# Solved (Free): In Baltimore City, the average number of deaths due to breast cancer reported each month is 7.7

#### ByDr. Raju Chaudhari

Mar 14, 2021

In Baltimore City, the average number of deaths due to breast cancer reported each month is 7.7. Assume that the number of breast cancer deaths follows a Poisson distribution.

(1) What is the probability that no breast cancer deaths will be reported during a given month?
(2) What is the probability that at least four breast cancer deaths will be reported during a given month?
(3) What is the standard deviation of the number of breast cancer deaths reported each month?

#### Solution

Let $X$ denote the number of deaths due to breast cancer reported each month. The average number of deaths due to breast cancer reported each month is 7.7, i.e., $E(X)=\lambda = 7.7$.

$X\sim P(7.7)$.

The probability mass function of Poisson distribution with $\lambda =7.7$ is

 \begin{aligned} P(X=x) &= \frac{e^{-7.7}(7.7)^x}{x!},\; x=0,1,2,\cdots \end{aligned}

1. The probability that no breast cancer deaths will be reported during a given month is

 \begin{aligned} P(X=0) &= \frac{e^{-7.7}7.7^{0}}{0!}\\ &= 5\times 10^{-4} \end{aligned}

1. The probability that at least four breast cancer deaths will be reported during a given month is

 \begin{aligned} P(X\geq 4) &= 1- P(X < 4)\\ & =1- P(X\leq 3)\\ &= 1- \sum_{x=0}^{3} P(X=x)\\ &= 1-\bigg(P(X=0) + P(X=1)\\ &\quad +P(X=2)+P(X=3)\bigg)\\ & = 1- \bigg[\frac{e^{-7.7}(7.7)^0}{0!}+\frac{e^{-7.7}(7.7)^1}{1!}\\ &\quad +\frac{e^{-7.7}(7.7)^2}{2!}+\frac{e^{-7.7}(7.7)^3}{3!}\bigg]\\ &=1- (5e-04+0.0035+0.0134+0.0345)\\ &= 1- 0.0518\\ & = 0.9482 \end{aligned}

1. The standard deviation of the number of breast cancer deaths reported each month is $V(X) = \lambda =7.7$.