Actuarial Outpost ASM 1st Edition (Exam P, Weishaus) Practice Exam 2 Q14, 16,24
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#1
12-21-2017, 02:44 AM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14
ASM 1st Edition (Exam P, Weishaus) Practice Exam 2 Q14, 16,24

Hi,

I'm preparing for the Jan Exam P (2nd attempt) and I was wondering if anyone could help me shed some light on these questions as I've been staring at the answer keys and yet I'm not understanding it clearly.

14) A machine fails when two of its components fail. The failure times of the first component are uniformly distributed on [3,6]. The failure time of the second component, given that the first component failed at time x, is uniformly distributed on [2x,3x]. Calculate the probability that the machine does not fail before time 10.

A) 0.52 B) 0.55 C) 0.59 D) 0.62 E)0.65

My thoughts: I draw a graph for y=2x and y=3x and use [3,6] for my x limit and y>10 for my y limit. Get the joint density by 1/(6-3) * 1/(3x-2x) = 1/3x. But I don't know how to use my limits for my integration to find the probability.

16) A charity receives 12 donations in a day. The size of each donation is normally distributed with mean 100 and variance 6400. The charity makes 3 grants each day. Each grant is normally distributed with mean 350 and variance 10,000. The donations and grants are mutually independent. Calculate the probability that the sum of the donations minus the sum of the grants for one day is positive.
A) 0.69 B) 0.7 C) 0.71 D) 0.72 E) 0.73

My thoughts: Let x be the sum of all donations and let y be sum of all grants. I found the E[X-Y] = 150 and Var[x-y] = 106800. So then I have to find P[x-y>0]. Set z= x-y for simplicity. P[z>-150/326.8] => p[z<0.46]. But this is the wrong answer.

24) The length of a table is measured. The measurement error is uniformly distributed over [-0.05, 0.05]. To get a more accurate measurement, the measurement is repeated n times, and the n measurements are then average. Each measurement is independent of each other measurement. Calculate the minimum number of measurements necessary so that the measurement error is no more than 0.005 in absolute value with probability 95%.
A) 90 B) 100 C) 115 D) 122 E) 128

My thoughts: I can set up P[-0.005<x<0005]=0.95 where x is measurement error but I don't know how to move forward from here...

#2
12-21-2017, 08:33 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,817

Quote:
 Originally Posted by dslee85 Hi, I'm preparing for the Jan Exam P (2nd attempt) and I was wondering if anyone could help me shed some light on these questions as I've been staring at the answer keys and yet I'm not understanding it clearly. 14) A machine fails when two of its components fail. The failure times of the first component are uniformly distributed on [3,6]. The failure time of the second component, given that the first component failed at time x, is uniformly distributed on [2x,3x]. Calculate the probability that the machine does not fail before time 10. A) 0.52 B) 0.55 C) 0.59 D) 0.62 E)0.65 My thoughts: I draw a graph for y=2x and y=3x and use [3,6] for my x limit and y>10 for my y limit. Get the joint density by 1/(6-3) * 1/(3x-2x) = 1/3x. But I don't know how to use my limits for my integration to find the probability.
You need a double integral, since there are two failures involved. The limits of the outer integral can be 3 to 6, since there are some combinations of first and second failure, for each such x, that give failure time > 10.

It gets more complicated because the limits (with nonzero density) for the second failure depend on when the first failure was. You can't use 10 to infinity for the inside limits, since there may be 0 density at (x,10).

So you need to combine (evaluate separately and add) two double integrals, to make sure that (for example) (4,9) and (4,25) are not included in either but (4,11) is included in one (and not both); similarly for (5.5,10.25), (5.5, 30) and (5.5,12). The limits for the inside integral will be different in the two double integrals. (The limits for the outside limits will probably be different, with no overlapping, but could be the same if you choose the limits on the inside integral to avoid overlapping.)

Quote:
 16) A charity receives 12 donations in a day. The size of each donation is normally distributed with mean 100 and variance 6400. The charity makes 3 grants each day. Each grant is normally distributed with mean 350 and variance 10,000. The donations and grants are mutually independent. Calculate the probability that the sum of the donations minus the sum of the grants for one day is positive. A) 0.69 B) 0.7 C) 0.71 D) 0.72 E) 0.73 My thoughts: Let x be the sum of all donations and let y be sum of all grants. I found the E[X-Y] = 150 and Var[x-y] = 106800. So then I have to find P[x-y>0]. Set z= x-y for simplicity. P[z>-150/326.8] => p[z<0.46]. But this is the wrong answer.
Is there any errata for the practice exam? What value do you get for your expression, and what is the answer key?

