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Homework answers / question archive / Pepperdine University FINC 655 Chapter 17 Multiple-Choice Solutions: 1)You are taking a multiple-choice test that awards you one point for a  answer and penalizes you 0

Pepperdine University FINC 655 Chapter 17 Multiple-Choice Solutions: 1)You are taking a multiple-choice test that awards you one point for a  answer and penalizes you 0

Finance

Pepperdine University

FINC 655

Chapter 17

Multiple-Choice Solutions:

1)You are taking a multiple-choice test that awards you one point for a  answer and penalizes you 0.25 points for an in answer. If you have to make a random guess and there are five possible answers, what is the expected value of guessing?

    1. 0.5 points [the expected value of a random variable that can take on two values (x1,x2) with probabilities (p,1-p) is E(X)= (p*(x1)) + ((1-p)*(x2)). In this case, p represent the chance of getting a  answer, while x1 and x2 refer to the possible points]
    2. 0.25 points [the expected value of a random variable that can take on two values (x1,x2) with probabilities (p,1-p) is E(X)= (p*(x1)) + ((1-p)*(x2)). In this case, p represent the chance of getting a  answer, while x1 and x2 refer to the possible points]
    3. –0.25 points [the expected value of a random variable that can take on two values (x1,x2) with probabilities (p,1-p) is E(X)= (p*(x1)) + ((1-p)*(x2)). In this case, p represent the chance of getting a  answer, while x1 and x2 refer to the possible points]

d. 0 points [((1/5)*(1)) + ((4/5)*(-0.25)) = 0.2 – 0.2 = 0]

 

  1. A franchise restaurant chain is considering a new store in an unserved part of town. Its finance group estimates an NPV of $10 million if the population growth is 10% (40% probability), and NPV of $4 million of the population does not grow (30% probability), and an NPV of -$4 million if the population shrinks 5% (30% probability). What is the expected value of NPV (to the nearest dollar) for the following situation?
    1. $3.4 million [E(x)=((p1)*(x1))+((p2)*(x2))….+((pn)*(xn)). The probabilities are 0.4, 0.3 & 0.3 and the potential outcomes are $10, $4 & -$4 million in this case]

b. $4.0 million [((0.4)*(10m))+((0.3)*(4m))+((0.3)*(-4m))= 4m+1.2m-1.2m = 4 million]

c. $4.6 million [E(x)=((p1*)(x1))+((p2)*(x2))….+((pn)*(xn)). The probabilities are 0.4, 0.3 & 0.3 and the potential outcomes are $10, $4 & -$4 million in this case]

d. $5.2 million [E(x)=((p1)*(x1))+((p2)*(x2))….+((pn)*(xn)). The probabilities are 0.4, 0.3 & 0.3 and the potential outcomes are $10, $4 & -$4 million in this case]

 

  1. You’ve just decided to add a new line to your manufacturing plant. Compute the expected loss/profit from the line addition if you estimate the following:

There’s a 50% chance that profit will increase by $100,000. There’s a 30% chance that profit will remain the same.

There’s a 20% chance that profit will decrease by $15,000.

    1. Gain of $100,000 [E(x)=((p1)*(x1))+((p2)*(x2))….+((pn)(xn)). The probabilities are 0.5, 0.3 & 0.2 and the profits are

$10,000, $0 & -$15,000 in this case]

    1. Gain of $70,000 [E(x)=((p1*)(x1))+((p2)*(x2))….+((pn)(xn)). The probabilities are 0.5, 0.3 & 0.2 and the profits are

$10,000, $0 & -$15,000 in this case]

    1. Loss of $53,000 [E(x)=((p1)*(x1))+((p2)*(x2))….+((pn)(xn)). The probabilities are 0.5, 0.3 & 0.2 and the profits are

$10,000, $0 & -$15,000 in this case]

d. Gain of $47,000 [((0.5)*($100,000))+((0.3)*(0))+((0.2)*(-$15,000))=$50,000+$0-$3000=$47,000]

curse

 

  1. Your software development company is considering investing in a new mobile app. If it goes viral (10% probability), you expect an NPV of $1,000,000; if it is moderately successful (20% probability), you expect an NPV of

$200,000; and if it fails (70% probability), you expect an NPV of $-200,000. What is the expected NPV of the product?

a. $0 [((0.1)*($1,000,000))+((0.2)*($200,000))+((0.7)*(-$200,000))= $100,000+$40,000-$140,000 = $0]

b. $280,000 [E(x)=((p1)(x1))+((p2)(x2))….+((pn)(xn)). The probabilities are 0.1, 0.2 & 0.7 and the NPVs are $1,00,000,

$200,000 & -$200,000 in this case]

c. $333,000 [E(x)=((p1)(x1))+((p2)(x2))….+((pn)(xn)). The probabilities are 0.1, 0.2 & 0.7 and the NPVs are $1,00,000,

$200,000 & -$200,000 in this case]

d. None of the above [It is one of the above-In cases where you have the probability of different occurrences and their outcomes, you can determine the total expected outcome]

 

  1. Suppose an investment project has an NPV of $75 million if it becomes successful and an NPV of –$25 million if it is a failure. What is the minimum probability of success above which you should make the investment?
    1. 0.5 [the expected value of a random variable that can take on two values (x1,x2) with probabilities (p,1-p) is E(X)= (p(x1)) + ((1-p)*(x2)). In this case, you are looking for the value of p that will allow you to breakeven (i.e., where E(X)

