Edexcel S2 2022 January — Question 5 14 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2022
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Binomial to the Poisson distribution
TypeState exact binomial distribution
DifficultyModerate -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.
Spec2.04b 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

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
  1. Write down a suitable model for the distribution of the number of applicants identified as colour blind from a total of \(n\) applicants.
  2. State one assumption necessary for this distribution to be a suitable model of this situation.
  3. Using a suitable approximation, find the probability that exactly 5 out of 120 applicants are identified as colour blind.
  4. 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.
  5. Using a suitable approximation, test Jaymini's claim at the \(5 \%\) level of significance. State your hypotheses clearly.

Question 5:
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
\(B(n, 0.045)\)B1 For binomial with correct parameters \(n\) and 0.045
Part (b)
AnswerMarks Guidance
AnswerMark Guidance
Applicants are independent (no identical twins) or the proportion/probability identified as colour blind does not change over timeB1 For one of the given reasons. Must have context. Allow equivalent statements. Do not allow number for proportion/probability
Part (c)
AnswerMarks Guidance
AnswerMark 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 0.173A1 awrt 0.173. NB: A correct answer with no incorrect working scores 3/3
Part (d)
AnswerMarks Guidance
AnswerMark Guidance
Binomial with large \(n\)B1 \(n\) is large (Allow "number of trials" for \(n\))
and very small \(p\)B1
Part (e)
AnswerMarks Guidance
AnswerMark 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 0.145A1
There is insufficient evidence to reject \(H_0\)dM1
Insufficient evidence against Jaymini's claimA1
ALT (using probability of failing):
AnswerMarks Guidance
AnswerMark 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 0.145A1
There is insufficient evidence to reject \(H_0\)dM1
Insufficient evidence against Jaymini's claimA1
Question (e):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(p\) is smallB1 Allow probability for \(p\)
Both hypotheses correct in terms of \(p\) or \(\pi\), e.g. \(H_0: p = \) value, \(H_1: p \neq\) valueB1 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 deviationM1 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 commentsdM1 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 A0A1 NB: Award M1 A1 for a correct contextual statement on its own
ALT:
AnswerMarks Guidance
Answer/WorkingMark 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 deviationM1 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 commentsdM1 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 A0A1 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]}}