Let X be the number of breakdowns of the photocopier in a year. Since the distribution of X follows a Poisson distribution, the probability of having exactly \(\lambda\) breakdowns in a year is given by:
\(P(X = x) = \dfrac{ \lambda^{x}}{x!} e^{-\lambda}\)
a) As the probabilities for the photocopier to have exactly 2 repairs and exactly 3 repairs in a year are the same.
\(\begin{align}
\dfrac{\lambda^{2}}{2!} e^{-\lambda} &= \dfrac{ \lambda^{3}}{3!} e^{-\lambda} \\
\dfrac{ \lambda^{2}}{2!} &= \dfrac{ \lambda^{3}}{3!} \\
\lambda &= 1.5 \\
\end{align}\)
b i) Expected repair charge in a year if paln II is adopted
\(\begin{align}
&= P(X = 0) \times 0 + P(X = 1) \times 1000 + P(X = 2) \times 1800 + P(X = 3) \times 2600 \\
&= 0 + 1000 \times \dfrac{ 1.5^{1}}{1!} e^{-1.5} + 1800 \times \dfrac{ 1.5^{2}}{2!} e^{-1.5} + 2600 \times \dfrac{ 1.5^{3}}{3!} e^{-1.5} \\
&= 0.3347 \times 1000 + 0.2510 \times 1800 + 0.1255 \times 2600\\
&= $1112.8
\end{align}\)
b ii) As the expected repair charge in a year for plan II is less than that for plan I, plan II should be chosen to minimize the repair charge.
問題來源: 中學數學功課交流區
