| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Overbooking probability problems |
| Difficulty | Standard +0.3 This is a standard S2 normal approximation question requiring identification of binomial parameters, application of continuity correction, and reverse normal table lookup. While multi-part, each step follows routine procedures taught in the specification with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(B(303, 0.01)\) | B1 | \(B(303, 0.01)\) stated, allow \(p = 0.99\) or 0.1. Allow Bin implied clearly by parameters |
| \(\approx Po(3.03)\) | B1 | \(Po(3.03)\) stated or implied, can be recovered from (ii) |
| (ii) \(e^{-1.03}(1 + 3.03 + \frac{3.03^2}{2}) = 0.4165\) | M1, A1 | Correct formula, ± 1 term or "1 −" or both. Convincingly obtain 0.4165(02542) [Exact: 0.41535] |
| AG | [AG line] | |
| (iii) \(302\) seats \(\Rightarrow \mu = 3.02\); \(e^{-3.02}(1 + 3.02) = 0.1962\) | M1, M1, A1, A1 | Try smaller value of \(\mu\). Formula, at least one correct term. Correct number of terms for their \(\mu\). 0.1962 [or 0.1947 from exact] |
| \(0.196 < 0.2\); So \(302\) seats | A1 | Answer 302 only |
(i) $B(303, 0.01)$ | B1 | $B(303, 0.01)$ stated, allow $p = 0.99$ or 0.1. Allow Bin implied clearly by parameters
$\approx Po(3.03)$ | B1 | $Po(3.03)$ stated or implied, can be recovered from (ii)
(ii) $e^{-1.03}(1 + 3.03 + \frac{3.03^2}{2}) = 0.4165$ | M1, A1 | Correct formula, ± 1 term or "1 −" or both. Convincingly obtain 0.4165(02542) [Exact: 0.41535]
AG | | [AG line]
(iii) $302$ seats $\Rightarrow \mu = 3.02$; $e^{-3.02}(1 + 3.02) = 0.1962$ | M1, M1, A1, A1 | Try smaller value of $\mu$. Formula, at least one correct term. Correct number of terms for their $\mu$. 0.1962 [or 0.1947 from exact]
$0.196 < 0.2$; So $302$ seats | A1 | Answer 302 only
SR: $B(303, 0.99): B1B0; M0; M1$ then $N(298.98, 2.9898)$ or equiv, standardise: 2 with $np$ & $\sqrt{npq}$, M1A0; $N(0.1n, 0.09n)$; standardise with $np$ & $\sqrt{npq}$; solve quadratic for $\sqrt{n}$; $n = 339$: M1M1M1A1, total 4/9
SR: $p = 0.1$: $B(303, 0.1), N(30.3, 27.27)$ B1B0; Standardise 2 with $np$ & $\sqrt{npq}$, M1A0; $N(0.1n, 0.09n)$; standardise with $np$ & $\sqrt{npq}$; solve quadratic for $\sqrt{n}$; $n = 339$: M1M1M1A1, total 4/9
SR: 6/9; $B(303, 0.01) = N(3.03, 2.9997)$: B1B0; M0A0; M1A0
---5 An airline has 300 seats available on a flight to Australia. It is known from experience that on average only $99 \%$ of those who have booked seats actually arrive to take the flight, the remaining $1 \%$ being called 'no-shows'. The airline therefore sells more than 300 seats. If more than 300 passengers then arrive, the flight is over-booked. Assume that the number of no-show passengers can be modelled by a binomial distribution.\\
(i) If the airline sells 303 seats, state a suitable distribution for the number of no-show passengers, and state a suitable approximation to this distribution, giving the values of any parameters.
Using the distribution and approximation in part (i),\\
(ii) show that the probability that the flight is over-booked is 0.4165 , correct to 4 decimal places,\\
(iii) find the largest number of seats that can be sold for the probability that the flight is over-booked to be less than 0.2.
\hfill \mbox{\textit{OCR S2 2006 Q5 [9]}}