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
$$ \begin{aligned} P(X=x) &= \frac{e^{-7.7}(7.7)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$
- The probability that no breast cancer deaths will be reported during a given month is
$$ \begin{aligned} P(X=0) &= \frac{e^{-7.7}7.7^{0}}{0!}\\ &= 5\times 10^{-4} \end{aligned} $$
- The probability that at least four breast cancer deaths will be reported during a given month is
$$ \begin{aligned} P(X\geq 4) &= 1- P(X < 4)\\ & =1- P(X\leq 3)\\ &= 1- \sum_{x=0}^{3} P(X=x)\\ &= 1-\bigg(P(X=0) + P(X=1)\\ &\quad +P(X=2)+P(X=3)\bigg)\\ & = 1- \bigg[\frac{e^{-7.7}(7.7)^0}{0!}+\frac{e^{-7.7}(7.7)^1}{1!}\\ &\quad +\frac{e^{-7.7}(7.7)^2}{2!}+\frac{e^{-7.7}(7.7)^3}{3!}\bigg]\\ &=1- (5e-04+0.0035+0.0134+0.0345)\\ &= 1- 0.0518\\ & = 0.9482 \end{aligned} $$
- The standard deviation of the number of breast cancer deaths reported each month is
$V(X) = \lambda =7.7$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
