Eco 251 Quantitative Business Analysis I

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ECO 251 QUANTITATIVE BUSINESS ANALYSIS I

SECOND EXAM

OCTOBER 29, 2002

NAME: _____KEY______

SECTION ENROLLED: MWF TR 10 11 12:30

(Circle both days and time separately or just write down days and time)

Part I. Multiple Choice (40 Points) Remember that 'At least one' means 'one or two or three or four or……'. It does not mean 'exactly one.'

1.For the joint probability table below, the probability of the union of (not) and is:

/ .20 / .20
/ .10 / .50

a. .42

b.* .80

c. .52

d. Above 1

e. None of the above. For credit, write in an answer, showing your work

Solution: By the addition rule . You could also say that .

In problem 1, and are

a. Complements

b. Mutually exclusive

c. Collectively exhaustive

d. Independent

e.* None of the above.

The following table shows the probabilities of the Whizbang Corporation’s stock movement:

Economic Conditions / Stock Price Rises / StockPriceFalls
Good / .30 / .03 / __
Fair / .20 / __ / __
Poor / __ / .22 / __
Total / __ / .45

Complete the table and then find the probability that the stock price goes up given that economic conditions are fair.

a. .550

b.* .500

c. .200

d. .220

e. .364

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Solution: Here is the completed table.

Economic Conditions / Stock Price Rises / StockPriceFalls / Total
Good / .30 / .03 / .33
Fair / .20 / .20 / .40
Poor / .05 / .22 / .27
Total / .55 / .45 / 1.00

Using the notation in the next problem,

Let us name the events in the previous problem as follows: Economic Conditions Good; Economic Conditions Fair; Economic Conditions Poor; Stock price goes up and Stock price goes down. Then the probability that the stock price is going down if conditions are poor is

b. *

e. None of the above.

In the two problems above, the joint probability of a falling stock price and fair economic conditions is:

a.*

If the probability that a child is a boy or a girl is equally likely and I have 3 children, what is the probability that at least one is a girl?

a. .750

b. .375

c.* .875

d. .625

e. .667

Solution: The probability of at least one girl could be the sum of the probabilities of 1, 2 and 3 girls. To see how to do it this way look at the examples where we found the probabilities of 1, 2 or three heads. To do it right, find .

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A factory has three veeblefetzers. Veeblefetzer A produces defective items 3 percent of the time and thus is only used for 10% of the output. Veeblefetzer B produces 20% of the output, and 2% of its output is defective. Because only 1% of the output of Veeblefetzer C is defective, 70% of the output is produced on Veeblefetzer C. If a defective item is found, the probability that it comes from Veeblefetzer C is closest to

a.* 50%

b. 40%

c. 30%

d. 20%

e. 10%

Solution: Here is the completed table.

/ / / Total
/ 0.3 / 0.4 / 0.7 / 1.4
/ 9.7 / 19.6 / 69.3 / 98.6
Total / 10 / 20 / 70 / 100

To get the ‘total’ row, take 100 and multiply it by the percents given for production by each machine.

To get the row, multiply the total by the percent defective. For example, 3% of 10 is 0.3. Get the

total defective by adding the row. Then the fraction of defective items that come from is 0.7 out

of 1.4 or 50%. Formally, you were given the following:

and . You were asked for . By Bayes’ rule,

, where we found using

In problem 7, the proportion of output that comes from Veeblefetzer C and is not defective is

a. Between 90% and 100%

b. Between 80% and 90%

c. Between 70% and 80%

d.* Between 60% and 70%

e. Between 50% and 60%

Solution: From the table, 69.3 out of 100 units. Formally, since . So

The event has a probability of; the event also has a probability of. If the two events are independent

c.*

Solution: From the addition rule . ,

and, since and are independent, . So

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If the events and each have probabilities above zero and below 1, and are not independent, and , then the following must be true:.

a. is below .5 and above zero.

b. is between .5 and .6.

c.* Either a. or b. could be true.

d. is 1

e. There is not enough information to answer this question.

Solution: From the multiplication rule, and

. Since and are both less than 1,

must be less than both and .

The following table shows the number of days absent among your employees in a month and the probability.

Number of days absent Probability

0 .60

1 .20

2 .12

3 .04

4 .04

5 0

What is the expected value (mean) of the number of days absent?

a. 0.20.

b.* 0.72

c. 2.50

d. 2.00

e. 1.00

Solution: Here is the completed table for both the definitional and the computational formula as generated by Minitab. There is no reason to bother with the definitional method.

row

1 0 0.60 0.00 0.00 -0.72 -0.4320 0.311040

2 1 0.20 0.20 0.20 0.28 0.0560 0.015680

3 2 0.12 0.24 0.48 1.28 0.1536 0.196608

4 3 0.04 0.12 0.36 2.28 0.0912 0.207936

5 4 0.04 0.16 0.64 3.28 0.1312 0.430336

sum 1.00 0.72 1.68 0.0000 1.1610

We have a valid distribution since all the probabilities are between one and zero and add to 1. The formula for a population variance for a distribution is , where and . So and .

