1. ## Probability Distributions

Currently we are learning gamma,exp, uniform, and normal distributions.
I am having trouble with the following homework problem:

The time to failure of a personal computer display tube is exponentially distributed with mean three years.

a) A company offers insurance on these tubes for the first year of usage. On what percentage of policies will they have to pay a claim?

I think for this I should do

1-e^(-1/3)=.2835

b) If an user order three such display tubes (two tubes as spare), what is the
probability that he need to order such tube within six years?

Here is what I did for this:

3*mean of 3 = 9

So,
e^(-x/9)
P{x<6}= 1-e^(-6/9)= .4866

Am I on the right track???
Any help is greatly appreciated.

2. Originally Posted by kid funky fried
Currently we are learning gamma,exp, uniform, and normal distributions.
I am having trouble with the following homework problem:

The time to failure of a personal computer display tube is exponentially distributed with mean three years.

a) A company offers insurance on these tubes for the first year of usage. On what percentage of policies will they have to pay a claim?

I think for this I should do

1-e^(-1/3)=.2835

b) If an user order three such display tubes (two tubes as spare), what is the
probability that he need to order such tube within six years?

Here is what I did for this:

3*mean of 3 = 9

So,
e^(-x/9)
P{x<6}= 1-e^(-6/9)= .4866

Am I on the right track???
Any help is greatly appreciated.
a) Correct, to four decimal places.

b) I don't think so.

$T = T_1 + T_2 + T_3$ (which is very different to $T = 3T_1$ )

$f(t) = \, .....$ (reading this will give yu a clue to what needs to be done here).

$\Pr(T < 6) = ....$

3. Should I calculate the prob of each tube lasting 6 years,then add together???

e^(-2)=.1353

.1353+.1353+.1353=.4059

4. Originally Posted by kid funky fried
Should I calculate the prob of each tube lasting 6 years,then add together???

e^(-2)=.1353

.1353+.1353+.1353=.4059
No, you need to calculate the pdf for T, the sum of the three random variables. Read the link I provided.

But using a moment generating function approach is probably the simplest way. If I have time I'll post some details of the calculation later.

5. Originally Posted by mr fantastic
No, you need to calculate the pdf for T, the sum of the three random variables. Read the link I provided.

But using a moment generating function approach is probably the simplest way. If I have time I'll post some details of the calculation later.
If $T_i$ has mean $\beta$ and follows an exponential distribution then the moment generatting function of $T_i$ is $m_{T_i}(t) = \frac{1}{1 - \beta t}$.

If $T = T_1 + T_2 + T_3$ and the $T_i$ are i.i.d. random variables then the moment generating function of T is

$m_{T}(t) = \left( \frac{1}{1 - \beta t} \right)^3 = \frac{1}{(1 - \beta t)^3}$.

This is recognised as the moment generating function of a random variable that follows a gamma distribution.

The pdf of T is therefore $f(t) = \frac{1}{\Gamma(3) \beta^3} \, t^{3 - 1} \, e^{-t/\beta} = \frac{1}{2 \beta^3} \, t^{2} \, e^{-t/\beta}$.

In your case $\beta = 3$.

$\Pr(T < 6) = \int_0^6 f(t) \, dt$.