Profit of an insurance company

[allalongthewatc]

Hello All,
I came across this problem while reviewing my quant notes. I am unable to solve this problem. I would appreciate any help.
An insurance company has calculated the probability of a major road accident = 0.025. Assume that driving behavior and accidents are independent. Each accident costs $65000 for repair (to the insurance company). Assume that the driver can have only claim per policy in a year. Annual premium = $2500. If the company sells 1000 such policies, calculate the probability that the company makes a loss.
Here’s what I did:
Let N = number of accidents in 1000 policies. This is a binomial random variable with E(N) = 1000*0.025 = 25 and SD(N) = sqrt(n*p*(1-p)) = 4.9371
Approximating normal distribution for binomial,
Profit (P) is a random variable that is normal.
P = (1000* $2.5K) -(N * $65K)
E(P) = $875K
SD(P) = $65K*SD(N) = $320,911
Therefore, Prob(P<0) = Prob[{(P-875K)/320,911} < (0-875K)/320,911)] =0.00319.
OA is 0.00423
Can someone please help me?
I would appreciate any help.
Thanks in advance.

 

[S2000magician]

The total annual premium collected is 1,000 × $2,500 = $2,500,000. The number of accidents that would cause the company to break even is $2,500,000 ÷ $65,000 = 38.36, so the company makes a profit if there are 38 or fewer accidents, and it has a loss if there are 39 or more accidents.
Thus, the probability that they make a profit is P(0) + P(1) + … + P(37) + P(38) = 99.49%. Therefore, the probability that they have a loss is 100% – 99.49% = 0.51%.
Using the normal approximation with μ = 25 and σ = 4.9371, we want the probability that a standard normal variable is greater than (38.36 – 25) / 4.9371 = 2.727. From a z-table, we get 0.32%.
I have no idea how they got 0.423%. Do they show a calculation?

 

[allalongthewatc]

S2000magician wrote:
The total annual premium collected is 1,000 × $2,500 = $2,500,000. The number of accidents that would cause the company to break even is $2,500,000 ÷ $65,000 = 38.36, so the company makes a profit if there are 38 or fewer accidents, and it has a loss if there are 39 or more accidents.
Thus, the probability that they make a profit is P(0) + P(1) + … + P(37) + P(38) = 99.49%. Therefore, the probability that they have a loss is 100% – 99.49% = 0.51%.
Using the normal approximation with μ = 25 and σ = 4.9371, we want the probability that a standard normal variable is greater than (38.36 – 25) / 4.9371 = 2.727. From a z-table, we get 0.32%.
I have no idea how they got 0.423%. Do they show a calculation?

Hello S2000magician,
(Note: RV = Random Variable in my response below)
Thank you so much for your response. After seeing your method, I believe the prep guys could have taken the floor value of 38.36 ~ 38. (I tried the ceil value of 38.36 value as well ~39, but I didn’t get the answer.) If Z = (38-25/SD), then we can get 0.423%. I think this is what the prep guys are doing. Unfortunately, I don’t have the solution.
Do you know how I can find this value (i.e. 38) when I am using RV as “Profit” and not “Number of accidents,” as I have done above ? If I use RV as “Profit,” (and then use the identity that in P=a+bN, P is normal if N is normal), I have no way of applying “ceil” or “floor” function to Profit RV, which is “P”, because “P” can take any continuos value. On the other hand, Number of accidents RV, which is “N”, is discrete. I would truly appreciate your thoughts and comments.
Best regards,
Allalongthewatchtower