A-Level Mathematics October/November 2024 Q6(c): An inspector waits at desk B. She wants to wait long enough to be 90% certain of seeing…

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  Let $t$ be the time in minutes that the inspector waits. Let $X$ be the number of customers arriving at desk B in this time. Assuming from previous parts of the question that the rate of arrival at desk B is 2.1 customers per 10 minutes, this is equivalent to a rate of $\lambda = 0.21$ customers per minute.

  Therefore, the number of customers arriving in $t$ minutes follows a Poisson distribution: $X \sim \text{Po}(0.21t)$.

  The inspector wants to be 90% certain of seeing at least one customer. This can be written as a probability statement: $P(X \ge 1) \ge 0.90$

  It is easier to work with the complement event, $P(X=0)$: $1 - P(X=0) \ge 0.90$

  For a Poisson distribution with mean $\mu = 0.21t$, the probability of zero events is $P(X=0) = e^{-\mu} = e^{-0.21t}$.

  Substituting this into the inequality: $1 - e^{-0.21t} \ge 0.90$

  Now, we solve for $t$: $0.10 \ge e^{-0.21t}$ $e^{-0.21t} \le 0.10$

  Take natural logarithms of both sides: $\ln(e^{-0.21t}) \le \ln(0.10)$ $-0.21t \le \ln(0.10)$

  To isolate $t$, divide by -0.21. Remember to reverse the inequality sign when dividing by a negative number: $t \ge \frac{\ln(0.10)}{-0.21}$ $t \ge \frac{-2.302585...}{-0.21}$ $t \ge 10.96469...$

  Since the inspector must wait long enough for the condition to be met, we must find the smallest integer number of minutes that is greater than or equal to 10.96469... . Waiting for 10 minutes would be insufficient.

  Therefore, the minimum time she should wait is 11 minutes.

  Final Answer: 11 minutes.

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  How the marks are awarded

  • M1 — Setting up the correct probability inequality for 'at least one' customer, $1 - e^{-\lambda} \ge 0.90$, using the Poisson distribution. In the model answer, this is shown as $1 - e^{-0.21t} \ge 0.90$.
  • M1 — Correctly rearranging the inequality to the form $e^{-0.21t} \le 0.1$ and demonstrating the intention to solve by taking logarithms.
  • A1 — Correctly solving the logarithmic inequality to find the boundary value for the time, $t \ge 10.96...$ minutes (or a more accurate value).
  • A1 — Stating the final answer as the next integer value, 11 minutes, correctly interpreting the context of 'minimum time' to satisfy the condition.

  Common mistakes

  • Incorrectly rounding the calculated time of 10.96... minutes down to 10 minutes. This would not satisfy the condition of being at least 90% certain.
  • Forgetting to reverse the inequality sign when dividing by a negative number (e.g., -0.21), leading to an incorrect conclusion like $t \le 10.96...$.
  • Setting up the wrong probability, such as $P(X=1) = 0.90$ instead of $P(X \ge 1) \ge 0.90$.
  • Using an incorrect rate for the Poisson distribution, for example using 2.1 per minute instead of 0.21 per minute, which would lead to a time of $t \ge 1.096...$ and an incorrect final answer of 2 minutes.

  Examiner tip: When solving for a minimum value to satisfy a probability condition like $P(X \ge 1) \ge k$, always round your calculated boundary value up to the next appropriate integer to ensure the condition is met.

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