| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson approximation to binomial with standard conditions (large n, small p). Parts (i) and (iii) are bookwork recall, part (ii) is routine calculation with continuity correction, and part (iv) is a standard hypothesis test. All techniques are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02n Sum of Poisson variables: is Poisson5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Max no. of passengers plane can take oe | B1 | oe e.g. No of passengers who bought tickets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = 3.2\) | B1 | |
| \(e^{-3.2}\left(\frac{3.2^3}{3!} + \frac{3.2^4}{4!} + \frac{3.2^5}{5!}\right)\) | M1 | Any \(\lambda\). Allow one end error |
| \(= 0.5146 = 0.515\) (3 sfs) | A1 | SR Use of \(\text{Bin}(640, 0.005)\) scores B1 (only) for 0.516 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n > 50\) | B1 | Accept n is large |
| \(np = 1.6\), which is \(< 5\) or \(p = 0.005\) which is \(< 0.1\) | B1 | Allow \(np = 3.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean (for 5 days) \(= 8\); \(H_1\): Pop mean (for 5 days) \(< 8\) | B1 | or Pop mean (for 1 day) \(= 1.6\); Pop mean (for 1 day) \(< 1.6\). Allow \(\lambda\) or \(\mu\) but not just 'mean' |
| \(e^{-8}\left(1 + 8 + \frac{8^2}{2!}\right)\) | M1 | Any \(\lambda\ (\neq 1.6)\). No end errors. Accept use of \(\text{Bin}(1600, 0.005)\ P(0,1,2) = 0.0136\) |
| \(= 0.0138\) | A1 | |
| Comp \(0.025\) | M1 | Valid comparison |
| Evidence that mean no. failing to arrive has decreased | A1 | FT their '0.0138' or '0.0136'. No contradictions |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Max no. of passengers plane can take oe | B1 | oe e.g. No of passengers who bought tickets |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = 3.2$ | B1 | |
| $e^{-3.2}\left(\frac{3.2^3}{3!} + \frac{3.2^4}{4!} + \frac{3.2^5}{5!}\right)$ | M1 | Any $\lambda$. Allow one end error |
| $= 0.5146 = 0.515$ (3 sfs) | A1 | SR Use of $\text{Bin}(640, 0.005)$ scores B1 (only) for 0.516 |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n > 50$ | B1 | Accept n is large |
| $np = 1.6$, which is $< 5$ or $p = 0.005$ which is $< 0.1$ | B1 | Allow $np = 3.2$ |
## Question 7(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean (for 5 days) $= 8$; $H_1$: Pop mean (for 5 days) $< 8$ | B1 | or Pop mean (for 1 day) $= 1.6$; Pop mean (for 1 day) $< 1.6$. Allow $\lambda$ or $\mu$ but not just 'mean' |
| $e^{-8}\left(1 + 8 + \frac{8^2}{2!}\right)$ | M1 | Any $\lambda\ (\neq 1.6)$. No end errors. Accept use of $\text{Bin}(1600, 0.005)\ P(0,1,2) = 0.0136$ |
| $= 0.0138$ | A1 | |
| Comp $0.025$ | M1 | Valid comparison |
| Evidence that mean no. failing to arrive has decreased | A1 | FT their '0.0138' or '0.0136'. No contradictions |7 All the seats on a certain daily flight are always sold. The number of passengers who have bought seats but fail to arrive for this flight on a particular day is modelled by the distribution $\mathbf { B } ( 320,0.005 )$.\\
(i) Explain what the number 320 represents in this context.\\
(ii) The total number of passengers who have bought seats but fail to arrive for this flight on 2 randomly chosen days is denoted by $X$. Use a suitable approximating distribution to find $\mathrm { P } ( 2 < X < 6 )$.\\
(iii) Justify the use of your approximating distribution.\\
After some changes, the airline wishes to test whether the mean number of passengers per day who fail to arrive for this flight has decreased.\\
(iv) During 5 randomly chosen days, a total of 2 passengers failed to arrive. Carry out the test at the 2.5\% significance level.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S2 2019 Q7 [11]}}