Monthly Archives: June, 2011

Exam P Practice Problem 10

Problem 10a
An individual is facing an outcome of an annual financial loss X (in tens of thousands of dollars) whose probability density function is given by

\displaystyle f(x)=0.003 x^2, \ \ \ \ 0<x<10

The probability of a loss in the next year is 0.08. If there is a loss, there is only one loss in any given year. An insurance policy is available to protect against the financial loss by paying in full when a loss occurs.

  1. What is the probability that the insurer’s payment to the insured will exceed $50,000?
  2. What is the mean payment made by the insurer to the insured?
  3. What is the variance of the amount of payment made by the insurer?

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Problem 10b
Suppose that instead of buying a policy that pays the loss in full, the individual buys a policy that has a 80/20 coinsurance provision, i.e., the insurance company pays 80% of the loss and the insured retains the remaining 20% of a loss. Answer the same three questions.

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Solution is found below.

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Solution to Problem 10a
Answers
\displaystyle 10a.1 \ \ \ \ \ 0.07

\displaystyle 10a.2 \ \ \ \ \ 0.6

\displaystyle 10a.3 \ \ \ \ \ 4.44

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Let X be the loss variable as described in the problem. Then the following is the probability P(X>t).

\displaystyle \begin{aligned}P(X>t)&=\int_t^{10} 0.003 x^2 \ dx \\&=1-0.001 t^3   \end{aligned}

One important thing to keep in mind is that the occurrence of a financial loss is not certain. So the answer to question #1 is not P(X>5). Let Y be the insurance payment to the insured. Note that Y is conditional on the occurrence of a loss. If the loss does not happen, Y=0. If the loss does happen, Y=X. Likewise, P(Y>t)=0 in case of no loss and P(Y>t)=P(X>t) in case of a loss. So we can use the law of total probability to obtain P(Y>5).

\displaystyle \begin{aligned}P(Y>5)&=0 \times 0.92+P(X>5) \times 0.08  \\&=(1-0.001 5^3) \times 0.08 \\&=0.07   \end{aligned}

The answers to the other two questions can also be obtained by using the law of total probability.

\displaystyle \begin{aligned}E(Y)&=0 \times 0.92+\int_0^{10} x 0.003 x^2 \ dx \times 0.08  \\&=\int_0^{10} 0.003 x^3 \ dx \times 0.08 \\&=7.5 \times 0.08 \\&=0.6 \\&=\$6000   \end{aligned}

\displaystyle \begin{aligned}E(Y^2)&=0 \times 0.92+\int_0^{10} x^2 0.003 x^2 \ dx \times 0.08  \\&=\int_0^{10} 0.003 x^4 \ dx \times 0.08 \\&=60 \times 0.08 \\&=4.8   \end{aligned}

\displaystyle Var(Y)=4.8-0.6^2=4.44

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Answers to Problem 10b
\displaystyle 10b.1 \ \ \ \ \ 0.06046875

\displaystyle 10b.2 \ \ \ \ \ 0.48

\displaystyle 10b.3 \ \ \ \ \ 2.8416

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Exam P Practice Problem 9

Problem 9a
Suppose that when a policyholder incurs a loss, the size of the loss (in thousands of dollars) is Y=X^{-1} where the probability density function of X is f(x)=2.5 x^{1.5} where 0<x<1. Find the variance of the size of loss.

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Problem 9b
Suppose that when a policyholder incurs a loss, the size of the loss (in thousands of dollars) is Y=X^{-1} where the probability density function of X is f(x)=3 x^{2} where 0<x<1. Suppose that the insurance company just receives a notification that the policyholder had incurred a loss over $2000, what is the probability that the loss exceeds $3000?

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Solution is found below.

