- A newspaper finds that the mean number of typographical errors per page is five. Find the probability that

a. exactly five typographical errors will be found on a page,

b. fewer than five typographical errors will be found on a page, and

c. no typographical errors will be found on a page.

#### Solution

Let $X$ denote the number of typographical errors per page. The mean number of typographical errors per page is five, i.e., $E(X)=\lambda = 5$.

The random variable $X$ follows Poisson distribution. That is, $X\sim P(5)$.

The probability mass function of Poisson distribution with $\lambda =5$ is

` $$ \begin{aligned} P(X=x) &= \frac{e^{-5}(5)^x}{x!},\\ &\quad \; x=0,1,2,\cdots \end{aligned} $$ `

a. The probability that exactly five typographical errors will be found on a page is

` $$ \begin{aligned} P(X=5) &= \frac{e^{-5}5^{5}}{5!}\\ &= 0.1755 \end{aligned} $$ `

b. The probability that fewer than five typographical errors will be found on a page is

` $$ \begin{aligned} P(X < 5) &=\sum_{x=0}^{4}P(X=x)\\ &=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\\ &= \frac{e^{-5}5^{0}}{0!}+\frac{e^{-5}5^{1}}{1!}+\frac{e^{-5}5^{2}}{2!}+\frac{e^{-5}5^{3}}{3!}+\frac{e^{-5}5^{4}}{4!}\\ &= 0.0067+0.0337+0.0842+0.1404+0.1755\\ &= 0.4405 \end{aligned} $$ `

c. The probability that no typographical errors will be found on a page is

` $$ \begin{aligned} P(X=0) &= \frac{e^{-5}5^{0}}{0!}\\ &= 0.0067 \end{aligned} $$ `

#### Further Reading

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