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15.Math-Review. Review 2. Exercises.

15.Math-Review

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15.Math-Review

Review 2

- P1. Historical data indicates that the starting salary for a new MBA graduate in a leading management consulting firm can be modeled as a normal distribution with mean $90,000 and standard deviation $20,000. Second year salaries increase by 20%. The bonus each year, for the first and following years, can be modeled as a normal distribution with mean $25,000 and standard deviation $5,000. We assume that the bonus is independent of the initial salary increase. For the purpose of this problem annual compensation is salary plus bonus.
(a) What is the expected annual compensation for a new hire?

(b) What is the standard deviation of the annual compensation for a new hire?

(c) What is the expected annual compensation after completing a year in the firm, I.e., just after the salary increase is announced?

(d) What is the probability that the annual compensation after completing a year in the firm, i.e., just after the salary increase is announced, will exceed $140,000?

(Hint: The annual compensation is also normally distributed).

- P2. Suppose 5% of the microchips produced by a leading microchip manufacturer are defective. An inspector inspects 10 microchips. Historically, given that a microchip is defective, the inspector accepts it 10% of the time thinking it has no defect. If a microchip is not defective, he always correctly accepts it.
(a) What is the probability that all 10 microchips in the sample are not defective?

(b) What is the probability that the inspector accepts the first microchip?

(c) What is the probability that the inspector accepts 9 (out of 10) microchips?

(d) Given that the inspector accepts a microchip, what is the probability that it has no defect?

- P3. To the best of our knowledge, with probability 0.8 Al is guilty of the crime for which he is about to be tried. Bo and Ci, each of whom knows whether or not Al is guilty, have been called to testify. Bo is a friend of Al’s and will tell the truth if Al is innocent, but will lie with probability 0.2 if Al is guilty. Ci hates every body but the judge and will tell the truth if Al is guilty but will lie with probability 0.3 if Al is innocent. Given this model:
(a) Determine the probability that the witnesses give conflicting testimony.

(b) Which witness is more likely to commit perjury?

(c) What is the conditional probability that Al is innocent, given that Bo and Ci gave conflicting testimony?

(d) Are the events “Bo tells a lie” and “Ci tells a lie” independent? Are these events conditionally independent to an observer who knows whether or not Al is guilty?

- P4. The probability that any particular bulb will burn out during its Kth month of use is given by the probability distribution function for K:
Does this distribution sum to 1?

- Four bulbs are life-tested simultaneously. Determine the probability that
(a) None of the four bulbs fails during its first month of use.

(b) Exactly two bulbs have failed by the end of the third month.

(c) Exactly one bulb fails during each of the first three months.

(d) Exactly one bulb has failed by the end of the second month, and exactly two bulbs are still working at the start of the fifth month.

- P5. Discrete random variable X is described by the probability distribution function:
- Let d1, d2,…, d20 represent 20 successive independent experimental values of random variable X.
- (a) Determine the numerical value of K.
- (b) Determine the probability that d1 > d2 .
- ( c) Determine the probability that d1+ d2+…+ d20 1.0

- P6. Back to stocks. We know that the return of different stocks are random variables. And for:
- Snowboard Inc.: mean return = 0.9, stand. deviation of return = 0.075.
- Skiboots Inc.: mean return = 0.9, stand. deviation of return = 0.27.

- If you decide to invest 30% of your capital in Snowboard Inc., and 10% in Skiboots Inc. What is the mean and variance of the return of the resulting portfolio?
- How much of each stock should you buy to obtain a portfolio with minimal variance? (minimal risk!)

- P7. Helen has an early class tomorrow morning. She knows that she needs to get to bed by 10 PM in order to be sufficiently rested to concentrate and participate in class. However, before she goes to bed, she must start and then complete a homework assignment which is due tomorrow morning. According to her experience, the time it takes her to complete a homework assignment for this class is Normally distributed with mean = 3.5 hours and standard deviation = 1.2 hours. Helen looks at her watch, and sees that it is now 6pm. What is the probability that she will be able to get to bed in time to be sufficiently rested?

- P8. According to the latest Census, 52.2% of Boston residents are female. Suppose a group of 100 Bostonians is selected at random.
- Find the mean and the standard deviation of the number of female members of the group.
- Use the Normal approximation to find the probability that less than one half of the members of the group are female.
- Use the Normal approximation to find the probability that the number of female members of the group is between 45 and 55.

- P1.(a) 115,000
(b) 20,615.53

(c) 133,000

(d) 0.3859

- P2.(a) 0.95^10
(b) 0.955

(c) 10(0.955)^9(0.45)

(d) 0.95/0.955

- P3.(a) 0.22
(b) Bo is more likely. 0.16 > 0.06.

(c) 3/11

(d) First question NO. Second YES.

- P4.Yes
(a) (4/5)^4

(b) 0.374568

(c) 0.030968

(d) 0.154538

- P5.(a) K=5/12
(b) 47/144=0.326389

(c) 0.4228*10^-7

- P6.(a) E(Z)=0.9, V(Z)=0.0298485
(b) 80.89% of Snowboard Inc.

V(Z*)=0.0016474

- P7. P(X<=4) = 0.66095
- P8.(a) E(X)=52.2, Std(X)=2.49516
(b) P(X<50) = 0.18805

(c) P(45<=X<=55) = 0.86775