| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | State exact binomial distribution |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard bookwork on binomial-Poisson approximation and normal approximation to binomial. Parts (a), (b), and (d) are pure recall of conditions. Part (c) is a routine calculation using Poisson approximation with λ=5.4. Part (e) is a standard hypothesis test using normal approximation. All steps are textbook procedures with no problem-solving or novel insight 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 binomial2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B(n, 0.045)\) | B1 | For binomial with correct parameters \(n\) and 0.045 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Applicants are independent (no identical twins) or the proportion/probability identified as colour blind does not change over time | B1 | For one of the given reasons. Must have context. Allow equivalent statements. Do not allow number for proportion/probability |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B(120, 0.045) \Rightarrow \text{Po}(5.4)\) | B1 | Using or writing Po(5.4) |
| \(P(X=5) = \frac{e^{-5.4} \times 5.4^5}{5!}\) | M1 | For \(\frac{e^{-\lambda}\lambda^5}{5!}\) with any value of \(\lambda\) |
| \(= 0.1728\ldots\) awrt 0.173 | A1 | awrt 0.173. NB: A correct answer with no incorrect working scores 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Binomial with large \(n\) | B1 | \(n\) is large (Allow "number of trials" for \(n\)) |
| and very small \(p\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = 0.75 \quad H_1: p \neq 0.75\) | B1 | |
| \(B(96, 0.75) \Rightarrow N(72, 18)\) | B1 | |
| \(Z = \frac{67.5 - 72}{\sqrt{18}}\) or \(\frac{x \pm 0.5 - 72}{\sqrt{18}}\) | M1 | |
| \(= -1.06066\ldots\) or \(\frac{x+0.5-72}{\sqrt{18}} < -1.96\) or \(\frac{x-0.5-72}{\sqrt{18}} > 1.96\) | A1 | |
| \(P(z < -1.06) = 0.1444\ldots/0.1446\) or CR \(< 63.2\) awrt 0.144 or 0.145 | A1 | |
| There is insufficient evidence to reject \(H_0\) | dM1 | |
| Insufficient evidence against Jaymini's claim | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = 0.25 \quad H_1: p \neq 0.25\) | B1 | |
| \(B(96, 0.25) \Rightarrow N(24, 18)\) | B1 | |
| \(Z = \frac{28.5-24}{\sqrt{18}}\) or \(\frac{x \pm 0.5 - 24}{\sqrt{18}}\) | M1 | |
| \(= 1.06066\ldots\) or \(\frac{x+0.5-24}{\sqrt{18}} < -1.96\) or \(\frac{x-0.5-24}{\sqrt{18}} > 1.96\) | A1 | |
| \(P(z > 1.06) = 0.1444\ldots/0.1446\) or CR \(> 32.8\) awrt 0.144 or 0.145 | A1 | |
| There is insufficient evidence to reject \(H_0\) | dM1 | |
| Insufficient evidence against Jaymini's claim | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p\) is small | B1 | Allow probability for \(p\) |
| Both hypotheses correct in terms of \(p\) or \(\pi\), e.g. \(H_0: p = \) value, \(H_1: p \neq\) value | B1 | Must be attached to \(H_0\) and \(H_1\) |
| For writing or using \(N(72, 18)\) | B1 | May be implied by a correct standardisation expression |
| Standardising using \(67.5\) or \(67\) or \(66.5\) or \(x \pm 0.5\) with their mean and standard deviation | M1 | Allow \(\pm\) |
| awrt \(-1.06\) (may be implied by awrt \(0.144\) or \(0.145\)) or a correct standardisation with \(\pm 1.96\) | A1 | Ignore incorrect inequality symbol and allow \(=\) |
| Using probability route: awrt \(0.144\) or \(0.145\), or critical value \(z = \pm 1.96\); Using critical region route: \(CR < 63.2\) | A1 | |
| Correct statement – no context needed but do not allow contradicting non-contextual comments | dM1 | Dependent on M1 A1. Ignore any comparisons |
| Correct conclusion in context. Must include the word claim. If answer refers to the claim must include words applicants (oe) and pilots. No hypotheses then A0 | A1 | NB: Award M1 A1 for a correct contextual statement on its own |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Both hypotheses correct in terms of \(p\) or \(\pi\) | B1 | Must be attached to \(H_0\) and \(H_1\) |
| For writing or using \(N(24, 18)\) | B1 | May be implied by a correct standardisation expression |
| Standardising using \(28.5\) or \(29\) or \(29.5\) or \(x \pm 0.5\) with their mean and standard deviation | M1 | Allow \(\pm\) |
| awrt \(1.06\) (may be implied by awrt \(0.144\) or \(0.145\)) or a correct standardisation with \(\pm 1.96\) | A1 | Ignore incorrect inequality symbol and allow \(=\) |
| Using probability route: awrt \(0.144\) or \(0.145\), or critical value \(z = \pm 1.96\); Using critical region route: \(CR < 32.8\) | A1 | |
| Correct statement – no context needed but do not allow contradicting non-contextual comments | dM1 | Dependent on M1 A1. Ignore any comparisons |
| Correct conclusion in context. Must include the word claim. Must include words applicants (oe) and pilots. No hypotheses then A0 | A1 | NB: Award M1 A1 for a correct contextual statement on its own |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $B(n, 0.045)$ | B1 | For binomial with correct parameters $n$ and 0.045 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Applicants are independent (no identical twins) or the proportion/probability identified as colour blind does not change over time | B1 | For one of the given reasons. Must have context. Allow equivalent statements. Do not allow number for proportion/probability |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $B(120, 0.045) \Rightarrow \text{Po}(5.4)$ | B1 | Using or writing Po(5.4) |
