# Weights of elephants approximately follow an exponential distribution with a mean of 2.5 tons....

## Question:

Weights of elephants approximately follow an exponential distribution with a 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?

## Gamma Distribution and the relationship between the CDFs of Gamma and Poisson Distribution

The density of {eq}gamma(\alpha,p) {/eq} distribution is given by

{eq}\begin{align*} g(x;\alpha,p)&= \left\{ \begin{array}{ll} \frac{\alpha^p e^{-\alpha x}x^{p-1}}{\Gamma(p)} & x>0 \\ 0 & \text{elsewhere} \\ \end{array} \right. \end{align*} {/eq}

If {eq}X_i\sim gamma(\alpha,p_i) {/eq} are independent, {eq}i=1,2,\cdots,k {/eq}, then {eq}\sum_{i=1}^k X_i\sim gamma(\alpha,\sum_{i=1}^k p_i) {/eq}.

Note that, {eq}exp(\alpha) {/eq} distribution is the same as {eq}gamma(\alpha,1) {/eq} distribution.

Now, if {eq}X\sim gamma(\alpha,k+1), Y\sim Poisson(\lambda), {/eq} then {eq}P(\alpha X>\lambda)=P(Y\leq k),k=0,1,2,\cdots {/eq}.

Weights of elephants approximately follow an exponential distribution with a mean of 2.5 tons. One hundred elephants are being transported on a ship that has a cargo limit of 300 tons. Let {eq}X_i {/eq} denote the weight of the {eq}i {/eq}th elephant, then {eq}X_i\stackrel{iid}{\sim}exp(\frac{1}{2.5})\ \text{i.e.}\ exp(0.4)\ \text{i.e.}\ gamma(0.4,1) {/eq}. So {eq}S=\sum_{i=1}^{100}X_i\sim gamma(0.4,100) {/eq}.

So the chance this ship will sink is,

{eq}\begin{align} P(S>300)&=P(0.4S>120)\\ &=P(Y\leq 99) \ \ \ \ \ (\text{where}\ Y\sim Poisson(120))\\ &\approx0.0279 \end{align} {/eq}