Exercises

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May 05'23

A machine consists of two components, whose lifetimes have the joint density function

[[math]] f(x,y) = \begin{cases} \frac{1}{50}, \,\, x\gt0, y \gt 0, x+y \lt 10 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

The machine operates until both components fail. Calculate the expected operational time of the machine.

1.7
2.5
3.3
5.0
6.7

AAdmin

May 05'23

A client spends [math]X[/math] minutes in an insurance agent’s waiting room and [math]Y[/math] minutes meeting with the agent. The joint density function of [math]X[/math] and [math]Y[/math] can be modeled by

[[math]] f(x,y) = \begin{cases} \frac{1}{800} e^{-x/40}e^{-y/20}, \,\, x \gt 0, y \gt 0\\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine which of the following expressions represents the probability that a client spends less than 60 minutes at the agent’s office.

[math]\frac{1}{800}\int_0^{40}\int_0^{20}e^{-x/40}e^{-y/20} dy dx[/math]
[math]\frac{1}{800} \int_0^{40}\int_0^{20-x}e^{-x/40}e^{-y/20} dy dx[/math]
[math]\frac{1}{800} \int_0^{20}\int_0^{40-x}e^{-x/40}e^{-y/20} dy dx[/math]
[math] \frac{1}{800} \int_0^{60}\int_0^{60}e^{-x/40}e^{-y/20} dy dx[/math]
[math]\frac{1}{800} \int_0^{60}\int_0^{60-x}e^{-x/40}e^{-y/20} dy dx [/math]

AAdmin

May 05'23

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to buy an extended warranty for the car. Let [math]X[/math] denote the number of luxury cars sold in a given day, and let [math]Y[/math] denote the number of extended warranties sold.

[[math]] \begin{align*} \operatorname{P}[X = 0, Y = 0] &= 1/6 \\ \operatorname{P}[X = 1, Y = 0] &= 1/12 \\ \operatorname{P}[X = 1, Y = 1] &= 1/6 \\ \operatorname{P}[X = 2, Y = 0] &= 1/12 \\ \operatorname{P}[X = 2, Y = 1] &= 1/3 \\ \operatorname{P}[X = 2, Y = 2] &= 1/6 \\ \end{align*} [[/math]]

Calculate the variance of [math]X[/math].

0.47
0.58
0.83
1.42
2.58

AAdmin

May 05'23

A company is reviewing tornado damage claims under a farm insurance policy. Let [math]X[/math] be the portion of a claim representing damage to the house and let [math]Y[/math] be the portion of the same claim representing damage to the rest of the property. The joint density function of [math]X[/math] and [math]Y[/math] is

[[math]] f(x,y) = \begin{cases} 6 [1 − ( x + y ) ], \,\, x \gt 0, y \gt 0, x+y \lt 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the probability that the portion of a claim representing damage to the house is less than 0.2.

0.360
0.480
0.488
0.512
0.520

AAdmin

May 05'23

Let [math]X[/math] represent the age of an insured automobile involved in an accident. Let [math]Y[/math] represent the length of time the owner has insured the automobile at the time of the accident. [math]X[/math] and [math]Y[/math] have joint probability density function

[[math]] f(x,y) = \begin{cases} \frac{10-xy^2}{64}, \,\, 2 \leq x \leq 10, 0 \leq y \leq 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected age of an insured automobile involved in an accident.

4.9
5.2
5.8
6.0
6.4

AAdmin

May 05'23

The table below shows the joint probability function of a sailor’s number of boating accidents and number of hospitalizations from these accidents this year.

	Number of Hospitalizations from Accidents
Number of Accidents		0	1	2	3
	0	0.700
	1	0.150	0.050
	2	0.060	0.020	0.010
	3	0.005	0.002	0.002	0.001

Calculate the sailor’s expected number of hospitalizations from boating accidents this year.

0.085
0.099
0.410
1.000
1.500

AAdmin

Jun 01'22

For what values of [math]a[/math] and [math]b[/math] is the function [math]f(x,y)[/math] given below a valid joint density function?

[[math]] f(x,y) = \begin{cases} ax + by, \quad 0 \leq x \leq y \leq 1 \\ 0 \quad \textrm{otherwise} \end{cases} [[/math]]

[math]2a + b = 6 [/math]
[math]2a + b = 6 [/math] and both [math]a [/math] and [math]b[/math] must be non-negative
[math]a + b = 2 [/math] and both [math]a [/math] and [math]b[/math] must be non-negative
For all non-negative [math]a[/math] and [math]b[/math]
There are no values [math]a[/math] and [math]b[/math] that give a valid joint density

AAdmin

May 05'23

An insurance company will cover losses incurred from tornadoes in a single calendar year. However, the insurer will only cover losses for a maximum of three separate tornadoes during this timeframe. Let [math]X[/math] be the number of tornadoes that result in at least 50 million in losses, and let [math]Y[/math] be the total number of tornadoes. The joint probability function for [math]X[/math] and [math]Y[/math] is

[[math]] p(x,y) = \begin{cases} c(x + 2y), \, \textrm{for} \, x = 0,1,2,3, \, y = 0,1,2,3, \, x \leq y \\ 0, \, \textrm{otherwise.} \end{cases} [[/math]]

where [math]c[/math] is a constant.

Calculate the expected number of tornadoes that result in fewer than 50 million in losses.

0.19
0.28
0.76
1.00
1.10

AAdmin

May 05'23

A device runs until either of two components fails, at which point the device stops running. The joint density function of the lifetimes of the two components, both measured in hours, is

[[math]] f(x,y) = \frac{x+y}{2}, \textrm{for} \,\, 0 \lt x \lt2 \,\, \textrm{and} \,\, 0 \lt y \lt 2. [[/math]]

Calculate the probability that the device fails during its first hour of operation.

0.125
0.141
0.391
0.625
0.875

AAdmin

May 05'23

A device contains two components. The device fails if either component fails. The joint density function of the lifetimes of the components, measured in hours, is [math]f(s,t)[/math], where [math]0 \lt s \lt 1[/math] and [math]0 \lt t \lt 1 [/math].

Determine which of the following represents the probability that the device fails during the first half hour of operation.

[math]\int_{0}^{0.5}\int_{0}^{0.5}f(s,t)\,ds\,dt[/math]
[math]\int_0^1\int_0^{0.5}f(s,t)\,ds\,dt[/math]
[math]\int_{0.5}^1\int_{0.5}^1f(s,t)\,ds\,dt[/math]
[math]\int_{0}^{0.5}\int_{0}^1f(s,t)\,ds\,dt + \int_{0}^1\int_{0}^{0.5}f(s,t)\,ds\,dt[/math]
[math]\int_{0}^{0.5}\int_{0.5}^1f(s,t)\,ds\,dt + \int_{0}^1\int_0^{0.5}f(s,t)\,ds\,dt[/math]