Actuarial Outpost Crazy Question #2
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#1
11-21-2010, 03:51 AM
 JasonT Member Join Date: Jun 2010 Posts: 36
Crazy Question #2

So I'm just impatient and would like to know where and how I got wrong.

Question: A company produces a single kind of remote control. The lifespan of the remote control follows normal distribution with mean = 3 years and standard deviation = 4.

3 identical remote controls were purchased. Find the probability that the total lifespan of the first two controls exceed the third control by 0.3 years.

Is it just Prob (x1 + x2 > 0.1 + x3)? If so, can I simplify it to Prob (x1+x2-X3> 0.1) and subsequently, Prob (x>0.1)?

#2
11-21-2010, 06:45 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,459

Quote:
 Originally Posted by JasonT So I'm just impatient and would like to know where and how I got wrong. Question: A company produces a single kind of remote control. The lifespan of the remote control follows normal distribution with mean = 3 years and standard deviation = 4. 3 identical remote controls were purchased. Find the probability that the total lifespan of the first two controls exceed the third control by 0.3 years. Is it just Prob (x1 + x2 > 0.1 + x3)? If so, can I simplify it to Prob (x1+x2-X3> 0.1)
Yes to both, if the 0.3 in the problem statement was a typo and should have been 0.1. All those blue constants should be the same. (If you intentionally converted 0.3 to 0.3/3 since there were 3 random variables, no)
Quote:
 and subsequently, Prob (x>0.1)?
No, assuming independence (which is surely the intent of the problem) you cannot do that. X1+X2-X3 is a normal random variable. You can compute its mean, variance and standard deviation. X1 is normal with mean 3 and standard deviation 4; X1+X2-X3 is normal with mean 3 but a different standard deviation. Thus the probability that X1 > 0.1 and then X2 > 0.1 is different.
#3
11-21-2010, 01:44 PM
 JasonT Member Join Date: Jun 2010 Posts: 36

You are right Gandalf, the question did mention independence of the controls. In this case, would the variance of X1+X2-X3 be 3*(16^2)? and standard deviation = sqrt(48)? If so, it would be prob[ Z> (0.1-3)/sqrt(48) ]. And the probability is 0.6844?

Thanks a lot!
#4
11-21-2010, 01:49 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,327

Quote:
 Originally Posted by JasonT You are right Gandalf, the question did mention independence of the controls. In this case, would the variance of X1+X2-X3 be 3*(16^2)? and standard deviation = sqrt(48)? If so, it would be prob[ Z> (0.1-3)/sqrt(48) ]. And the probability is 0.6844? Thanks a lot!
If $Var \, = \, 3 \cdot 16^2$ and $\sigma \, = \, \sqrt{Var} \, = \, \sqrt{3 \cdot 16^2} \, = \, 16 \sqrt{3}$

How are you getting $\sqrt{48}$?
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.
#5
11-21-2010, 02:05 PM
 JasonT Member Join Date: Jun 2010 Posts: 36

Shoot! Double mistake.

If standard deviation of a control is 4, its variance is 16.

The total variance for X1+X2-X3 should be 3*4^2 = 48 and std dev = sqrt(48). Right?
#6
11-21-2010, 02:20 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 4,988

Yes.

BTW, you don't need independence, pairwise uncorrelated will do.

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