distributions, with sufficiently large
    samples. Can approximate Binomial distribution if np,nq>5
    It is symmetrical, bell-shaped in appearance, its interquartile range is from
    -0.67 standard deviations to +0.67 std.devs.
    Normal Distribution from Z-score
    For
      a normal distribution, we can find thee probability of the variable being below a
    certain value by using the Z-table.
    Calculate Z=, the corresponding value in the table shows the probability of the
    variable being less than or equal to that value.
    To find the probability of the variable being between X1 and X2, P(X1<X<X2)=
    P(Z(X2))-P(Z(X1)).
    P(X<X1)=P(Z(X1)
    P(X>X1)= 1-P(Z(X1)
    Normal curve of mean 200, stdev 50, what is P(X<=168)? Z(168)=-0.64, P(Z(168)=
    0.2611
    What is P(X>300)? Z(300) = 2. P(Z(300)) = 0.9772, P(X>300) = 1-0.9772=0.0228
    What is P(168<=X<=240) = P(X<=240)-P(X<=168) = 0.7881-0.2611=0.527
    The annual household income of 300 surveyed families has a mean of 16 lakhs
    with stdev 90,000. How many families have an income between 10 and 15
    lakh?
     n=300, Mean=16, Stdev = 0.90
    No. of houses between 10 and 15 = n*P(10<=x<=15) = n*[P(x<=15)-
    P(x<=10)]
    P(x<=15)=P(Z(15))=P((15-16)/0.9)=P(-1.11) = 0.1335
    P(x<=10) = P(Z(10))=P((10-16)/0.9)=P(-6.67)=0
    No. of houses = 300*(0.1335 – 0)=40.05 or 40 approx.
    How many families do you expect to have income above 17.5 lakhs?
    P(x>17.5) = 1 – P(x<=17.5)
    P(x<=17.5) = P(z(17.5)) = P(1.67) = 0.9525
    No. of families = 300*(1 – 0.9525) = 14.25 or 14 approx
    Excel commands for Normal Distribution
    NORM.DIST(x, Mean, stdev, 1): Returns the probability of a random variable
    being less than or equal to x, for a given Mean and Stdev.
    NORM.INV(p, Mean, Stdev): Finds the value x for which (PX<=x)=p, for a
    given mean and stdev.
    NORM.S.DIST(z,1): Returns the corresponding probability value for a certain
    Z-score (can act as substitute for Z-table)
    NORM.S.INV(p): Returns the Z-score for a certain probability (can act as
    substitute for Z-table)

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