In a college, the number of accidents to students requiring hospitalization in one year of $30$ weeks is recorded as follows:
| Number of accidents requiring hospitalization each week | Frequency |
|---|---|
| $0$ | $25$ |
| $1$ | $4$ |
| $2$ | $1$ |
| $3$ or more | $0$ |
This data is used to assesses the risk of such accidents.
Assuming that the Poisson distribution is a suitable model calculate the probability that
- In any one week there will be $3$ accidents requiring hospitalization,
- In a term of $8$ weeks there will be no accidents.
Frequency table to illustrate the number of the number of accidents to students requiring hospitalization in each week:
| Number of accidents requiring hospitalization each week $\left(x_i\right)$ | Frequency $\left(f_i\right)$ | $x_i \cdot f_i$ |
|---|---|---|
| $0$ | $25$ | $0$ |
| $1$ | $4$ | $4$ |
| $2$ | $1$ | $2$ |
| $\geq 3$ | $0$ | $0$ |
For this set of data,
$n = \sum f_i = 30$; $\;$ $\sum x_i \cdot f_i = 6$
$\therefore \;$ Mean number of accidents requiring hospitalization each week
$= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{6}{30} = 0.2$
Mean $=$ Parameter of Poisson distribution $= \lambda = 0.2$
- P(in any one week there will be $3$ accidents requiring hospitalization) $= P \left(X = 3\right)$
$\begin{aligned} P \left(X = 3\right) & = \dfrac{e^{- \lambda} \times \lambda^{3}}{3!} \\\\ & = \dfrac{e^{- 0.2} \times \left(0.2\right)^{3}}{6} \\\\ & = \dfrac{0.8187 \times 0.008}{6} = 0.0011 \end{aligned}$
- P(in $1$ week there will be no accidents) $= P \left(X = 0\right)$
$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!} \\\\ & = e^{- 0.2} = 0.8187 \end{aligned}$
$\therefore \;$ P(in a term of $8$ weeks there will be no accidents)
$= \left[P \left(X = 0\right)\right]^{8} = \left(0.8187\right)^8 = 0.2018$