= 0)]

    1. 1/3 [the expected value of a random variable that can take on two values (x1,x2) with probabilities (p,1-p) is E(X)= (p(x1)) + ((1-p)*(x2)). In this case, you are looking for the value of p that will allow you to breakeven (i.e., where E(X)

= 0)]

    1. 0.25 [you are looking for the breakeven- the minimum probability at which you would invest ((p)*($75m))+((1-p)*(-$25m)=0 Therefore 75p-25+25p=0 or 100p=25, so p=0.25]
    2. 0.1 [the expected value of a random variable that can take on two values (x1,x2) with probabilities (p,1-p) is E(X)= (p(x1)) + ((1-p)*(x2)). In this case, you are looking for the value of p that will allow you to breakeven (i.e., where E(X)

= 0)]

 

  1. To test the effectiveness of a two web advertising agencies, you increase your ad purchase with agency A by 50% without changing your purchase through agency B. The referrals to your website from agency A increased by only 34% but the referrals from agency B fell by 21%. What is the difference-in-difference estimate of the referrals per dollar through agency A?
    1. 1.2 referrals per dollar [To construct a "difference-in-difference" estimate of referrals per dollar, measure the difference in referral percent changes from Agency A to Agency B over the difference in percent purchase amount changes]
    2. 1.1 referrals per dollar [To construct a "difference-in-difference" estimate of referrals per dollar, you measure the change in referrals from Agency A (+34%) relative to the control group of Agency B (less -21%), for an estimate per dollar of ((55%)change in referrals/(50%)change in dollars) or 1.1. The difference between treatment (agency A) and control (agency B) referrals was 55% (=34%-(-21%). The difference in purchase amounts was 50% (=50%-0%). The diff-in-diff in referrals per dollar was 55%/50% or 1.1]
    3. 1.0 referrals per dollar [To construct a "difference-in-difference" estimate of referrals per dollar, measure the difference in referral percent changes from Agency A to Agency B over the difference in percent purchase amount changes]

 

    1. 0.9 referrals per dollar [To construct a "difference-in-difference" estimate of referrals per dollar, measure the difference in referral percent changes from Agency A to Agency B over the difference in percent purchase amount changes]

 

  1. Your company has a customer list that includes 3000 people. Your market research indicates that 90 of them responded to the coupon. If you send a coupon to one customer at random, what’s the probability that he or she will use the coupon?
    1. 0.03 [the probability of using the coupon is 90/3000=0.03 or 3%]
    2. 0.09 [the probability the coupon will be used is equal to the chance a customer will respond, in other words, the number of potential positive responses over total customers]
    3. 0.30 [the probability the coupon will be used is equal to the chance a customer will respond, in other words, the number of potential positive responses over total customers]
    4. 0.90 [the probability the coupon will be used is equal to the chance a customer will respond, in other words, the number of potential positive responses over total customers]

 

  1. Your production line has recently been producing a serious defect. One of two possible processes, A and B, could be the culprit. From past experience you know that the probability that A is causing the problem is 0.8 but investigating A costs $100,000 while investigating B costs only $20,000. What are the expected error costs of shutting down process B first?
    1. $80,000 [this is the expected cost of shutting down process A first]
    2. $20,000 [the expected error cost in this case comes from investigating B and having A be the culprit. In other words, it is the probability the total money spent on B will be wasted]
    3. $16,000 [the expected error cost in this case comes from investigating B and having A be the culprit, in other words the probability the total money spent on B will be wasted, so (0.8)*($20,000)=$16,000.
    4. $4,000 [this is the expected cost of shutting down process B first]

 

  1. You have two types of buyers for your product. Forty percent of buyers value your product at $10 and sixty percent value it at $6. What price maximizes your expected revenue?
    1. $10 [In this case, only forty percent of buyers will purchase your product for expected revenue per customer of $4]
    2. $6 [in this case, all 100% of buyers will purchase your product]
    3. $7.60 [This price still only appeals to the top 40% of buyers, and for less than they are willing to pay for expected revenue per customer of $3.04]
    4. $8 [This price still only appeals to the top 40% of buyers, and for less than they are willing to pay for expected revenue per customer of $3.2]

 

  1. You are considering entry into a market in which there is currently only one producer (incumbent). If you enter, the incumbent can take one of two strategies, price low or price high. If they price high, then you expect a $60k profit per year. If they price low, then you expect a $20k loss per year. You should enter if:
  1. You believe demand is inelastic. [if demand is inelastic, price will not impact the demand for your product]
  2. You believe the probability that the incumbent will price low is greater than 0.75. [To solve this, you need to find the minimum probability needed to breakeven. E(X)= (p(x1)) + ((1-p)*(x2)). Once you solve for p, determine what p is of the incumbent pricing low]

 

  1. You believe the probability that the incumbent will price low is less than 0.75. [To solve this, you need to find the minimum probability needed to breakeven. E(X)= (p(x1)) + ((1-p)*(x2)). Using ((p)*($60k))+((1-p)*($-20k))=0, you find 60p-20+20p=0 or 80p=20. Therefore, p (the probability of them pricing high) needs to be a minimum of

0.25 for you to breakeven. Therefore, you would only enter if you believed the probability of them pricing low (1-p) is less than 0.75]

  1. You believe the market-size is growing. [If market size is growing, demand may accommodate the addition of a competitor]

 

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