| $P(X=5) = \frac{e^{-5.4} \times 5.4^5}{5!}$ | M1 | For $\frac{e^{-\lambda}\lambda^5}{5!}$ with any value of $\lambda$ |
| $= 0.1728\ldots$ awrt 0.173 | A1 | awrt 0.173. NB: A correct answer with no incorrect working scores 3/3 |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Binomial with large $n$ | B1 | $n$ is large (Allow "number of trials" for $n$) |
| and very small $p$ | B1 | |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.75 \quad H_1: p \neq 0.75$ | B1 | |
| $B(96, 0.75) \Rightarrow N(72, 18)$ | B1 | |
| $Z = \frac{67.5 - 72}{\sqrt{18}}$ or $\frac{x \pm 0.5 - 72}{\sqrt{18}}$ | M1 | |
| $= -1.06066\ldots$ or $\frac{x+0.5-72}{\sqrt{18}} < -1.96$ or $\frac{x-0.5-72}{\sqrt{18}} > 1.96$ | A1 | |
| $P(z < -1.06) = 0.1444\ldots/0.1446$ or CR $< 63.2$ awrt 0.144 or 0.145 | A1 | |
| There is insufficient evidence to reject $H_0$ | dM1 | |
| Insufficient evidence against Jaymini's claim | A1 | |
**ALT (using probability of failing):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.25 \quad H_1: p \neq 0.25$ | B1 | |
| $B(96, 0.25) \Rightarrow N(24, 18)$ | B1 | |
| $Z = \frac{28.5-24}{\sqrt{18}}$ or $\frac{x \pm 0.5 - 24}{\sqrt{18}}$ | M1 | |
| $= 1.06066\ldots$ or $\frac{x+0.5-24}{\sqrt{18}} < -1.96$ or $\frac{x-0.5-24}{\sqrt{18}} > 1.96$ | A1 | |
| $P(z > 1.06) = 0.1444\ldots/0.1446$ or CR $> 32.8$ awrt 0.144 or 0.145 | A1 | |
| There is insufficient evidence to reject $H_0$ | dM1 | |
| Insufficient evidence against Jaymini's claim | A1 | |
# Question (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p$ is small | B1 | Allow probability for $p$ |
| Both hypotheses correct in terms of $p$ or $\pi$, e.g. $H_0: p = $ value, $H_1: p \neq$ value | B1 | Must be attached to $H_0$ and $H_1$ |
| For writing or using $N(72, 18)$ | B1 | May be implied by a correct standardisation expression |
| Standardising using $67.5$ or $67$ or $66.5$ or $x \pm 0.5$ with their mean and standard deviation | M1 | Allow $\pm$ |
| awrt $-1.06$ (may be implied by awrt $0.144$ or $0.145$) or a correct standardisation with $\pm 1.96$ | A1 | Ignore incorrect inequality symbol and allow $=$ |
| Using probability route: awrt $0.144$ or $0.145$, or critical value $z = \pm 1.96$; Using critical region route: $CR < 63.2$ | A1 | |
| Correct statement – no context needed but do not allow contradicting non-contextual comments | dM1 | Dependent on M1 A1. Ignore any comparisons |
| Correct conclusion in context. Must include the word claim. If answer refers to the claim must include words applicants (oe) and pilots. No hypotheses then A0 | A1 | NB: Award M1 A1 for a correct contextual statement on its own |
**ALT:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Both hypotheses correct in terms of $p$ or $\pi$ | B1 | Must be attached to $H_0$ and $H_1$ |
| For writing or using $N(24, 18)$ | B1 | May be implied by a correct standardisation expression |
| Standardising using $28.5$ or $29$ or $29.5$ or $x \pm 0.5$ with their mean and standard deviation | M1 | Allow $\pm$ |
| awrt $1.06$ (may be implied by awrt $0.144$ or $0.145$) or a correct standardisation with $\pm 1.96$ | A1 | Ignore incorrect inequality symbol and allow $=$ |
| Using probability route: awrt $0.144$ or $0.145$, or critical value $z = \pm 1.96$; Using critical region route: $CR < 32.8$ | A1 | |
| Correct statement – no context needed but do not allow contradicting non-contextual comments | dM1 | Dependent on M1 A1. Ignore any comparisons |
| Correct conclusion in context. Must include the word claim. Must include words applicants (oe) and pilots. No hypotheses then A0 | A1 | NB: Award M1 A1 for a correct contextual statement on its own |
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5 Applicants for a pilot training programme with a passenger airline are screened for colour blindness. Past records show that the proportion of applicants identified as colour blind is 0.045
\begin{enumerate}[label=(\alph*)]
\item Write down a suitable model for the distribution of the number of applicants identified as colour blind from a total of $n$ applicants.
\item State one assumption necessary for this distribution to be a suitable model of this situation.
\item Using a suitable approximation, find the probability that exactly 5 out of 120 applicants are identified as colour blind.
\item Explain why the approximation that you used in part (c) is appropriate.
Jaymini claims that 75\% of all applicants for this training programme go on to become pilots.
From a random sample of 96 applicants for this training programme 67 go on to become pilots.
\item Using a suitable approximation, test Jaymini's claim at the $5 \%$ level of significance. State your hypotheses clearly.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2022 Q5 [14]}}