Probability

The numbers of customers entering a shop in forty consecutive periods of one minute are given below:

$3, \; 0, \; 0, \; 1, \; 0, \; 2, \; 1, \; 0, \; 1, \; 1$
$0, \; 3, \; 4, \; 1, \; 2, \; 0, \; 2, \; 0, \; 3, \; 1 $
$1, \; 0, \; 1, \; 2, \; 0, \; 2, \; 1, \; 0, \; 1, \; 2 $
$3, \; 1, \; 0, \; 0, \; 2, \; 1, \; 0, \; 3, \; 1, \; 2 $

  1. Calculate values of the mean and variance of the number of customers entering the shop in a one minute period.
  2. Fit a Poisson distribution to the data and comment on the degree of agreement between the calculated and observed values.


  1. Frequency table to illustrate the number of customers entering a shop in forty consecutive periods of one minute:


    Table 1

    Number of customers $\left(x_i\right)$ Number of samples $\left(f_i\right)$ $x_i \cdot f_i$ $x_i^2$ $x_i^2 \cdot f_i$
    $0$ $13$ $0$ $0$ $0$
    $1$ $13$ $13$ $1$ $13$
    $2$ $8$ $16$ $4$ $32$
    $3$ $5$ $15$ $9$ $45$
    $\geq 4$ $1$ $4$ $\geq 16$ $16$


    For this set of data,

    $n = \sum f_i = 40$; $\;$ $\sum x_i \cdot f_i = 48$; $\;$ $\sum x_i^2 \cdot f_i = 106$

    $\therefore \;$ Mean number of customers entering the shop in a one minute period

    $= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{48}{40} = 1.2$

    Variance of number of customers entering the shop in a one minute period

    $= s^2 = \dfrac{\sum x_i^2 \cdot f_i - n \cdot \overline{x}^2}{n - 1}$

    i.e. $\;$ $s^2 = \dfrac{106 - \left(40 \times 1.2^2\right)}{40 - 1} = 1.2410$

  2. From the cumulative probability tables with $\lambda = 1.2$ we have


    Table 2

    $\left(x_i\right)$ Probability
    $0$ $P \left(X = 0\right) = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!}$
    $ = e^{-1.2} = 0.3012$
    $\leq 1$ $P \left(X \leq 1\right) = \sum \limits_{0}^{1} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
    $= 0.3012 + \dfrac{e^{-1.2} \times 1.2^1}{1!} = 0.3012 + 0.3614 = 0.6626$
    $\leq 2$ $P \left(X \leq 2\right) = \sum \limits_{0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
    $= 0.6626 + \dfrac{e^{-1.2} \times 1.2^2}{2!} = 0.6626 + 0.2169 = 0.8795$
    $\leq 3$ $P \left(X \leq 3\right) = \sum \limits_{0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
    $= 0.8795 + \dfrac{e^{-1.2} \times 1.2^3}{3!} = 0.8795 + 0.0867 = 0.9662$
    $\leq 4$ $P \left(X \leq 4\right) = \sum \limits_{0}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
    $= 0.9662 + \dfrac{e^{-1.2} \times 1.2^4}{4!} = 0.9662 + 0.0260 = 0.9922$


    $\therefore \;$ The probabilities are


    Table 3

    $\left(x_i\right)$ Probability Expected frequency $=$ Probability $\times \; n$
    $0$ $0.3012$ $12.048$
    $1$ $0.6626 - 0.3012 = 0.3614$ $14.456$
    $2$ $0.8795 - 0.6626 = 0.2169$ $8.676$
    $3$ $0.9662 - 0.8795 = 0.0867$ $3.468$
    $4$ $0.9922 - 0.9662 = 0.0260$ $1.040$


    Comparing the given frequency of number of customers (Table 1) with the frequency of customers calculated using Poisson distribution (Table 3) we have,


    Table 4

    Number of customers Actual frequency Theoretical frequency (1 d.p)
    $0$ $13$ $12.0$
    $1$ $13$ $14.5$
    $2$ $8$ $8.7$
    $3$ $5$ $3.5$
    $4$ $1$ $1.0$


    As seen from Table 4, the fit is very good.

    This is because the mean and the variance are very close to one another.

at
Labels: