15.MathReview

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15.Math-Review
Wednesday 8/23/00
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Binomial Distribution

• Let us consider the following experiment
• We will flip a coin n times.
• Heads can come up with probability p for this
  coin.
• Xi is 1 if the i-th flip came up heads, 0 if it
  was tails.

• It represents k successful outcomes out of n
  independent tries.

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Binomial Distribution

• Graphically, for n40, p 0.3

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Binomial Distribution

• Mean of Y

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Overbooking

• Ontario Gateway Airlines first class cabin have
  10 seats in each plane. Ontarios overbooking
  policy is to sell up to 11 first class tickets,
  since cancellations and no-shows are always
  possible (and indeed are quite likely)...

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Overbooking

• ... For a given flight on Ontario Gateway, there
  were 11 first class tickets sold. Suppose that
  each of the 11 persons who purchased tickets has
  a 20 chance of not showing up for the flight,
  and that the likelihood of different persons
  showing up for the flight are independent.

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Overbooking

• (a) What is the probability that at most 5 of the
  11 persons who purchased first class tickets show
  up for the flight?
• Let X denote the number of people who show up for
  the flight.
• P(XP(X4) P(X5)

P(person not show up for a particular flight)
0.2P(person shows up for a particular flight)
0.8
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Overbooking

• ...(a) continued
• Definition Binomial B(n,p)

• P(X?5) P(X 0) P(X 1) P(X 2) P(X
  3) P(X 4) P(X 5)

P(person not show up for a particular flight)
0.2P(person shows up for a particular flight)
0.8
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Overbooking

• (b) What is the probability that exactly 10 of
  the persons who purchased first class tickets
  show up for the flight?
• P(X10)11!/ (10! 1!)(0.8)10 (0.2)10.236

P(person not show up for a particular flight)
0.2P(person shows up for a particular flight)
0.8
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Overbooking

• (c) Suppose that there are 10 seats in first
  class available and that the cost of each first
  class ticket is 1,200.(This 1,200 contributes
  entirely to profit since the variable cost
  associated with a passenger on a flight is close
  to zero.) Suppose further that any overbooked
  seat costs the airline 3,000, which is the cost
  of the free ticket issued the passenger plus some
  potential cost in damaged customer relations.
  (First class passenger expect not to be
  bumped!)...

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Overbooking

• (c) ... Thus, for example, if 10 of the first
  class passengers show up for the flight, the
  airlines profit is 12,000. If 11 first class
  passengers show up, the profit is 9,000. What is
  the expected profit from first class passengers
  for this flight?

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Overbooking

• (c) Expected profit Z if 11 tickets were sold
  E(1200X) - P(X 11)(12003000) 1200 E(X)
  
• 120011(0.8) - (0.8)11(4200) 10560
  257.70 10,302.30

Orig. ticket free ticket good will
P(person not show up for a particular flight)
0.2P(person shows up for a particular flight)
0.8
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Overbooking

• (d) Suppose that only 10 first class tickets had
  been sold. What would be the expected profit from
  first class passengers for this flight?
• Expected profit Z if only 10 tickets were sold
  E(1200X) 1200 E(X) 1200 10 (0.8)
  9,600

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Overbooking

• (e) (Thought Exercise) People often travel in
  groups of two or more. Does this affect the
  independence assumption about passenger behavior?
  Why or why not?
• Yes, traveling in groups of two or more affects
  the independence assumption given in the
  problem...

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Overbooking

• (e) ... The probability that the whole group
  shows up is 0.8 and that the group does not show
  up is 0.2. The probability of overbooking, i.e.
  P(X 11), is originally (0.8)11 to account for
  independence. In the case of group traveling, the
  probability is increased by a factor of 1.25
  (reciprocal of 0.8) for each person after the
  first person in the group. For example, if we
  know that there is a group of 2 people, P(X 11,
  group of 2) is 1.25 times the original P(X 11,
  independent), that is, (0.8)11 vs. (0.8)10.

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Uniform Distribution

• If X is equally likely to take on any value in
  the range (a,b) where ba, it is a uniform r.v.
• This is noted XU(a,b).
• Its probability density function is

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Uniform Distribution

• Graphically

• Mean is (ab)/2
• Variance is (b-a)2/12

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Uniform Distribution

• Example On November 15, 1991, Ursula
  hypothesizes that, at a randomly-chosen gas
  station in Massachusetts, the price of a gallon
  of unleaded gasoline is equally likely to be
  anywhere from 1.00 to 1.35. Minerva, however,
  believes that the price is equally likely to be
  anywhere over the range from 1.25 to 1.50.
  (They treat the price per gallon as a continuous
  variable).

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Uniform Distribution

• (A)Suppose that four Massachusetts gas stations
  are chosen at random. Assume that Ursula is
  correct, find the probability that
• (a) the first one chosen has a price between
  1.00 and 1.25
• (b) all four have prices between 1.00 and 1.25
• (c) none have prices between 1.00 and 1.25
• (d) at least one has a price between 1.00 and
  1.25
• (B)Find the probabilities of (a) (c) assuming
  that Minerva is correct.

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Normal Distribution

• Arguably the most important probability
  distribution.
• This family of continuous distributions follows a
  bell-shaped density curve and is called normal
  (or Gaussian).
• The normal distribution is an excellent
  representation of many physical processes and of
  numerous economic and social processes that are
  not literally continuous. This due to some
  powerful theorems in statistics.

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Normal Distribution

• If X is a normal r.v. with mean ? and standard
  deviation ? we write XN(?, ?)
• The density function for X has the form

• And it looks like a bell shaped curve

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Normal Distribution

• If Z N(0,1) it is called a standard normal r.v.
• Given XN(?, ?), the random variable Z(X- ?)/?
  is a standard normal r.v.
• The normal table enables us to find F(z)P(Z ?
  z), when ZN(0,1). This enables us to obtain
  values for any XN(?, ?).
• Example X N(2,3)
• F(3)?
• What x is such that F(x).95?

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Normal Distribution

• Example During a bull market the weekly price
  change of a share of stock X is normally
  distributed with mean 0.05P and variance 1, where
  P is the price at the beginning of the week.
• (a) If a share of stock X costs 24 at the
  beginning of a week, what is the probability the
  stock goes up that week?
• (b) Given that the stock goes up that week, what
  is the probability it reaches 27?
• (c) Given that the stock goes up that week, is
  the probability of further increase the next week
  more or less than the quantity calculated in (a)?
  Explain. (No calculations are necessary).

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Normal Distribution

• Example Mendel hypothesizes that a stock-market
  crash is imminent, with the time until the crash
  normally distributed with mean 27 (business) days
  and standard deviation 4 days. Until the crash,
  stocks will gain in value at an average of 2 per
  day (i.e. if a share sells at price V on one day,
  it will sell on average at 1.02V the next).. On
  the day of the crash, the market will drop by
  50, and it will stay at that level for a long
  while thereafter.

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Normal Distribution

• Mendel has an investment of A dollars in a
  diverse portfolio of stocks. Assume that the
  value of his portfolio changes each day by the
  same percentage as does the entire stock market.
  Assume also that his hypothesis about the
  stock-market crash is correct.
• (a) For each x from 20 to 25, find the
  probability that the market crash occurs on the
  xth day from now.
• (b) Given that the crash occurs 24 days from now
  and that Mendel has not sold his stocks before
  then, what will be the value of his investment
  after the crash? (Answer in terms of A.)