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Solution to Problem 9a

More Direct Solution

    \displaystyle E(Y)=E(X^{-1})=\int_0^1 x^{-1} \ 2.5 x^{1.5} \ dx =\frac{5}{3}

    \displaystyle E(Y^2)=E(X^{-2})=\int_0^1 x^{-2} \ 2.5 x^{1.5} \ dx =5

    \displaystyle Var(Y)=5- \biggl(\frac{5}{3}\biggr)^2=\frac{20}{9}

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The original solution

Note that the support of Y is y>1 (the values with positive probabilities). When 0<y^{-1}<1, we have y>1. First derive the cdf of X and then derive the cdf of Y=X^{-1}. Upon differentiation, the density function of Y is obtained, which is then used to compute the variance. The following shows the derivations.

\displaystyle F_X(x)=\left\{\begin{matrix}0& \ \ \ \ \ \ x \le 0 \\{\text{ }}& \\{x^{2.5}}&\ \ \ \ \ \ 0 < x <1 \\{\text{ }}& \\{1}&\ \ \ \ \ \ x \ge 1  \end{matrix}\right.

\displaystyle \begin{aligned}F_Y(y)&=P(Y \le y) \\&=P(X^{-1} \le y) \\&=P(X>y^{-1}) \\&=1-F_X(y^{-1}) \\&=1-y^{-2.5},y>1   \end{aligned}

\displaystyle f_Y(y)=2.5 y^{-3.5}.

\displaystyle E(Y)=\int \limits_{1} ^\infty 2.5 y^{-2.5} \ dy=\frac{5}{3}

\displaystyle E(Y^2)=\int \limits_{1} ^\infty 2.5 y^{-1.5} \ dy=5

\displaystyle Var(Y)=5-\frac{25}{9}=\frac{20}{9}

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Answer to Problem 9b
\displaystyle \frac{8}{27}=0.2963

Hint. It is not necessary to derive the distribution of Y before computing the probabilities concerning Y. For example, P(Y > 2)=P(X^{-1} > 2)=P(X < 2^{-1}).

Exam P Practice Problem 8

Problem 8a
Suppose that the number of parking tickets issued in a calendar year to a certain driver follows a Poisson distribution with mean of 4.7. The fine for each ticket is $75. What is the probability that this driver will pay more than $350 in fine in the upcoming calendar year?

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Problem 8b
Suppose a certain tour operator offers excursion tours and the number of tourists served by this tour operator follows a Poisson distribution with mean of 12 per day. The cost of the excursion tour is $125 per person. What is the probability that the total daily revenue will not exceed $700?

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Solution is found below.

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Solution to Problem 8a
The probability function of the Poisson distribution in question is:

\displaystyle P(N=n)=\frac{e^{-4.7} 4.7^n}{n!}

Having 5 or more tickets would lead to a total fine of more than $350. Thus the answer is:

\displaystyle \begin{aligned}P(N \ge 5)&=1-P(N \le 4) \\&=1-e^{-4.7}\biggl(1+4.7+\frac{4.7^2}{2!}+\frac{4.7^3}{3!}+\frac{4.7^4}{4!}\biggr) \\&=1-54.3808375 e^{-4.7} \\&=0.5053912139   \end{aligned}

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Answer to Problem 8b
0.0203410294

Exam P Practice Problem 7

Problem 7a
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by

      \displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0

The property owner purchased an insurance policy that pays the amount of the damage in full during the next year.

  1. What is the probability that the insurer’s payment to the owner will exceed $17,500?
  2. What is the mean payment made by the insurer to the owner of the property?
  3. What is the variance of the amount of payment made by the insurer?

Problem 7b
The probability that a property will not be damaged in the upcoming year is 0.80. When there is a damage to the property, the probability density function of the amount of the damage (in thousands of dollars) is given by

      \displaystyle f(x)=0.05 e^{-0.05x} \ \ \ \ \ x>0

The property owner purchased an insurance policy with a coinsurance provision that pays 80% of the amount of the damage during the next year. The remaining 20% of the amount of the damage is retained by the property owner.

