The time until the next car accident for a particular driver is exponentially distributed with a mean of 200 days. Calculate the probability that the driver has no accidents in the next 365 days, but then has at least one accident in the 365-day period that follows this initial 365-day period.
- 0.026
- 0.135
- 0.161
- 0.704
- 0.839
The annual profit of a life insurance company is normally distributed.
The probability that the annual profit does not exceed 2000 is 0.7642. The probability that the annual profit does not exceed 3000 is 0.9066.
Calculate the probability that the annual profit does not exceed 1000.
- 0.1424
- 0.3022
- 0.5478
- 0.6218
- 0.7257
The profits of life insurance companies A and B are normally distributed with the same mean. The variance of company B's profit is 2.25 times the variance of company A's profit. The 14th percentile of company A’s profit is the same as the [math]p^{\textrm{th}}[/math] percentile of company B’s profit. Calculate [math]p[/math].
- 5.3
- 9.3
- 21.0
- 23.6
- 31.6
Insurance companies A and B each earn an annual profit that is normally distributed with the same positive mean. The standard deviation of company A’s annual profit is one half of its mean. In a given year, the probability that company B has a loss (negative profit) is 0.9 times the probability that company A has a loss.
Calculate the ratio of the standard deviation of company B’s annual profit to the standard deviation of company A’s annual profit.
- 0.49
- 0.90
- 0.98
- 1.11
- 1.71
An insurance policy covers losses incurred by a policyholder, subject to a deductible of 10,000. Incurred losses follow a normal distribution with mean 12,000 and standard deviation [math]c.[/math] The probability that a loss is less than [math]k[/math] is 0.9582, where [math]k[/math] is a constant. Given that the loss exceeds the deductible, there is a probability of 0.9500 that it is less than [math]k[/math].
Calculate [math]c[/math].
- 2045
- 2267
- 2393
- 2505
- 2840
A device that continuously measures and records seismic activity is placed in a remote region. The time, [math]T[/math], to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is [math]X = \max(T, 2)[/math].
Calculate [math]\operatorname{E}(X)[/math].
- [math]2 + \frac{e^{-6}}{3}[/math]
- [math]2-2e^{-2/3} + 5e^{-4/3}[/math]
- 3
- [math]2+3e^{-2/3}[/math]
- 5
A piece of equipment is being insured against early failure. The time from purchase until failure of the equipment is exponentially distributed with mean 10 years. The insurance will pay an amount x if the equipment fails during the first year, and it will pay 0.5x if failure occurs during the second or third year. If failure occurs after the first three years, no payment will be made.
Calculate x such that the expected payment made under this insurance is 1000.
- 3858
- 4449
- 5382
- 5644
- 7235
Automobile claim amounts are modeled by a uniform distribution on the interval [0, 10,000]. Actuary A reports [math]X[/math], the claim amount divided by 1000. Actuary B reports [math]Y[/math], which is [math]X[/math] rounded to the nearest integer from 0 to 10.
Calculate the absolute value of the difference between the 4th moment of [math]X[/math] and the 4th moment of [math]Y[/math].
- 0
- 33
- 296
- 303
- 533
An insurance company’s annual profit is normally distributed with mean 100 and variance 400. Let Z be normally distributed with mean 0 and variance 1 and let F be the cumulative distribution function of Z.
Determine the probability that the company’s profit in a year is at most 60, given that the profit in the year is positive.
- [math]1 – F(2)[/math]
- [math]F(2)/F(5)[/math]
- [math][1 – F(2)]/F(5)[/math]
- [math][F(0.25) – F(0.1)]/F(0.25)[/math]
- [math][F(5) – F(2)]/F(5)[/math]