S5 2004 7 Edexcel

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Remember that if X \sim \mathrm{Po}(\lambda), we have:

\displaystyle \mathrm P(X = k) = \frac {e^{-\lambda} \lambda^k} {k!}

(ai)

Since X \sim \mathrm {Po}(3) we have:

\displaystyle \mathrm P(X = 2) = \frac {e^{-3} \times 3^2} {2!} = \frac 9 2 e^{-3} = 4.5 e^{-3}

(aii)

We have:

\begin{align*}\mathrm P(X \ge 4) & = 1 - \mathrm P(X \le 3) \\ & = 1 - \left(\mathrm P(X = 0) + \mathrm P(X = 1) + \mathrm P(X = 2) + \mathrm P(X = 3)\right) \\ & = 1 - \left(\frac {e^{-3} 3^0} {0!} + \frac {e^{-3} 3^1} {1!} + \frac {e^{-3} 3^2} {2!} + \frac {e^{-3} 3^3} {3!}\right) \\ & = 1 - e^{-3} \left(1 + 3 + \frac 9 2 + \frac 9 2\right) \\ & = 1 - 13e^{-3}\end{align*}

(b)

Since there’s only 4 cleaners, we assume that if there are more than 4 cleaning requests in a day, all 4 cleaners get sent out. So if X \ge 4 then Y = 4 and vice versa. So \mathrm P(Y = 4) = \mathrm P(X \ge 4) = 1 - 13e^{-3}.

If less than 4 requests are received, we can give exactly one job to each cleaner (and hence send out as many cleaners as jobs) so \mathrm P(Y = 0) = \mathrm P(X = 0), \mathrm P(Y = 1) = \mathrm P(X = 1), \mathrm P(Y = 2) = \mathrm P(X = 2) and \mathrm P(Y = 3) = \mathrm P(X = 3). We have:

\mathrm P(X = 0) = e^{-3}

\mathrm P(X = 1) = 3 e^{-3}

\mathrm P(X = 2) = \dfrac {3^2} 2 e^{-3} = 4.5 e^{-3}

\mathrm P(X = 3) = \dfrac {3^3} {3!} e^{-3} = 4.5 e^{-3}

So we have:

\displaystyle G_Y(t) = e^{-3} + 3e^{-3} t + 4.5 e^{-3} t^2 + 4.5 e^{-3} t^3 + (1 - 13 e^{-3})t^4 \\= e^{-3} (1 + 3t + 4.5t^2 + 4.5t^3 - 13t^4) + t^4

[we make the assumption that workload is spread evenly between the cleaners and it’s not the case that all jobs are tended to by one cleaner]

©

We have:

\begin{align*} G'_Y(t) &= 3e^{-3} + 2 \times 4.5 e^{-3} t + 3 \times 4.5 e^{-3} t^2 + 4\left(1 - 13e^{-3}\right)t^3 \\ & = 3e^{-3} + 9e^{-3} t + 13.5 e^{-3} t^2 + 4 \left(1 - 13e^{-3}\right) t^3\end{align*}

so:

\begin{align*}\mathrm E(Y) & = G'_Y(1) \\ & = 3e^{-3} + 9e^{-3} + 13.5 e^{-3} + 4 - 4 \times 13 e^{-3} \\ & = 4 - 26.5 e^{-3}\end{align*}

Differentiating again we have:

\begin{align*}G''_Y(t) & = 9e^{-3} + 13.5 \times 2 e^{-3} t + 3 \times 4\left(1 - 13e^{-3}\right) t^2 \\ & = 9e^{-3} + 27 e^{-3} t + 12 \left(1 - 13e^{-3}\right) t^2\end{align*}

so:

\begin{align*}G''_Y(1) & = 9e^{-3} + 27 e^{-3} + 12 - 12 \times 13 e^{-3} \\ & = 12 - 120 e^{-3}\end{align*}

Then:

\begin{align*}\mathrm {var}(Y) & = 4 - 26.5 e^{-3} + 12 - 120 e^{-3} - \left(4 - 26.5 e^{-3}\right)^2 \\ & = 16 - 146.5 e^{-3} - \left(16 - 2 \times 4 \times 26.5 e^{-3} + 702.25e^{-6}\right) \\ & = 65.5 e^{-3} - 702.25 e^{-6}\end{align*}

So the standard deviation of Y is:

\displaystyle \sqrt {65.5e^{-3} - 702.25 e^{-6}} = 1.23302 \ldots