The coach arriving into Station A each morning at 0800 has a capacity of $54$ passengers.
On this route standing passengers are not allowed.
Over a period of time the number of passengers alighting at Station A on $20$ days was recorded as follows:
$50, \; 50, \; 50, \; 50, \; 50, \; 51, \; 51, \; 51, \; 51, \; 51$
$51, \; 51, \; 52, \; 52, \; 52, \; 52, \; 52, \; 52, \; 53, \; 54$
$50$ of the passengers were regular commuters. The random variable, $X$, is the number of passengers arriving at Station A on this coach other than the $50$ regulars.
Show that the Poisson distribution is a reasonable fit for these data.
Use this distribution with your mean for the data above to estimate the probability that on any given day someone has been left behind because the coach was full.
The random variable, $X$, is the number of passengers arriving at Station A on this coach other than the $50$ regulars.
$\therefore \;$ Number of passengers (other than regulars) arriving $= X - 50$
i.e. $\;$ $0, \; 0, \; 0, \; 0, \; 0, \; 1, \; 1, \; 1, \; 1, \; 1, \;$
$1, \; 1, \; 2, \; 2, \; 2, \; 2, \; 2, \; 2, \; 3, \; 4$
Frequency table to illustrate the number of passengers arriving at station A (other than the $50$ regulars):
| Number of passengers $\left(x_i\right)$ | Frequency $\left(f_i\right)$ | $x_i \cdot f_i$ | $x_i^2$ | $x_i^2 \cdot f_i$ |
|---|---|---|---|---|
| $0$ | $5$ | $0$ | $0$ | $0$ |
| $1$ | $7$ | $7$ | $1$ | $7$ |
| $2$ | $6$ | $12$ | $4$ | $24$ |
| $3$ | $1$ | $3$ | $9$ | $9$ |
| $4$ | $1$ | $4$ | $16$ | $16$ |
For this set of data,
$n = \sum f_i = 20$; $\;$ $\sum x_i \cdot f_i = 26$; $\;$ $\sum x_i^2 \cdot f_i = 56$
$\therefore \;$ Mean number of passengers arriving at station A (other than the $50$ regulars)
$= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{26}{20} = 1.3$
Variance of number of passengers arriving at station A (other than the $50$ regulars)
$= s^2 = \dfrac{\sum x_i^2 \cdot f_i - n \cdot \overline{x}^2}{n - 1}$
i.e. $\;$ $s^2 = \dfrac{56 - \left(20 \times 1.3^2\right)}{20 - 1} = 1.168$
$\because \;$ The mean and the variance of the modified data are close to one another, Poisson distribution is a reasonable fit to the modified data (passengers other than the $50$ regulars).
$\therefore \;$ Mean of data $=$ Parameter of Poisson distribution $= \lambda = 1.3$
P(someone is left behind on a given day) $= P \left(X > 4\right)$
$P \left(X > 4\right) = 1 - P \left(X \leq 4\right)$
$\begin{aligned} P \left(X \leq 4\right) & = \sum \limits_{x = 0}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} \right] \\\\ & = e^{- 1.3} \times \left[1 + \dfrac{1.3}{1} + \dfrac{1.3^2}{2} + \dfrac{1.3^3}{6} + \dfrac{1.3^4}{24} \right] \\\\ & = 0.2725 \times 3.6302 = 0.9892 \end{aligned}$
$\therefore \;$ $P \left(X > 4\right) = 1 - 0.9892 = 0.0108$