  1. What is the probability that the insurer’s payment to the owner will exceed $17,500?
  2. What is the mean payment made by the insurer to the owner of the property?
  3. What is the variance of the amount of payment made by the insurer?

Solution is found below.

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Solution to Problem 7a
Let X be the loss amount (i.e. the pdf is the one given in the problem). One important thing to keep in mind is that a loss to the property is not certain. So the answer for #1 is not P(X>17.5). Let Y be the payment made by the insurer to the property owner. The answer for #1 is P(Y>17.5). One way to look at the problem is through the law of total probability.

\displaystyle \begin{aligned}P(Y>17.5)&=P(Y>17.5 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +P(Y>17.5 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+e^{-0.05 (17.5)} \times 0.2 \\&=0.2 \times e^{-0.875} \\&=0.08337  \end{aligned}

The following provides the answers to the rest of the problem:

\displaystyle \begin{aligned}E(Y)&=E(Y \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{1}{0.05} \times 0.2 \\&=4 \\&=\$4000  \end{aligned}

\displaystyle \begin{aligned}E(Y^2)&=E(Y^2 \lvert \text{ No Damage}) \times P(\text{ No Damage}) \\&\ \ \ +E(Y^2 \lvert \text{ Damage}) \times P(\text{ Damage}) \\&=0 \times 0.8+\frac{2}{0.05^2} \times 0.2 \\&=160  \end{aligned}

\displaystyle \begin{aligned}Var(Y)&=E(Y^2)-E(Y)^2 \\&=160-4^2 \\&=144  \end{aligned}

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Answer to Problem 7b
\displaystyle 7b.1 \ \ \ \ \ \ 0.2 e^{-1.09375}=0.0669916086

\displaystyle 7b.2 \ \ \ \ \ \ \$3200

\displaystyle 7b.3 \ \ \ \ \ \ 92.16

Exam P Practice Problem 6

Problem 6a
An auto insurer offers collison coverage to two large groups of policyholders, Group 1 and Group 2. On the basis of historical data, the insurer has determined that the loss due to collision for a policyholder in Group 1 has an exponential distribution with mean 5. On the other hand, the loss due to collision for a policyholder in Group 2 has an exponential distribution with mean 10.

Considering the two groups as one block, about 75% of the losses are from Group 1.

  1. Given a randomly selected loss in this block, what is the probability that the loss is greater than 15?
  2. If a randomly selected loss is greater than 15, what is the probability that it is a from a policyholder in Group 1?

Problem 6b
An auto insurer has two groups of policyholders – those considered good risks and those considered bad risks. On the basis of historical data, the insurer has determined that the number of car accidents during a policy year for a policyholder classified as good risk follows a binomial distribution with n=2 and p=\frac{1}{10}. The number of car accidents for a policyholder classified as bad risk follows a binomial distribution with n=2 and p=\frac{3}{10}. In this block of policies, 75% are classified as good risks and 25% are classified as bad risks. A new customer, whose risk class is not yet known with certainty, has just purchased a new policy.

  1. What is the probability that this new policyholder is not accident-free in the upcoming policy year?
  2. By the end of the policy year, it is found that this policyholder is not accident-free, what is the probability that the policyholder is a “good risk” policyholder?

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Soultion is found below.

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Solution to Problem 6a
Let X be the loss amount of a randomly selected policyholder. The conditional probabilities of a loss greater than 7.5 are:

\displaystyle P(X>15 \lvert \text{ Group 1 Policyholder})=e^{-\frac{15}{5}}
\displaystyle P(X>15 \lvert \text{ Group 2 Policyholder})=e^{-\frac{15}{10}}

By the law of total probability, the unconditional probability is:

