# Thread: Exponential Distribution

1. ## Exponential Distribution

Weights of elephants approximately follow an exponential distribution with the mean of 2.5 tons. One hundred elephants are being transported on a ship that has a cargo limit of 300 tons. What is the chance this ship will sink?

I think i kind of understand this problem but unsure.

lambda = 1/2.5 = 0.40
exp. dist. x < 300

so 1- e^-(0.40)(100) = 1 so it will sink?

2. Originally Posted by aaronrj
Weights of elephants approximately follow an exponential distribution with the mean of 2.5 tons. One hundred elephants are being transported on a ship that has a cargo limit of 300 tons. What is the chance this ship will sink?

I think i kind of understand this problem but unsure.

lambda = 1/2.5 = 0.40
exp. dist. x < 300

so 1- e^-(0.40)(100) = 1 so it will sink?
You're told the pdf of X, the random variable 'weight of a single elephant'. I suggest you think about what the pdf of Y = 100X is and then calculate Pr(Y > 300).

3. I would use the normal approximation, with n=100.

By the Central Limit Theorem

$P\left(\sum_{i=1}^{100}X_i>300\right)=P\left(\bar X>3\right)$

$\approx P\left(Z>{3-\mu\over \sigma/\sqrt{100}}\right)$

NOW use your distribution..
You have a mean of 2.5 and a variance of 2.5 via the exponential distribution...

$\approx P\left(Z>3.16227766\right)$

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

If you don't want to approximate this, then use that fact that a sum of iid exponentials is a gamma.

Either way you need to calculate $P\left(\sum_{i=1}^{100}X_i>300\right)$

4. Originally Posted by matheagle
I would use the normal approximation, with n=100.

By the Central Limit Theorem

$P\left(\sum_{i=1}^{100}X_i>300\right)=P\left(\bar X>3\right)$

$\approx P\left(Z>{3-\mu\over \sigma/\sqrt{100}}\right)$

NOW use your distribution..
You have a mean of 2.5 and a variance of 2.5 via the exponential distribution...

$\approx P\left(Z>3.16227766\right)$

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

If you don't want to approximate this, then use that fact that a sum of iid exponentials is a gamma.

Either way you need to calculate $P\left(\sum_{i=1}^{100}X_i>300\right)$
You're too kind (I know some would beg to differ!!). I came back to fix my basic error (and I blame my physio for having made it) but was glad to see your reply.

5. so if (Z > 3.16) do find P(Z > 3.16) = 1-.9992 according to the normal table?

Using the gamma: (2.5 x 1< 3) givs approx. 0.7576 ? Im confused.

6. no problem.....
The only reason I write any post is to help others
wait, who says I'm mean?

7. Come on please just point me in the right direction. I am studying for a midterm and I would not ask if I understood.

8. Originally Posted by aaronrj
so if (Z > 3.16) do find P(Z > 3.16) = 1-.9992 according to the normal table?

Using the gamma: (2.5 x 1< 3) givs approx. 0.7576 ? Im confused.
To confirm the answer using the normal approximation, note that if X1, X2, .... Xn ~ Exp(lambda) then Y = X1 + X2 + ... + Xn ~ Gamma(n, 1/lambda).