The time to failure of a component in an electronic device has an exponential distribution with a median of four hours.

Calculate the probability that the component will work without failing for at least five hours.

• 0.07
• 0.29
• 0.38
• 0.42
• 0.57

The waiting time for the first claim from a good driver and the waiting time for the first claim from a bad driver are independent and follow exponential distributions with means 6 years and 3 years, respectively.

Calculate the probability that the first claim from a good driver will be filed within 3 years and the first claim from a bad driver will be filed within 2 years.

• [math]\frac{1}{18}(1 − e^{−2/3} − e^{−1/2} + e^{−7/6} )[/math]
• [math]\frac{1}{18} e^{−7/6} [/math]
• [math]1 − e^{−2/3} − e^{−1/2} + e^{−7/6}[/math]
• [math]1 − e^{−2/3} − e^{−1/2} + e^{−1/3} [/math]
• [math]1 − \frac{1}{3}e^{−2/3} − \frac{1}{6}e^{−1/2} + \frac{1}{18}e^{−7/6} [/math]

A company has two electric generators. The time until failure for each generator follows an exponential distribution with mean 10. The company will begin using the second generator immediately after the first one fails.

Calculate the variance of the total time that the generators produce electricity.

• 10
• 20
• 50
• 100
• 200

Losses covered by a flood insurance policy are uniformly distributed on the interval [0, 2]. The insurer pays the amount of the loss in excess of a deductible d.

The probability that the insurer pays at least 1.20 on a random loss is 0.30.

Calculate the probability that the insurer pays at least 1.44 on a random loss.

• 0.06
• 0.16
• 0.18
• 0.20
• 0.28

The lifespan, in years, of a certain computer is exponentially distributed. The probability that its lifespan exceeds four years is 0.30.

Let [math]f(x)[/math] represent the density function of the computer’s lifespan, in years, for [math]x \gt 0[/math].

Determine which of the following is an expression for [math]f(x)[/math].

• [math]1 − (0.3)^{x/4}[/math]
• [math]1 − (0.7)^{x/4}[/math]
• [math]1 − (0.3)^{x/4}[/math]
• [math]-\frac{\ln 0.7}{4}(0.7)^{x/4}[/math]
• [math]-\frac{\ln 0.3}{4}(0.3)^{x/4}[/math]

Losses covered by an insurance policy are modeled by a uniform distribution on the interval [0, 1000]. An insurance company reimburses losses in excess of a deductible of 250.

Calculate the difference between the median and the 20th percentile of the insurance company reimbursement, over all losses.

• 225
• 250
• 300
• 375
• 500

For a certain health insurance policy, losses are uniformly distributed on the interval [0, 450]. The policy has a deductible of [math]d[/math] and the expected value of the unreimbursed portion of a loss is 56. Calculate [math]d[/math].

• 60
• 87
• 112
• 169
• 224

The lifetime of a certain electronic device has an exponential distribution with mean 0.50. Calculate the probability that the lifetime of the device is greater than 0.70, given that it is greater than 0.40.

• 0.203
• 0.247
• 0.449
• 0.549
• 0.861

A loss under a liability policy is modeled by an exponential distribution. The insurance company will cover the amount of that loss in excess of a deductible of 2000. The probability that the reimbursement is less than 6000, given that the loss exceeds the deductible, is 0.50. Calculate the probability that the reimbursement is greater than 3000 but less than 9000, given that the loss exceeds the deductible.

• 0.28
• 0.35
• 0.50
• 0.65
• 0.72

Losses due to accidents at an amusement park are exponentially distributed. An insurance company offers the park owner two different policies, with different premiums, to insure against losses due to accidents at the park.

Policy A has a deductible of 1.44. For a random loss, the probability is 0.640 that under this policy, the insurer will pay some money to the park owner.

Policy B has a deductible of d. For a random loss, the probability is 0.512 that under this policy, the insurer will pay some money to the park owner.

Calculate d.

• 0.960
• 1.152
• 1.728
• 1.800
• 2.160