Transcript 15.060 Data, Models and Decisions

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.
n
Y   X i follows a Binomial distribution.
i 1
The Binomial distribution is given by:
 n k
P(Y  k )    p (1 p) nk for all k  0,1,2,,n
k
It represents k successful outcomes out of n independent tries.
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Binomial Distribution
 n k
P(Y  k )    p (1 p) nk for all k  0,1,2,,n
k
Graphically, for n=40, p =0.3:
0.15
0.1
0.05
0
1
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7
10 13
16 19 22
25
28 31
34
37 40
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Binomial Distribution
Mean of Y:
n
n
i1
i1
Y  E (Y )   E ( X i )  1 p  0(1 p)  np
Variance of Y:
n
 VAR(Y )  VAR( X i ) 
2
Y
i1
n
 (1 p ) 2 p  (0  p ) 2 (1 p )  np (1 p )
i1
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Overbooking
Ontario Gateway Airlines’ first class cabin have 10
seats in each plane. Ontario’s 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(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) = 0.2;
P(person shows up for a particular flight) = 0.8
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Overbooking
...(a) continued
Definition: Binomial B(n,p)
 n x
n!
n x
P( X  x)    p (1 p) 
p x (1 p) nx
x!(n  x)!
 x
 P(X5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
11!
11! 1 10 11! 2 9 11! 3 8
0.80 0.211 
0.8 0.2 
0.8 0.2  0.8 0.2 
0!11!
1!10!
2!9!
3!8!
11! 4 7 11! 5 6

0.8 0.2 
0.8 0.2
4!7!
5!6!
 0.0117

P(person not show up for a particular flight) = 0.2;
P(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(X=10)
=[11!/ (10! 1!)](0.8)10 (0.2)1
=0.236
 n
P ( X  x)    p x (1 p ) n x
 x
P(person not show up for a particular flight) = 0.2;
P(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 airline’s 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)(1200+3000)
11!
0.8110.2 0 (4200)
= $ 1200 E(X) 
11!0!
= $ 1200*11*(0.8) = $ 10560 – 257.70
= $ 10,302.30
(0.8)11(4200)
Orig. ticket
+ free ticket
+ good will
 n
P ( X  x)    p x (1 p ) n x
 x
P(person not show up for a particular flight) = 0.2;
P(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 b>a, it is a uniform r.v.
This is noted X~U(a,b).
Its probability density function is:

 1 (b  a) x[a,b]
f ( x)  

o/w
 0
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Uniform Distribution
Graphically:
f(x)
F(x)
1
1/(b-a)
a
b
x
a
b
x
Mean is (a+b)/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 bellshaped 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 X~N(, )
 The density function for X has the form:
( x  ) 2

1
2 2
f ( x) 
e
2 
 And it looks like a bell shaped curve:
1.2
1
0.8
0.6
0.4
0.2
0
-3
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-2
-1
0
-0.2
1
2
3
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Normal Distribution
 If Z ~N(0,1) it is called a standard normal r.v.
 Given X~N(, ), the random variable Z=(X- )/ is a
standard normal r.v.
 The normal table enables us to find F(z)=P(Z  z), when
Z~N(0,1). This enables us to obtain values for any X~N(, ).
 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.)
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