Introduction to Management Science

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Introduction to Management Science
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Probability Statistics Ch. 11

• Random variables are functions based on the
  outcome of a random process.
• There are two basic types of r.v.
  discrete e.g. roll a die 1,2,3,4,5,6continu
  ous e.g. daily rainfall 0,1000.

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Discrete random variable

• Roll a fair (unbiassed) die and observe number on
  uppermost face.
• The possible events are 1, 2, 3, 4, 5, 6.
• These events aremutually exclusive
  (m.e.)collectively exhaustiveequally likely.

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Die rolling

• Probability of each event is 1/6.Because ..
• Probs of m.e. events can be added P(1 or 2)
  P(1) P(2).
• Total prob of collectively exhaustive events is
  1.
• Equally likely means equal probs.

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Toss a fair coin

• The possible events are head (H) and tail (T).
• These events aremutually exclusivecollectively
  exhaustiveequally likely.
• P(H) P(T) 0.5 (or ½)

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Text page 445

• The text is a little misleading.It just says
  that the probs of m.e. events sum to 1, without
  mentioning that they also need to be collectively
  exhaustive.
• E.g. die roll P(1) P(2) 1/6 1/6 1/3

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More generally

• For events A and B, P(A or B) P(A) P(B)
  P(A and B), or, sometimesP(AB) P(A) P(B)
  P(AB).
• If A and B are m.e. then P(AB)0, so P(AB)
  P(A) P(B) 0 P(A) P(B) as before.(Note
  typo on page 447).

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More generally II

• For three events A, B and C, P(A or B or C)
  P(A) P(B) P(C) P(AB) P(BC) P(AC)
  P(ABC)(Draw a Venn diagram!)

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Independent Events

• A and B are independent if and only if
  P(AB) P(A)P(B).This is quite different
  from mutually exclusive, where P(AB)0.

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Independent Events

• For example, draw a card at random from a pack of
  52 playing cards.P(card is a King) 4/52
  1/13. P(card is a Diamond) 13/52 1/4. P(K
  and ?) P(K)P(D) 1/52.

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Independent Events

• Independence often applies when a process
  consists of a number of steps.
• E.g. (roll a die, toss a coin, pick a
  card).P(3,T,King) (1/6)(1/2)(1/13)
  (Multiplication Principle).

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Conditional Probability

• P(AB) P(AB) / P(B)is the probability of
  event A, with the condition that event B must
  also occur, or is already known to have
  occurred.
• e.g., P(card is ? card is red) ½.

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Bayes Rule

• If A and B are mutually exclusive and
  collectively exhaustive, and D is just another
  event, then P(D) P(DA) P(DB)
  P(DA)P(A)P(DB)P(B), so .

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Bayes Rule

• If A and B are mutually exclusive and
  collectively exhaustive, then P(AD) P(AD)
• P(D) P(DA)P(A)
• P(DA)P(A)P(DB)P(B)

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Bayes Rule

• Example from textbookMachine is set up
  correctly (C) or incorrectly (IC), each with prob
  ½. It may make defective parts (D).P(DC)
  0.1, P(DIC) 0.4.If the machine produces a
  defective part, whats the probability its set
  up incorrectly?

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Bayes Rule

• P(C) ½, P(IC) ½ P(DC) 0.1, P(DIC)
  0.4P(ICD) P(DIC)P(IC)
• P(DIC)P(IC)P(DC)P(C) 0.2 0.8 0.2
  0.05

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Bayes Rule

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Bayesian Exercise

• Seventy percent of all men over a certain age
  have prostate cancer. There is a test that
  detects cancer in 95 of patients who have it,
  but also falsely detects cancer in 10 of
  patients who dont have it.
• When the test indicates that a patient has
  cancer, what is the probability of the diagnosis
  being correct?
• Answer at end of this show.

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Binomial Distribution

• Bernoulli trialsuccess prob pfailure prob q
  1-p.
• In a sequence of n such trials, the probability
  of r successes is P(r) C(n,r) pr qn-rwhere
  C(n,r) n! / (r!(n-r)!).
• See text page 451.

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Binomial Distribution

• E.g. roll a die, success is a 1. Do it 3 times
  and count successes.P(0) 1(1/6)0(5/6)3
  125/216P(1) 3(1/6)1(5/6)2 75/216P(2)
  3(1/6)2(5/6)1 15/216P(3) 1(1/6)3(5/6)0
  1/216

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Mean and Variance

• The mean (µ), or expected value, of a discrete
  numerical random variable is given by the sum
  over all values, of the value multiplied by its
  probabilityE(x) x1P(xx1) x2P(xx2)

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Mean and Variance

• The variance is the expected value of the square
  of the difference between the value of the random
  variable and its meanvar(x) s2
  E(x-E(x)2) x1-E(x)2 P(xx1) x2-E(x)2
  P(xx2) E(x2) E(x)2

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Mean and Variance

• Binomial examplemean 0125/216 175/216
  215/216 31/216 108/216 ½.
• Generally, mean np.
• s2 E(x2) E(x)2 0125/216 175/216
  415/216 91/216 ¼144/216 ¼ 5/12.
• Generally, variance npq.

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Standard deviation

• The std devn, s, is just the square root of the
  variance. s v s2 .
• Note that if the random variable has any units,
  like breakdowns per month, then the mean and
  the std devn have the same units, but the units
  of the variance would be (breakdowns per month)2.

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Averages

• There are several measures of the location of a
  distribution. They can all be called averages.
• Among them are the meanmode the most
  frequently occurring valuemedian the middle
  value.
• Variance is a measure of spread.

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Continuous Distributions

• Prob of any particular value is 0.P(xa) 0.
• Prob of being between two different values is at
  least zero.P(a x b) 0 if a lt b.
• P(a x b) area under prob density function
  curve between a and b.

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Continuous Distributions

• P(a x b) found by evaluating an integral
  (where possible) rather than a sum.
• Mean and variance calculated using integrals
  instead of sums.

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Normal Distribution

• Is a continuous distribution.
• So called because if we take means of a group of
  samples from any distribution then these means
  will have a normal distribution.
• Cant be directly integrated.

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Standard Normal Distribution

• Mean 0
• Std devn 1
• Look up probabilities in a table (Appendix A,
  page 733).
• Any normal distribution can be converted to a
  standard normal by subtracting the mean and
  dividing by the std devnZ (x-µ)/s

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Sample Mean and Variance

• Calculated from a sample of the population.
• Approximate true population mean and variance.

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Chi-square test

• Tests based on the chi-square distribution can be
  used to check validity of several types of
  statistical assumptions.
• In particular, can test the assumption that data
  comes from a normal distribution.See example in
  text, pages 465-468.

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Chi-square test of normality

• There is an exercise on this in Activity F2
  (number 40).
• In the assignment 4 question Valley Swim Club
  Case Problem, just assume a normal distribution
  in the data, and use this in part 1. There is no
  need to test the normality.

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Bayesian Exercise solution

• P(C) 0.7, P(notC) 0.3
• P(positiveC) 0.95
• P(positivenot C) 0.1
• P(Cpos) 0.665 / (0.6650.03) 0.665/0.695
  0.957.