A-Level Mathematics October/November 2024 Q7(d): Using λ = 0.6 and a suitable approximating distribution, find the probability that ther…

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  Let $X$ be the number of accidents in a period of 30 years. The rate of accidents is given as $\lambda = 0.6$ per year. For a 30-year period, the parameter for the Poisson distribution is $30 \times 0.6 = 18$. So, $X \sim \text{Po}(18)$.

  Since the parameter $\lambda = 18$ is large (e.g., $>10$), a Normal distribution can be used as an approximation. The parameters for the Normal approximation are $\mu = 18$ and $\sigma^2 = 18$. So, $X \approx N(18, 18)$.

  We want to find the probability of more than 10 accidents, which is $P(X > 10)$. Applying a continuity correction for this discrete-to-continuous approximation, we find the probability for the continuous variable: $P(X > 10.5)$

  Now, we standardise this value to find the Z-score: $Z = \frac{X - \mu}{\sigma} = \frac{10.5 - 18}{\sqrt{18}}$ $Z = \frac{-7.5}{4.2426...} = -1.7677... \approx -1.768$

  So we need to find $P(Z > -1.768)$. Using the symmetry of the Normal distribution, $P(Z > -1.768) = P(Z < 1.768)$. This is equivalent to looking up the value of $\Phi(1.768)$ in the standard Normal distribution tables.

  $\Phi(1.768) = 0.96147...$

  Rounding to 3 significant figures, the probability is: $P(X > 10) = 0.961$

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  How the marks are awarded

  • B1 — The B1 mark is awarded for correctly identifying the parameters of the Normal approximation. The mean and variance are both equal to the new Poisson parameter, $30 \times 0.6 = 18$, so the distribution $N(18, 18)$ is stated or used.
  • M1 — The first M1 is a method mark for standardising the value, including the essential continuity correction. The calculation $\frac{10.5 - 18}{\sqrt{18}}$ demonstrates this. The continuity correction correctly changes the discrete value of $10$ to a continuous boundary of $10.5$ for a 'greater than' probability.
  • M1 — The second M1 is for finding the correct probability area under the Normal curve corresponding to the calculated Z-score. For a 'greater than' probability of a negative Z-value, this involves using the property $P(Z > -z) = \Phi(z)$, as shown by finding $\Phi(1.768)$.
  • A1 — The final A1 mark is for the correct probability, rounded to 3 significant figures. The calculated value of $0.961$ (or $0.962$ due to rounding differences) earns this mark.

  Common mistakes

  • Forgetting to apply the continuity correction, i.e., standardising 10 instead of 10.5. This leads to an incorrect Z-score of -1.886 and loses method and accuracy marks.
  • Using an incorrect continuity correction, such as 9.5 instead of 10.5. This shows a misunderstanding of how to adjust the boundary for a P(X > a) inequality.
  • Using the wrong variance for the Normal approximation, for example N(18, 18²) or N(18, 0.6). The variance for a Normal approximation to a Poisson(μ) distribution is always μ.
  • Calculating the wrong tail of the distribution, for example finding $1 - \Phi(1.768)$ instead of $\Phi(1.768)$, which would give the probability for $P(Z < -1.768)$.

  Examiner tip: When approximating a discrete distribution with a continuous one, always remember to apply a continuity correction to the discrete value before standardising.

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