Crazy Question #2

[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) and subsequently, Prob (x>0.1)?

Thanks for all your help!

[Gandalf]

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)

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.

[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!

[Actuarialsuck]

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 and

How are you getting ?

[JasonT]

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?

[Colymbosathon ecplecticos]

Yes.

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