| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 This is a standard S2 question testing routine application of binomial distribution and normal approximation. Part (a) requires direct calculator use for binomial probabilities, part (b)(i) is textbook normal approximation with continuity correction, and part (b)(ii) tests recall of np>5 and nq>5 conditions. All steps are procedural with no problem-solving or novel insight required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(P(X = 26) = 0.9686 - 0.9427 = \text{awrt } 0.0259\) or \(^{50}C_{26}(0.4)^{26}(0.6)^{24}\) | M1 A1 | Allow alternative notations for \(^{50}C_{26}\). A1 awrt 0.0259 (correct answer scores 2 out of 2) |
| (a)(ii) \(P(X \geqslant 26) = 1 - P(X \leqslant 25) = 1 - 0.9427 = \text{awrt } 0.0573\) | M1 A1 | M1 writing or using \(1 - P(X \leqslant 25)\); A1 awrt 0.0573 (calc 0.0573437...) (correct answer scores 2 out of 2) |
| (a)(iii) \(k = 19\) from tables | B1 | 19 cao. \(k < 19\) or \(k \geqslant 19\) is B0 |
| (b)(i) \(J \sim N(240, 144)\) | M1A1 | 1st M1 For writing or using \(N(240, \ldots)\) (May be seen in standardisation). 1st A1 For writing or using \(N(240, 144)\) (May be seen in standardisation) |
| \(P(X \leqslant 222) \sim P(J < 222.5) = P\left(Z < \frac{222.5 - 240}{\sqrt{144}}\right)\) | M1M1 | 2nd M1 use of continuity correction \(222 \pm 0.5\) |
| \(P(Z < -1.46) = 1 - 0.9279 = \text{awrt } 0.0721 - 0.0724\) | A1 | 2nd A1 awrt 0.0721 through to awrt 0.0724 (calc 0.0723743...). Answer in the range implies all previous marks unless clearly comes from wrong method. [NB: Use of binomial distribution gives 0.0719] |
| (b)(ii) \(n\) is large (oe) and \(p\) is close to 0.5 | B1 | Both conditions required. For \(n\) is large allow in words e.g. 'sample is large'. Allow 0.4 in place of \(p\). Condone '\(n > 30\)' (or any number \(> 30\)). Ignore comments about \(np\) |
**(a)(i)** $P(X = 26) = 0.9686 - 0.9427 = \text{awrt } 0.0259$ or $^{50}C_{26}(0.4)^{26}(0.6)^{24}$ | M1 A1 | Allow alternative notations for $^{50}C_{26}$. A1 awrt 0.0259 (correct answer scores 2 out of 2)
**(a)(ii)** $P(X \geqslant 26) = 1 - P(X \leqslant 25) = 1 - 0.9427 = \text{awrt } 0.0573$ | M1 A1 | M1 writing or using $1 - P(X \leqslant 25)$; A1 awrt 0.0573 (calc 0.0573437...) (correct answer scores 2 out of 2)
**(a)(iii)** $k = 19$ from tables | B1 | 19 cao. $k < 19$ or $k \geqslant 19$ is B0
**(b)(i)** $J \sim N(240, 144)$ | M1A1 | 1st M1 For writing or using $N(240, \ldots)$ (May be seen in standardisation). 1st A1 For writing or using $N(240, 144)$ (May be seen in standardisation)
$P(X \leqslant 222) \sim P(J < 222.5) = P\left(Z < \frac{222.5 - 240}{\sqrt{144}}\right)$ | M1M1 | 2nd M1 use of continuity correction $222 \pm 0.5$
$P(Z < -1.46) = 1 - 0.9279 = \text{awrt } 0.0721 - 0.0724$ | A1 | 2nd A1 awrt 0.0721 through to awrt 0.0724 (calc 0.0723743...). Answer in the range implies all previous marks unless clearly comes from wrong method. [NB: Use of binomial distribution gives 0.0719]
**(b)(ii)** $n$ is large (oe) and $p$ is close to 0.5 | B1 | Both conditions required. For $n$ is large allow in words e.g. 'sample is large'. Allow 0.4 in place of $p$. Condone '$n > 30$' (or any number $> 30$). Ignore comments about $np$\begin{enumerate}
\item In a large population $40 \%$ of adults use online banking.
\end{enumerate}
A random sample of 50 adults is taken.\\
The random variable $X$ represents the number of adults in the sample that use online banking.\\
(a) Find\\
(i) $\mathrm { P } ( X = 26 )$\\
(ii) $\mathrm { P } ( X \geqslant 26 )$\\
(iii) the smallest value of $k$ such that $\mathrm { P } ( X \leqslant k ) > 0.4$
A random sample of 600 adults is taken.\\
(b) (i) Find, using a normal approximation, the probability that no more than 222 of these 600 adults use online banking.\\
(ii) Explain why a normal approximation is suitable in part (b)(i)
\hfill \mbox{\textit{Edexcel S2 2023 Q1 [11]}}