Question

The mortality rate for a certain disease is 0.007. Using Poisson distribution, calculate the probability for 2 deaths in a group of 400 people. [Use $e^{-2.8} = 0.0608$]

Hint:

In Poisson problems, $\lambda = n.p$. Use given $e^{-\lambda}$ and compute $\lambda^r / r!$.

Answer 1

Poisson distribution is applicable to rare events. The expected number of deaths, denoted by $\lambda$, is calculated as $400 \times 0.007 = 2.8$. The Poisson formula is given by: \[P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}\] Substituting $\lambda = 2.8$ and $r = 2$ into the formula yields: \[P(X = 2) = \frac{e^{-2.8}.(2.8)^2}{2!} = \frac{0.0608.7.84}{2} = \frac{0.476672}{2} = 0.238336 \approx \boxed{0.238}\]