Quote:
 24) The length of a table is measured. The measurement error is uniformly distributed over [-0.05, 0.05]. To get a more accurate measurement, the measurement is repeated n times, and the n measurements are then average. Each measurement is independent of each other measurement. Calculate the minimum number of measurements necessary so that the measurement error is no more than 0.005 in absolute value with probability 95%. A) 90 B) 100 C) 115 D) 122 E) 128 My thoughts: I can set up P[-0.005
Then do set up set up P[-0.005<x<0005] as a formula. It involves n. Figure out for what values of n your formula will evaluate to .95.
#3
12-21-2017, 02:46 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

I only have time to help with #14 right now.

Double integrals are not necessary.

Let $U\sim\text{Uniform}(3,6)$ and $X \mid U \sim\text{Uniform}(2U,3U)$

We want $P(X\geq10)$

There are 3 cases:
Case 1
$U \in (3,\frac{10}{3})$ - in this case, X can not exceed 10 so the conditional probability is 0.
Note: $P(U \in (3,\frac{10}{3}))=\frac{\frac{10}{3}-3}{6-3}=\frac{1}{9}$

Case 2
$U \in (\frac{10}{3},5)$ - in this case, X could exceed 10 and its interval contains 10 so the conditional probability can be found with the continuous version of the law of total probability. Keep in mind that $U\mid U\in(\frac{10}{3},5) \sim \text{Uniform}(\frac{10}{3},5)$ so our integral in this case will be reduced to that uniform PDF for U.
Note: $P(U \in (\frac{10}{3},5))=\frac{5-\frac{10}{3}}{6-3}=\frac{5}{9}$

Case 3
$U \in (5,6)$ - in this case, X is certain to exceed 10 since its lower bound does hence the conditional probability is 1.
Note: $P(U \in (5,6))=\frac{6-5}{6-3}=\frac{1}{3}$

SOLUTION

\begin{align*}
P(X\geq10)&=\frac{1}{9}\cdot0+\frac{5}{9}\int_{10/3}^{5} \frac{3x-10}{3x-2x}\cdot\frac{1}{5-\frac{10}{3}} \, dx + \frac{1}{3}\cdot 1\\
&=\frac{5}{9}\cdot\frac{3}{5}\int_{10/3}^{5} 3-\frac{10}{x} \, dx + \frac{1}{3}\\
&=\frac{1}{3} \left(3x-10\ln{x} \right)|_{10/3}^{5}+\frac{1}{3}\approx 0.6484
\end{align*}

That integral uses $\frac{3x-10}{3x-2x}$ because that's the conditional probability that X is at least 10 given the value of U and the other factor is the PDF of U (both conditional on case 2).

Does that make sense to you?
#4
12-21-2017, 05:16 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

You are correct for #16. As Gandalf suggested: errata

You'll find the quote "On page 480, in the solution to question 16, change Φ(0.459)=0.7057 to Φ(0.459)=0.6768. None of the five answer choices is correct."

---------------------------------------------------------------------------

For #24 the mean of the average is computed as 0. The variance of the average is computed as $\frac{1}{1200n}$... so the standard deviation is $\frac{1}{\sqrt{1200n}}$

So you want:
$0.95=P(-0.005

or

$0.975=\Phi(0.005\sqrt{1200n})$

look up 0.975 on the normal table and you see it corresponds to z=1.96 so

$0.005\sqrt{1200n}=1.96$

$n\approx 128$

Last edited by Z3ta; 12-21-2017 at 05:21 PM..
#5
12-27-2017, 06:08 PM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14

Quote:
 Originally Posted by Gandalf You need a double integral, since there are two failures involved. The limits of the outer integral can be 3 to 6, since there are some combinations of first and second failure, for each such x, that give failure time > 10. It gets more complicated because the limits (with nonzero density) for the second failure depend on when the first failure was. You can't use 10 to infinity for the inside limits, since there may be 0 density at (x,10). So you need to combine (evaluate separately and add) two double integrals, to make sure that (for example) (4,9) and (4,25) are not included in either but (4,11) is included in one (and not both); similarly for (5.5,10.25), (5.5, 30) and (5.5,12). The limits for the inside integral will be different in the two double integrals. (The limits for the outside limits will probably be different, with no overlapping, but could be the same if you choose the limits on the inside integral to avoid overlapping.) Is there any errata for the practice exam? What value do you get for your expression, and what is the answer key? Then do set up set up P[-0.005
Thank you very much Gandalf for your response, my apologies for the late reply. I just returned from a camping trip.

I'm still having trouble understanding the limits. I have attached an image that demonstrates how I'm seeing the question. But I don't know how to set the limits near the ? so I can find the probability.
Let me know if the link doesn't work or if you don't understand my question.

