# Suppose that the expected number of deaths due to bladder cancer for all workers at a tire plant on January 1, 1964, over the next 20 years

#### ByDr. Raju Chaudhari

Mar 29, 2021

Suppose that the expected number of deaths due to bladder cancer for all workers at a tire plant on January 1, 1964, over the next 20 years (1/1/64-12/31/83) based on US mortality rates is 1.8. If the Poisson distribution is assumed to hold and there are 6 reported deaths due to bladder cancer among the tire workers, then how unusual is this event?

#### Solution

Let $X$ denote the number of deaths due bladder cancer over 20 year period 1/1/64-12/31/83.

Given that $E(X) = \lambda = 1.8$. $X\sim Po(1.8)$.

The pmf of Poisson distribution with $\lambda =1.8$ is

\begin{aligned} P(X=x) &= \frac{e^{-1.8}(1.8)^x}{x!},\ &\quad x=0,1,2,\cdots \end{aligned}

The probability that there are $6$ deaths due to bladder cancer among the tire workers is

 \begin{aligned} P(X=6) &= \frac{e^{-1.8}1.8^{6}}{6!}\\ &= 0.00781 \end{aligned}

As the $P(X=6) = 0.00781$ is very close to zero, it is an unusual event.