\displaystyle \begin{aligned}P(X>15)&=P(X>15 \lvert \text{ Group 1 Policyholder}) \times P(\text{ Group 1 Policyholder}) \\&\ \ \ +P(X>15 \lvert \text{ Group 2 Policyholder}) \times P(\text{ Group 2 Policyholder}) \\&=\frac{3}{4} \times e^{-\frac{15}{5}}+\frac{1}{4} \times e^{-\frac{15}{10}} \\&=\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5} \\&=0.0931228413  \end{aligned}

The above calculation indicates that the unconditional probability is the weighted average of the conditional probabilities. The answer to the second question is obtained by applying the Bayes’ theorem:

\displaystyle \begin{aligned}P(\text{Group 1 Policyholder } \lvert X>15)&=\frac{P[(\text{Group 1 Policyholder}) \cap (X>15)]}{P(X>15)} \\&=\frac{\frac{3}{4} \times e^{-3}}{\frac{3}{4} \times e^{-3}+\frac{1}{4} \times e^{-1.5}}  \\&=0.400978973 \end{aligned}

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Answer to Problem 6b
\displaystyle 6b.1 \ \ \ \ \frac{108}{400}=0.27

\displaystyle 6b.2 \ \ \ \ \frac{57}{108}=0.5278

Exam P Practice Problem 5

Problem 5a
Suppose that X and Y are independent random variables with the following moment generating functions:

    \displaystyle M_X(t)=\frac{4}{9}+\frac{4}{9}e^t+\frac{1}{9}e^{2t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_Y(t)=e^{\displaystyle -3+3e^t}

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Find the probability \displaystyle P(\frac{Y}{2}=X).

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Problem 5b
Suppose that X and Y are independent random variables with the following moment generating functions:

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    \displaystyle M_X(t)=\frac{1}{3-2e^t} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_Y(t)=\biggl(\frac{1+3e^t}{4}\biggr)^2

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Find the probability P(Y=X-2).

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Solution is found below.

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Solution to Problem 5a
The mgf M_X(t) is that of a binomial distribution with n=2 and p=\frac{1}{3}. The mgf M_Y(t) is that of a Poisson distribution with parameter \lambda=3.

    \displaystyle \begin{aligned}P(Y=2X)&=P(X=0,Y=0)+P(X=1,Y=2)+P(X=2,Y=4) \\&=\biggl(\frac{2}{3}\biggr)^2 \ e^{-3}+2 \biggl(\frac{1}{3}\biggr) \biggl(\frac{2}{3}\biggr) \ e^{-3} \ \frac{3^2}{2}+\biggl(\frac{1}{3}\biggr)^2 \ e^{-3} \ \frac{3^4}{4!} \\&=\frac{203}{72} \ e^{-3} \\&=0.1403718733  \end{aligned}

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Answer to Problem 5b

    \displaystyle \frac{1}{12}

Exam P Practice Problem 4

Problem 4a
Suppose that the lifetime of an electronic device follows an exponential distribution. The device is sold with a t-year warranty. The manufacturer estimated that at the expiration of the warranty, about 85% of the devices are still working. The manufacturer is considering offering an extended warranty for an additional t years. What proportion of all devices that are found to be working at the expiration of the regular warranty will be working at the end of the extended warranty?

Problem 4b
Suppose that the lifetime of an electronic device follows an exponential distribution. The device is sold with a t-year warranty. The manufacturer estimated that at the expiration of the warranty, about 80% of the devices are still working. The manufacturer is considering offering an extended warranty for an additional \displaystyle \frac{t}{2} years. What proportion of all devices that are found to be working at the expiration of the regular warranty will be working at the end of the extended warranty?

Solution is found below.

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Solution to Problem 4a
0.85. Hint: the exponential distribution is memoryless.

Answer to Problem 4b
0.894427191.

Exam P Practice Problem 3

Problem 3a
Suppose that the lifetime (in years) of an electronic device has the following probability density function:

\displaystyle f(x)=0.0625 \ x \ e^{-0.25 x} \text{where } x>0

The device is sold with a one-year warranty. The manufacturer is considering offering an extended warranty for an additional one year. What proportion of all devices that are found to be working at the expiration of the regular warranty will be working at the end of the extended warranty?