Thank you for identifying the errata. The answer key according to the manual was (E).
#6
12-27-2017, 06:09 PM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14

https://ibb.co/cYi3Ew
#7
12-27-2017, 06:24 PM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14

Quote:
 Originally Posted by Z3ta You are correct for #16. As Gandalf suggested: errata You'll find the quote "On page 480, in the solution to question 16, change Φ(0.459)=0.7057 to Φ(0.459)=0.6768. None of the five answer choices is correct." --------------------------------------------------------------------------- For #24 the mean of the average is computed as 0. The variance of the average is computed as $\frac{1}{1200n}$... so the standard deviation is $\frac{1}{\sqrt{1200n}}$ So you want: $0.95=P(-0.005 or $0.975=\Phi(0.005\sqrt{1200n})$ look up 0.975 on the normal table and you see it corresponds to z=1.96 so $0.005\sqrt{1200n}=1.96$ $n\approx 128$
Thank you very much Zeta for your response and my apologies for the late response, I just returned from a camping trip.

The answers to the second and third question are good. I really like the idea of not doing the double integrals for the first question. However, I'm don't understand how you broke it up in 3 pieces and how you got those critical points.

Also when you say, U E (3, 10/3). I'm not too sure what you mean by that notation. Are you saying U ranging from 3 ~ 10/3?

I'm really sorry but could you elaborate a little bit more on why/how you broke this case into 3? I haven't been successful at understanding your explanation.

Thank you very much Zeta, and wish you happy holidays!
#8
12-27-2017, 06:36 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,817

OK, that picture is the right way to start, but I'm not sure how the picture works. The line y=2x, y=3x, x=3, x=6 and y=10 are all good lines to have.

It looks like the x=3 line intersects the y=10 line between where it intersects the y=2x line and the y=3x line (even though the intersection with the y=3x line is not shown).

Since you have a point labeled (10/3, 10) (a good point to know), I suggest you add the line x=10/3 to your picture. Since you have a point labeled (5, 10) (a good point to know), I suggest you add the line x=5 to your picture, and label all points of intersection among the various lines.

Then think about three double integrals. Maybe not the most efficient thing to do, but might help you get to the right result.

First, write the double integral that includes all points you want to include between x=5 and x=6. Second, write the double integral that includes all points you want to include between x=10/3 and x=5. Then, write the double integral that includes all points you want to include between x=3 and x=10/3.
#9
12-27-2017, 08:05 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by dslee85 Thank you very much Zeta for your response and my apologies for the late response, I just returned from a camping trip. The answers to the second and third question are good. I really like the idea of not doing the double integrals for the first question. However, I'm don't understand how you broke it up in 3 pieces and how you got those critical points. Also when you say, U E (3, 10/3). I'm not too sure what you mean by that notation. Are you saying U ranging from 3 ~ 10/3? I'm really sorry but could you elaborate a little bit more on why/how you broke this case into 3? I haven't been successful at understanding your explanation. Thank you very much Zeta, and wish you happy holidays!
That's correct. $U \in (3,\frac{10}{3})$ reads "U is in the interval between 3 and 10/3". It is just another way of writing $3 < U < \frac{10}{3}$

I worked backwards to determine the numbers 10/3 and 5. I knew that there would be certainty of $X\geq10$ if $2U\geq10 \rightarrow U \geq 5$ since 2U is its lower bound.

Similarly, there is certainty that $X\not\geq10$ if $3U\leq10 \rightarrow U\leq\frac{10}{3}$ since 3U is its upper bound.

The problem becomes a weighted sum of the conditional probabilities in each of those cases (weighted by the unconditional probability of each case) using the law of total probability. That's the discrete version.

The integral itself is an application of the continuous version of the law of total probability (you integrate $P(X\ge10\mid U)$ times the PDF of $U$ over the possible values of $U$). Note: here $\color{red}U$ is really $\color{red}U \mid U \in (\frac{10}{3},5)$ since we're within a specific case of our first use of the law of total probability... but I thought it might be confusing if I wrote that.

Does that make sense?

Last edited by Z3ta; 12-27-2017 at 08:12 PM..
#10
12-28-2017, 03:37 PM
 dslee85 SOA Join Date: Apr 2017 College: Carnegi Posts: 14

Quote:
 Originally Posted by Gandalf OK, that picture is the right way to start, but I'm not sure how the picture works. The line y=2x, y=3x, x=3, x=6 and y=10 are all good lines to have. It looks like the x=3 line intersects the y=10 line between where it intersects the y=2x line and the y=3x line (even though the intersection with the y=3x line is not shown). Since you have a point labeled (10/3, 10) (a good point to know), I suggest you add the line x=10/3 to your picture. Since you have a point labeled (5, 10) (a good point to know), I suggest you add the line x=5 to your picture, and label all points of intersection among the various lines. Then think about three double integrals. Maybe not the most efficient thing to do, but might help you get to the right result. First, write the double integral that includes all points you want to include between x=5 and x=6. Second, write the double integral that includes all points you want to include between x=10/3 and x=5. Then, write the double integral that includes all points you want to include between x=3 and x=10/3.
So in summary, I should have 3 integrations where one ranges from 3 ~ 10/3, and another one between 10/3 ~ 5, and the last one between 5 ~ 6. Got it!

Thank you very much for your explanation! Happy holidays and a happy new year!