If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.
Solution
Let $X$ denote the number of electricity failures per week. The average number of electricity failures per twenty weeks is 3. Thus the average number of electricity failures per week $E(X) = \lambda = 3/20 = 0.15$.
$X\sim P(0.15)$.
The probability mass function of Poisson distribution with $\lambda =0.15$ is
$$ \begin{aligned} P(X=x) &= \frac{e^{-0.15}(0.15)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$
The probability that there will not be more than one failure during a particular week is $P(X\leq 1)$
$$ \begin{aligned} P(X\leq1) &= P(X=0)+ P(X=1)\\ &= \frac{e^{-0.15}0.15^{0}}{0!}+\frac{e^{-0.15}0.15^{1}}{1!}\\ &= 0.8607+0.1291\\ &= 0.9898 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
