Commonly-Used Discrete Distributions

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Commonly-Used Discrete Distributions

• Stat 700 Lecture 06
• 9/25/2001

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Overview of Lecture

• Commonly-used discrete distributions and when
  they are appropriate
• Discrete Uniform
• Bernoulli
• Binomial
• Hypergeometric
• Poisson

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Commonly-Used Discrete Distributions

• A discrete random variable U is uniformly
  distributed over the set 1, 2, 3, , N, denoted
  U ? UNIF1,2,,N, where N is a positive integer,
  if its probability function is given by
• p(u) 1/N for u 1, 2, 3, , N.
• This is the model which assigns equal
  probabilities to the possible values of U.
• Applicability Random allocation

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Uniform continued

• For U ? UNIF1,2,,N
• Mean ? (N 1)/2.
• Variance ?2 (N2 - 1)/12.
• For example, suppose the number of students in
  Stat 700 who will be taking Stat 701 is uniformly
  distributed over 1,2,,12. Then N 12, so
• ? (12 1)/2 6.5
• ?2 (122 - 1)/12 143/12 11.92.

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Bernoulli Trials

• An experiment is a Bernoulli trial if its
  outcomes could be classified into two categories
  Success (S) or Failure (F) we could also use
  other labels such as Good or Defective,
  Female or Male, etc..
• We denote by p the probability of a Success and
  by q 1 - p the probability of a Failure.
• If X 1 denotes Success and X 0 denotes
  Failure, X is called a Bernoulli random
  variable.
• Its probability function is p(0) q and p(1)
  p.

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Examples of Bernoulli Trials

• Observing the outcome of a surgery Success or
  Failure. How to determine p?
• Observing whether for an insured person the event
  insured against occurs or not. Again, what will
  be p?
• Observing whether a machine will function for a
  specified period.
• Observing whether a genetic mutation has
  occurred.
• Determining if a biological organism will survive
  for a specified time.
• Observing whether it rains or not for a given day.

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Mean and Variance for a Bernoulli Random Variable

• If X has a Bernoulli distribution with success
  probability p, we write X ? BER(p).
• For X ? BER(p)
• Mean ? p since ? (0)(q) (1)(p) p.
• Variance ?2 pq since by the definitional
  formula,
• ?2 (0 - p)2(q) (1 - p)2(p) p2q q2p
  pq(pq) pq because p q 1.

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Binomial Experiments and Variables

• Consider now an experiment with the following
  characteristics
• a) it consists of n Bernoulli trials n
  replications
• b) each of the n Bernoulli trials has
  probability p of Success constant probability
  of success
• c) the n Bernoulli trials are independent.
  trials dont affect each other
• Then such an experiment is called a binomial
  experiment with parameters n and p.
• If X is the random variable that counts the
  number of successes out of the n trials, then X
  is said to be a binomial random variable with
  parameters n and p.

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Examples of Binomial Expts/Variables

• Example 1 Draw a sample of size n 10, with
  replacement, from a box containing 20 red and 30
  blue balls, and let X denote the number of red
  balls in the sample.
• Note the following
• success is getting a red
• n 10 (number of draws)
• p 20/50 .4 (probability of red per trial)
• independent trials hold because sampling is with
  replacement
• X counts the number of reds (successes)
• X is binomial with n 10 and p .4.

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Examples continued

• Example 2 A very large population of people is
  such that 5 have a certain type of disease, and
  95 does not have the disease. A sample of size
  n 100 is drawn from this population (without
  replacement), and X denotes the number of people
  in the sample who has the disease.
• Note that
• trial is picking a person and determining if
  he/she has the disease (success)
• independence is approximately satisfied because
  of the large size of the population
• p 0.05 (approximately) for each of the trials
• Therefore, X is (approximately) binomial with n
  100 and p 0.05.

