A company produces optical-fiber cable with a mean of 0.7 flaws per 100 feet. What is the probability that there will be exactly 3 flaws in 1100 feet of cable?

#### Solution

Let $X$ denote the number of flaws in the optical-fiber cable.

A mean number of flaws per 100 feet cable is $0.7$.

Thus the mean number of flaws per 1100 feet of cable is $\lambda = 0.7 *\frac{1100}{100}= 7.7$.

$X\sim P(7.7)$.

The probability mass function of $X$ (number of flaws in the 1100 feet cable) is Poisson distribution with $\lambda =7.7$ is

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

The probability that exactly 3 flaws in 1100 feet optical fiber cable is

` $$ \begin{aligned} P(X=3) &= \frac{e^{-7.7}7.7^{3}}{3!}\\ &= 0.0345 \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators