The number of fire engine call-outs each day from a station in a city follows a Poisson distribution with a mean of 1.5. If there are more than 4 call-outs on a given day, then the firefighters at the station are given additional pay.

i) Find the probability that on a given day the firefighters are given additional pay.

ii) Find the probability that the firefighters are given additional pay on fewer than two of 10

working days.

#### Solution

Let $X$ denote the number of fire engine call-outs each day from a station in a city. Given that $X$ follows a Poisson distribution with a mean of 1.5, i.e. `$\lambda = 1.5$`

.

That is `$X\sim P(1.5)$`

.

The probability mass function of Poisson distribution with `$\lambda =1.5$`

is

` $$ \begin{aligned} P(X=x) &= \frac{e^{-1.5}(1.5)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$ `

i) If there are more than 4 call-outs on a given day, then the firefighters at the station are given additional pay.

The probability that on a given day the firefighters are given additional pay (i.e., there are more than 4 call-outs on a given day) is

` $$ \begin{aligned} P(X>4) &=1-P(X\leq 4)\\ & =1-\sum_{x=0}^{4}P(X=x)\\ &=1-\bigg[P(X=0)+P(X=1)+P(X=2)\\ &\quad +P(X=3)+P(X=4)\bigg]\\ &= 1- \bigg[\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}+\frac{e^{-1.5}1.5^{2}}{2!}\\ &\quad +\frac{e^{-1.5}1.5^{3}}{3!}+\frac{e^{-1.5}1.5^{4}}{4!}\bigg]\\ &= 1-\bigg[0.2231+0.3347+0.251\\ &\quad +0.1255+0.0471\bigg]\\ &= 0.0186 \end{aligned} $$ `

ii) The probability that on a given day the firefighters are given additional pay is `$p = P(X>4)=0.0186$`

.

Let $Y$ denote the number of days out of 10 days on which the firefighters are given additional pay.

Thus `$Y\sim B(10, 0.0186)$`

.

The probability mass function of $X$ is

` $$ \begin{aligned} P(Y=y) &= \binom{10}{y} (0.0186)^y (1-0.0186)^{10-y},\\ &\quad y=0,1,\cdots, 10 \end{aligned} $$ `

The probability that the firefighters are given additional pay on fewer than two of 10 working days is

` $$ \begin{aligned} P(Y< 2) &= P(Y\leq 1)\\ &=\sum_{y=0}^{1} P(Y=y)\\ &= \bigg[P(Y=0)+P(Y=1)\bigg]\\ &=0.8288+0.1571\\ & = 0.9859 \end{aligned} $$ `

#### Further Reading

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