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Probability Function of a Binomial Random Variable

• Let X be a binomial random variable with
  parameters n and p, denoted X ? BIN(n,p). Then,
  the range of X is R 0,1,2,,n.
• Its probability function is given by recall that
  q 1 - p

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Parameters of Binomial Distribution

• For X a binomial random variable with parameters
  n and p, its mean, variance, and standard
  deviation are given by

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Applications of the Binomial Distribution

• Example 1 continued In this example, X, which
  counts the number of red balls in the sample of
  size 10 is binomial with n 10 and p 0.40.
  Therefore, the probability function is given by

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Example 1 continued

• Mean ? np 10(.4) 4
• Variance ?2 npq (10)(.4)(.6) 2.4
• Std. Dev. ? (2.4)(1/2) 1.549.
• Could also compute probabilities such as
• P(X 3) 10C3(.4)3(.6)10-3 (120)(.064)(.028)
  0.21504.
• P(X lt 1) 10C0(.4)0(.6)10 10C1(.4)1(.6)9
  0.0060 0.0403 0.0463.
• Such cumulative probabilities could be obtained
  using binomial tables or Minitab.

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The Probability Function and the Cumulative
Probabilities (obtained using Minitab)
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Graph of the Probability Function
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Binomial Examples continued

• Example 2 continued In this situation, the
  variable X which is the number of diseased
  individuals in a sample of size 100 is binomial
  with n 100 and p 0.05. Its probability
  function is therefore

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Graph of the Probability Function of a
BIN(100,0.05) Variable
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Graph of the Cumulative Distribution Function for
BIN(100,0.05)
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Example continued

• From the graph of Bin(100,0.05), notice even
  though it is very right-skewed, when we focus on
  the shape for the small values (from 0 to 10),
  the shape is approximately mound-shaped. This is
  a manifestation of the normal approximation to
  the binomial.
• For this X ? BIN(100,0.05)
• Mean ? (100)(0.05) 5
• Variance ?2 (100)(0.05)(0.95) 4.75
• Standard Deviation ? (4.75)(1/2) 2.18

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Probabilities in Certain Intervals about the Mean
(Probs were calculated using computer)

• P? - ? lt X lt ? - ? P5 - 2.18 lt X lt 5 2.18
  P2.82 lt X lt 7.18 p(3) p(4) p(5) p(6)
  p(7) .1396 .1781 .1800 .1500 .1060
  0.7537
• P? - 2? lt X lt ? - 2? P5 - 2(2.18) lt X lt 5
  2(2.18) P0.64 lt X lt 9.36 p(1) p(2)
  p(3) p(4) p(5) p(6) p(7) p(8) p(9)
  0.9659
• P? - 3? lt X lt ? - 3? P5 - 3(2.18) lt X lt 5
  3(2.18) P-1.54 lt X lt 11.54 p(0) p(1)
  p(2) p(3) p(4) p(5) p(6) p(7) p(8)
  p(9) p(10) p(11) 0.9957.
• Compare the agreement of these values with what
  are expected under the empirical rule. Quite
  Good!

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Using the Binomial Tables

• The binomial tables in the back of the book
  provide the probabilities for certain values of n
  and p.
• It provides p(k) PX k for k 0, 1, 2, ,
  n.
• To compute Pa lt X lt b we use
• Pa lt X lt b p(a1) p(b), provided that
  the values of n and p are in the table.
• Otherwise, computer programs (Minitab) or
  calculators could be used to calculate the
  probabilities.

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Using the Binomial Tables

• Example Assume that the proportion of all adult
  Americans who does not approve of President
  Bushs handling of the current crisis is 0.20.
  Suppose that a sample of size n 10 is taken
  from this population, and we let X denote the
  number in this sample who does not approve of
  Bushs handling of the crisis.

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Example continued

• Since the population sampled is so large and
  sampling is random, the variable X will be
  binomial with parameters n 20 and p 0.20.
• ? 20(.2) 4 and ? 20(.2)(.8)(1/2)
  (3.2)(1/2) 1.79.
• Using the binomial table with n 20 and p
  0.20, we also obtain PX 3 .2054.
• Also, PX lt 3 p(0) p(1) p(2) p(3)
  .0115 .0576 .1369 .2054 .4114.
• If X is BIN(n,p) with p gt .5, then Y n - X is
  BIN(n,1-p), so convert problem into Y.

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From Binomial to Poisson

• A box contains many, many balls some of which are
  red and others blue. Let p be the proportion of
  red balls in the box. Suppose we draw n balls
  (with replacement) from this box, and X denotes
  the number of red balls in the sample.
• Then, from what we have studied so far, X has a
  binomial distribution with parameters n and p.
  That is,
• p(x) (nCx)pxqn-x, x0,1,2,,n.

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Continued ...

• Imagine the situation where we increase the
  sample size n but at the same time decrease the
  proportion of red balls in the box such that the
  mean ? np remains equal to some constant ?,
  that is, ? np ?.
• Notice that as we increase the value of n, then
  the set of possible values of X also becomes
  larger.
• Clearly, the distribution of X as n increases
  still remains Binomial(n,p).
• But the question is
• What happens to the binomial(n,p) when we let n
  approach infinity with the constraint np ??

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Here comes the Poisson!

• It turns out that under such a situation, we
  obtain a new distribution, which is a limiting
  distribution of the binomial.
• This new distribution, called the Poisson
  distribution, is usually an excellent model for
  modeling rare events.
• The relation with the binomial is as follows

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Poisson Distribution

• A discrete random variable X taking values in the
  set 0, 1, 2, 3, is said to have a Poisson
  distribution with parameter or intensity rate ? gt
  0, and written X ? POI(?), if its probability
  function is given by the formula

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Some Properties of the Poisson Distribution

• Provides an approximation to the binomial
  probabilities when the parameters in BIN(n,p) are
  such that p is small and n is large. The Poisson
  approximation is the one with intensity ? np.
• It is a model for counting the occurrence of
  rare events, such as the occurrence of
  epidemics, getting cancer, accidents, terrorist
  acts, etc.
• For X ? POI(?), the mean of X is ? ? and at the
  same time, the variance is ?2 ?. These results
  can be seen by recalling the mean and variance
  for the binomial distribution.

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Applications of the Poisson

• Problem Suppose that the proportion in the
  population who are infected with a deadly virus
  is 0.0005. If 10000 people are chosen at random,
  what is the probability that exactly 4 out of
  these 10000 people will be infected?
• Solution 1 Since X ? BIN(10000,.0005), we could
  calculate the probability by making use of the
  binomial probability to obtain
• PX 4 (10000C4)(.0005)4(.9995)9996
  0.1754936.

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Solution continued

• Solution 2 Since n 10000 is very large, and p
  .0005 is very small, we could also approximate
  this probability by using the Poisson
  distribution with parameter or intensity rate ?
  np (10000)(.0005) 5. Consequently, the
  approximate probability is
• PX 4 ? (e-5)54/4! 0.175467.
• Comparing this approximate value with the exact
  value obtained from the binomial, notice that it
  is very good.

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Graph of the Probability Function of a
Poisson(?2) Distribution
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Sampling Without Replacement

• Let there be a group consisting of N objects, K
  of which are of Type I and (N-K) of Type II.
• Sample n objects, with replacement, and let X
  denote the number of Type I objects, then X has a
  binomial distribution with parameters n and p
  K/N.
• In this case, the mean is ? n(K/N) and the
  variance is ?2 n(K/N)(1 - K/N).
• What happens to the distribution of X when
  sampling is without replacement?

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Hypergeometric Distribution

• When sampling is without replacement, then the
  probability function of X, the number of Type I
  objects in the sample, is
• p(x) (KCx)(N-KCn-x)/(NCn), x0,1,2,,n.
• This is called the hypergeometric distribution.
• Mean of X ? n(K/N), same as in binomial.
• Var(X) ?2 n(K/N)(1-K/N)(N-n)/(N-1).
• This distribution is approximately BIN(n,K/N)
  when N is large compared to n.