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Lesechka [4]
2 years ago
11

Assume that a driver faces the following loss distribution:Outcome Probability12,000 0.15 0 0.851. What is the expected loss?2.

What is the standard deviation of loss?Now assume that this driver pools her losses with another driver facing the same loss distribution, and that the losses are not correlated.1. What is the expected loss for each member of the pool?2. What is the standard deviation of loss for each member of the pool?These two drivers decide to pool their losses with two other drivers with the same loss distribution, and all losses are not correlated. (This four-person group will be called Group A)1. What is the expected loss for each member of the pool?2. What is the standard deviation of loss for each member of the pool?These four drivers are considering pooling their losses with another group of four drivers (Group B) facing a different loss distribution (for each driver in the group):Outcome Probability28,000 0.0210,000 0.23 0 0.751. What is the expected loss for each member of this new pool of four drivers?2. What is the standard deviation of loss for each member of this new pool?Now, Group A and Group B decide to pool their risks, becoming a single pool of eight drivers (Group C).1. What is the expected loss for each member of this new 8-person pool of drivers?2. The standard deviation for each member of this new 8-person pool is fairly complicated to calculate. Directionally, though, how would you expect the standard deviation for the members of Group C to compare with the standard deviations of Group A and Group B (Higher/Lower/No Change). Why?If each driver were required to contribute an equal share to the Group C pool to cover losses, would the people in Group A choose to enter Group C? Would the drivers in Group B want to join the pool? Why or why not?If each driver were required to contribute an amount proportional to their own expected loss (this means that if the actual losses of Group C were equal to the Group C expected loss, then Group A drivers would contribute an amount equal to the Group A expected loss and Group B would make a contribution equal to the Group B expected loss), would the people in Group A choose to join the pool? Would the people in Group B want to join the pool? Why or why not?Finally, there is another group of four drivers (Group D) with the yet another loss distribution:Outcome Probability60,000 0.0125,000 0.1010,000 0.300 0.591. What is the expected loss for each member of Group D?2. What is the standard deviation for each member of Group D?3. Under what conditions, if any, would the Group A drivers be willing to pool with the members of Group D to become a single pool of eight drivers?
Mathematics
1 answer:
xxMikexx [17]2 years ago
3 0

Answer:

- 1800 ; 4284.8579 ;

Step-by-step explanation:

Given the table :

Outcome _____ probability

-12000 ________ 0.15

__ 0 __________ 0.85

Expected loss, m: Σx*p(x)

(-12000 * 0.15) + (0 * 0.85)

-1800 + 0 = - 1800

Standard deviation = sqrt(Var(x))

Var(x) = Σx²*p(x) - m²

(-12000^2 * 0.15) + (0^2 * 0.85)] - 1800^2

21,600,000 - 3240000

= 18360000

Standard deviation = sqrt(18360000)

Standard deviation = 4284.8579

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Leto [7]
Opposite angles are supplementary, the sum is 180 degrees.
6x+10+81.5=180
Combine like terms:
6x+91.5=180
Subtract 91.5 to both sides:
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x=14.75
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Hope this helped!
3 0
2 years ago
So I’m doing my homework and I don’t know 3 2/3 + 5/4
vivado [14]

Answer:

4 11/12

Step-by-step explanation:

4 x 2/3 = 8/12

3 x 5/4 = 15/ 12

8/12 + 15/12 = 23/12 = 1 11/12 + 3 = 4 11/12

3 0
3 years ago
Find the mean of the data. give your answer in decimal form
aleksandr82 [10.1K]
Mean = (1+1+2+3+3+3+4+5)/8 = 22/8 = 2.75

answer
mean = 2.75
8 0
3 years ago
If a 24-kg mass stretches a spring 15 cm, what mass will stretch the spring 10 cm
natulia [17]

Answer:

<h2>16kg</h2>

Step-by-step explanation:

 This problem is borers on elasticity of materials.

according to Hooke's law,<em> "provided the elastic limit of an elastic material is not exceeded the the extension e is directly proportional to the applied force."</em>

F= ke

where F is the applied force in N

          k is the spring constant N/m

          e is the extension in meters

Given data

mass m= 24kg

extensnion=15cm in meters= \frac{15}{100}= 0.15m

we can solve for the spring constant k

we also know that the force F = mg

assuming g=9.81m/s^{2}

therefore

24*9.81=k*0.15\\235.44=k*0.15\\k=\frac{235.44}{0.15} \\k=1569.6N/m\\

We can use this value of k to solve for the mass that will cause an extension of  10cm= 0.1m

x*9.81=1569.6*0.1\\\\x= \frac{156.96}{9.81} \\\x= 16kg

5 0
3 years ago
Suppose that you believe that contractor 1 is twice as likely to win as contractor 3 and that contractor 2 is three times as lik
Ray Of Light [21]

Answer:

probability that contractor 1,2 and 3 win is 33%,50% and 17% respectively

Step-by-step explanation:

assuming that there are no other contractors then:

probability that 1 , 2 or 3 win = 1

denoting as X= probability that contractor 3 wins , then assuming that only one wins , we have

probability that 1 , 2 or 3 win = 1

probability that contractor 1 wins + probability that contractor 2 wins + probability that contractor 3 wins = 1

2*P(X) + 3*P(X) + P(X) = 1

6*P(X) = 1

P(X) =1/6

then

-probability that contractor 1 wins = 2/6 (33%)

-probability that contractor 3 wins = 3/6 (50%)

-probability that contractor 3 wins = 1/6 (17%)

3 0
2 years ago
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