15.Math-Review Wednesday 8/23/00
follows a Binomial distribution. • The Binomial distribution is given by: 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.
0.15 0.1 0.05 0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 Binomial Distribution • Graphically, for n=40, p =0.3:
Variance of Y: Binomial Distribution • Mean of Y:
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)...
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.
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
Overbooking • ...(a) continued • Definition: Binomial B(n,p) • P(X5) = 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
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 P(person not show up for a particular flight) = 0.2;P(person shows up for a particular flight) = 0.8
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!)...
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?
Overbooking • (c) Expected profit Z if 11 tickets were sold= E(1200X) - P(X = 11)(1200+3000) = $ 1200 E(X) = $ 1200*11*(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) = 0.2;P(person shows up for a particular flight) = 0.8
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
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...
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.
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:
f(x) F(x) 1 1/(b-a) a b x a b x Uniform Distribution • Graphically: • Mean is (a+b)/2 • Variance is (b-a)2/12
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).
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.
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.
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: • And it looks like a bell shaped curve:
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?
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).
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.
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.)