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.