Aa Aa 2. The expected value of illegal parking stelios has difficulty finding pa

Question

Aa Aa 2. The expected value of illegal parking stelios has difficulty finding parking in his neighbArhood and, thus, is considering the gamble of illegally parking on the sidewalk because of the opportunity cost of the time he spends searching for parking. On any given day, Stelios knows he may or may not get a ticket, but he also expects that if he were to do it even day, the average amount he would pay for parking tickets should converge to the expected value. If the expected value is positive, then in the long run, it will be optimal for him to park on the sidewalk and occasionally pay the tickets, in exchange for the benefits of not searching for parking. suppose that stelios knows that the fine for parking this way is $100 and his opportunity cost (oc) of searching for parking is $15 per day. That is, if he parks on the sidewalk and does not get a ticket, he gets a positive payoff worth $15; if he does get a ticket, he ends up with a payoff of Given that he still does not know the probability of getting caught, compute his expected payoff from parking on the sidewalk when the probability of getting a ticket is 10% and then when the probability is 50%. Probability of Ticket 10% 50% EV of Sidewalk Parking (OC- $15) Based on the values you found in the previous table, u the blue line (circle symbols) on the following graph to plot the expected value of sidewalk parking when the opportunity cost of time is $15. Now, suppose Stelios gets a new job that requires him to work longer hours. As a result, the opportunity cost of his time rises, and he now values the time saved from not having to look for parking at $30 comput the expected value of the payoff from parking on the sidewalk, given the two different probabilities of getting a ticket. Probability of Ticket

Explanation / Answer

2. Payoff when he parks on sidewalk and doesn't get a ticket = $15

Payoff when he parks and gets a ticket = -($100-$15) = -$85

SO, expected value when prob of ticket is 10% : 0.1*(-85)+0.9*15 = $5

SO, expected value when prob of ticket is 50% : 0.5*(-85)+0.5*15 = -$35

Now the OC is $30.

SO, payoff when he gets ticket is $30 and when he doesn't get ticket is -(100-30) = -$70

SO, expected value when prob of ticket is 10% : 0.1*(-70)+0.9*30 = $20

SO, expected value when prob of ticket is 50% : 0.5*(-70)+0.5*30 = -$20

When probability is 0.275 and OC is $15, EV = 15*0.715-85*0.275 = -$12.5. So, he loses $12.5

When probability is 0.275 and OC is $30, EV = 30*0.725-70*0.275 = $2.5. SO, he gains $2.5

3. Let break even probability be p

SO, (1-p)*18-82*p <0

100p>18, p >0.18

So, 19,24,25% are the answers

Let parking fine be x

Probability is 15%.

So, 0.85*18 - 0.15*(x-18)=0

0.15x = 18

x = 18/0.15 = $120

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