Probability

The number of bacteria in $50$ $100cc$ samples of water are given in the following table.

Number of bacteria per sample Number of samples
$0$ $23$
$1$ $16$
$2$ $9$
$3$ $2$
$4 \;$ or more $0$

  1. Find the mean number and the variance of bacteria in a $100cc$ sample.
  2. State whether the conditions for using the Poisson distribution as a model apply.
  3. Using the Poisson distribution with the mean found in part (1), estimate the probability that another $100cc$ sample will contain
    1. no bacteria,
    2. more than $4$ bacteria.


  1. For the given set of data,

    Number of bacteria per sample $\left(x_i\right)$ Number of samples $\left(f_i\right)$ $x_i \cdot f_i$ $x_i^2$ $x_i^2 \cdot f_i$
    $0$ $23$ $0$ $0$ $0$
    $1$ $16$ $16$ $1$ $16$
    $2$ $9$ $18$ $4$ $36$
    $3$ $2$ $6$ $9$ $18$
    $\geq 4$ $0$ $0$ $\geq 16$ $0$


    $n = \sum f_i = 50$; $\;$ $\sum x_i \cdot f_i = 40$; $\;$ $\sum x_i^2 \cdot f_i = 70$

    $\therefore \;$ Mean number of bacteria $= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{40}{50} = 0.8000$

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

    i.e. $\;$ $s^2 = \dfrac{70 - \left(50 \times 0.8^2\right)}{50 - 1} = 0.7755$

  2. The number of bacteria present in a given sample are random and are independent of one another.

    The mean is close to the variance.

    Hence, Poisson distribution as a model may be applied to the given problem.

  3. $\because \;$ Mean $= \overline{x} = 0.80$,

    parameter of Poisson's distribution $= \lambda = 0.8$

    1. $P \left(\text{sample contains no bacteria}\right) = P \left(X = 0\right)$

      $\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!}\\\\ & = e^{- \lambda} \\\\ & = e^{- 0.80} = 0.4493 \end{aligned}$


    2. $P \left(\text{sample contains more than 4 bacteria}\right) = P \left(X > 4\right)$

      Now, $P \left(X > 4\right) = 1 - P \left(X \leq 4\right)$

      $\begin{aligned} P \left(X \leq 4\right) & = \sum \limits_{x = 0}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} \right] \\\\ & = e^{- 0.8} \times \left[1 + \dfrac{0.8}{1} + \dfrac{0.8^2}{2} + \dfrac{0.8^3}{6} + \dfrac{0.8^4}{24} \right] \\\\ & = 0.4493 \times 2.2224 = 0.9985 \end{aligned}$

      $\therefore \;$ $P \left(X > 4\right) = 1 - 0.9985 = 0.0015$

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