Problem 3b
Losses under the two policies (policy 1 and policy 2) are independent and each of the losses follows an exponential distribution with mean 2. if the total loss is greater than 3, what is the probability that the loss from policy 1 is greater than 3?

Solution is found below.

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Solution to Problem 3a
The problem is to find the probability that a device will survive two years given that it has already survived one year. Let X be the lifetime. Then we need to find P(X>2 \lvert X>1). Note that the pdf given in the problem is a Gamma distribution with parameters 0.25 and 2. We need to find the following right tail of this distribution:

\displaystyle \int_t^{\infty} 0.0625 \ x \ e^{-0.25 x} \ dx

The integral can be evaluated by the method of integration by parts. We can also evaluate using a Poisson distribution. The integral is equivalent to P(N \le 1) where N has the Poisson distribution with parameter 0.25t (see Evaluating the Gamma right tail).

\displaystyle \begin{aligned}\int_t^{\infty} 0.0625 \ x \ e^{-0.25 x} \ dx&=e^{-0.25t}+e^{-0.25t} \ (0.25t) \\&=e^{-0.25t} \ (1+0.25t)  \end{aligned}

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\displaystyle \begin{aligned}P(X>2 \lvert X>1)&=\frac{e^{-0.5} \ 1.5}{e^{-0.25} \ 1.25} \\&\text{ } \\&=\frac{1.5}{1.25} \ e^{-0.25} \\&\text{ }  \\&=0.93456  \end{aligned}

Answer to Problem 3b
0.4

Exam P Practice Problem 2

Problem 2a
The lifetime of an electronic device follows an exponential distribution. The device is sold with a one-year warranty. The manufacturer estimated that at the expiration of the warranty, about 85% of the devices are still working. For such a device that was recently purchased, what is the probability that it will still be working in 5 years?

Problem 2b
The lifetime of an electronic device follows an exponential distribution. The device is sold with a one-year warranty. The manufacturer estimated that at the expiration of the warranty, about 90% of the devices are still working. An extended warranty for an additional 2 years is available, covering any needed repair during this period. What proportion of the devices sold will still be working at the end of the extended warranty period?

Solution is found below.

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Solution to Problem 2a
The survival function for these devices is S(t)=e^{-\alpha t}. Since 85% of the devices are still working at the end of the warranty period, we have:

\displaystyle e^{-\alpha}=0.85 \ \ \ \ \ \Longrightarrow \ \ \ \ \ \alpha=ln(0.85).

Thus S(5)=e^{-5ln(0.85)}=0.85^5=0.4437

Answer to Problem 2b
0.729

Evaluating the Gamma right tail

We present an alternative way of evaluating the following integral, which takes advantage of the connection between Gamma distribution and Poisson process.

\displaystyle (0) \ \ \ \ \ \ \ \ \ \ \ \ \int_t^{\infty} \frac{\alpha^n}{(n-1)!} \ x^{n-1} \ e^{-\alpha x} \ dx

The integrand in (0) is the probability density function (pdf) of a Gamma distribution with parameters \alpha (scale parameter) and n (shape parameter) where \alpha>0 and n is a positive integer. Thus the integral is the right tail of a Gamma distribution. If the pdf is to model the lifetime of a biological life, the integral is the survival function (i.e. the probability that the life will survice beyond time t).

The integral (0) can be evaluated using the method of integration by parts and the following provides the alternative method.

\displaystyle (1) \ \ \ \ \ \ \ \ \ \ \ \int_t^{\infty} \frac{\alpha^n}{(n-1)!} \ x^{n-1} \ e^{-\alpha x} \ dx=\sum \limits_{j=0}^{n-1}\ \frac{(\alpha t)^j \ e^{-\alpha t}}{j!}

The goal is not to memorize (1), but rather to understand the thought process behind the formula. The thought process involves identifying the Poisson process that is associated with the Gamma pdf in the integral. The integral is equivalent to a probability calculation in the Poisson process. The right-hand-side of (1) is the probability that there are less than n changes in the Poisson process from time zero to time t. The left-hand-side of (1) is the probability that the n^{th} change in the Poisson process occurrs after time t. The description of the algorithm is followed by examples and a concluding remark.

The Algorithm

  1. From the Gamma pdf in the left-hand-side of (1), pick out the three numbers \alpha, n and t. The first two are the parameters of the Gamma pdf and the third one is the lower limit of the integral.
  2. Consider the Poisson process with parameter \alpha. One thing to keep in mind here is that in this Poisson process, the number of changes in a unit time interval is modeled by \displaystyle P(N_1=j)=\frac{\alpha^j e^{-\alpha}}{j!} and the number of changes in a time interval of length t is modeled by \displaystyle P(N_t=j)=\frac{(\alpha t)^j e^{-\alpha t}}{j!}
  3. The right-hand-side of (1) is the probability that there are less than n changes in the time interval (0,t) (or at most n-1 changes).

Example 1

Evaluate \displaystyle \int_{6}^{\infty} x \ e^{-\frac{x}{3}} \ dx.

The three parameters in the algorithm are \alpha=\frac{1}{3}, n=2 and t=6. Consider the Poisson process with parameter \alpha=\frac{1}{3}. We need to find the probability that there are less than 2 changes in the time interval (0,6). The Poisson probability is \displaystyle P(N_6=j)=\frac{(2)^j e^{-2}}{j!}. Thus we have:

\displaystyle \begin{aligned}\int_{6}^{\infty} x \ e^{-\frac{x}{3}} \ dx&=9 \int_{6}^{\infty} \frac{1}{9} \ x^{2-1} \ e^{-\frac{x}{3}} \ dx \\&\text{ } \\&=9 \ \sum \limits_{j=0}^{1}\ \frac{(2)^j \ e^{-2}}{j!} \\&\text{ } \\&=9 \ e^{-2}(1+2)\\&\text{ } \\&=27e^{-2} \end{aligned}

Example 2

Evaluate \displaystyle \int_{1.5}^{\infty} x^3 \ e^{-2 x} \ dx.

The three parameters in the algorithm are \alpha=2, n=4 and t=1.5. We are interested in the Poisson process with parameter \alpha=2 and in finding the probability that there are less than 4 changes in time interval (0,1.5). The Poisson probability is \displaystyle P(N_{1.5}=j)=\frac{(3)^j e^{-3}}{j!}. The integral, after adjusting for a multiplicative constant, is the probability that the fourth change occurs after time 1.5.

\displaystyle \begin{aligned}\int_{1.5}^{\infty} x^3 \ e^{-2x} \ dx&=\frac{6}{16} \int_{1.5}^{\infty} \frac{2^4}{3!} \ x^{4-1} \ e^{-2x} \ dx \\&\text{ } \\&=\frac{3}{8} \ \sum \limits_{j=0}^{3}\ \frac{(3)^j \ e^{-3}}{j!} \\&\text{ } \\&=\frac{3}{8} \ e^{-3}(1+\frac{3}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!})\\&\text{ } \\&=\frac{39}{8} e^{-3} \end{aligned}

Remark
The reason that the summation in the right-hand-side of (1) works as an answer to the integral on the left-hand-side is that the summation is how the Gamma pdf is derived (where the shape parameter is a positive integer). Let X be the time until the occurrence of the n^{th} change in the Poisson process that has parameter \alpha. Then the right-hand-side of (1) is S_X(t)=P(X>t). The distribution function is then F_X(t)=1-S_X(t). Taking derivative of F_X(t)=1-S_X(t) produces the integrand in the left-hand-side of (1). Thus the alternative approach indicated here is simply going back to the root of the Gamma pdf.