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What is the value closest to the standard deviation of the number of days absent in problem 11?

a.* 1.08

b. 1.87.

c. 2.50

d. 1.16

e. 1.68

What is using the table in problem 10?

a. 1.00

b. .08.

c. .20

d.* .92

e. None of the above. For credit, write in an answer, showing your work.

Solution: You have been asked for the probability that x is less than or equal to 2.

If I know that eight workers are available to fill four openings, in how many ways can I fill them?

a. 4098

b.* 70

c. 32

d. 1680

e. 65536

Solution: Since order is not important and ,

The number of ways that 4 items can be taken from 13 if order is important and replacement is not allowed is

a. 0

b. 1

c. 15444

d. 715

e.* None of the above. For credit, write in an answer, showing your work.

Solution: Since order is important and ,

If is a standardized variable (z-score) and , the expected value of is

a.* 2

b. 8

c. -6

d. -8

e. None of the above. For credit, write in an answer, showing your work

Solution: Since if and and a standardized variable has a mean of zero and a standard deviation of 1, let and and compute and

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If is a standardized variable (z-score) and , the standard deviation of is

a. 36

b. -8

c. -6

d.* 8

e. None of the above. For credit, write in an answer, showing your work

Solution: From the above , so .

As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that as long as one of the two components is working, the jorcillator works. If the probability of the phillinx failing in the first month of service is .5, and the probability of the phillinx failing in the second month is .4, while the probability of the flubberall failing in the first month is .4 and probability of the flubberall failing in the second month is .4, what is the probability that both components fail in the first month? Assume that the failure of one component is independent of failure of the other.

a. 0

b.* .20

c. .90

d. .70

e. None of the above. For credit, write in an answer, showing your work.

Solution: The problem asks nothing about the jorcillator. Let be the probability that the phillinx fails in the first period, let be the probability that the phillinx fails in the second period, and let be the probability that the phillinx fails in the third period (after the first and second period). Let be the probability that the flubberall fails in the first period, let be the probability that the flubberall fails in the second period, and let be the probability that the flubberall fails in the third period. The events are independent. Then we have been asked for

As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that as long as one of the two components is working, the jorcillator works. If the probability of the phillinx failing in the first month of service is .5, and the probability of the phillinx failing in the second month is .4, while the probability of the flubberall failing in the first month is .4 and probability of the flubberall failing in the second month is .4, what is the probability that the jorcillator will last beyond 2 months? Assume that the failure of one component is independent of failure of the other.

a. .02

b. .30

c.* .28

d. .32

e. None of the above. For credit, write in an answer, showing your work

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Solution: The problem asks for the probability that the jorcillator lasts beyond 2 months. We already know that . Make the following table.

/ / / Total
/ .20 / .16 / .04 / 0.40
/ .20 / .16 / .04 / 0.40
/ .10 / .08 / .02 / 0.20
Total / .50 / .40 / .10 / 1.00

The event , for example will down the jorcillator in the third period because as long as one of the two components is working, the jorcillator works. If we add together all the events that down the jorcillator in the third period, we find . This turns out to be the same as

20. If and are complementary events:

d.* All of the above.

e. None of the above

Part II. Show your work!

The number of days of absence last month for a sample of 6 employees appears below. (This is a sample not a probability distribution!) Compute the sample standard deviation (5 Points - 2 Point Penalty for not trying.)

Number of
Days /
0 / 0
4 / 16
0 / 0
5 / 25
16 / 256
1 / 1
26 / 298

So and .

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Part III. Do at least 1 (5+ Points) of the two following problems: Show your work! You receive extra credit for extra work!

1. You are dealt four cards from a deck. Remember that there are 4 cards of each denomination (Aces. 2's, etc.) but 13 cards of each suit (Hearts, Clubs etc). Let be the event that you get three diamonds and the event that you get one heart.

a. How many possible hands are there? (2) Work out answer to a) . You may leave the answers to the remaining sections of this question in factorial form.

Solution: ,

b. (2)

Solution:

c. (2)

Solution:

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2. We are trying to sell residential real estate. Let be the event that the selling price of the property is at least 98% of the asking price. . . Let be the event that the property takes more than 3 months to sell. If the selling price is at least 98% of the asking price, there is a 30% probability that the property takes more than 3 months to sell. If the selling price is less than 98% of the asking price, there is a 40% probability that the property takes more than 3 months to sell.

a. Find

Solution: Since, and , it must be true that and for 100 properties, we can make the following table

/ 7.5 / 17.5 / 25
/ 30 / 45 / 75
37.5 / 62.5 / 100

, , so .

b. If the property is on the market for more than 3 months, what is the probability that it will sell for more than 98% of the asking price? (5)

Solution: From the table, , or by Bayes’ rule,