| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Challenging +1.8 This question requires understanding of uniform distribution properties (mean and variance), applying CLT, then working backwards from a confidence interval bound to find sample size. It combines multiple statistical concepts and requires algebraic manipulation of the confidence interval formula, going beyond routine application. The uniform distribution context and reverse-engineering n from the CI bound elevates this above standard S3 questions. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(D) = x+2\) | M1 | \(E(D)\) correct |
| \(\text{Var}(D) = \frac{((x+5)-(x-1))^2}{12} [=3]\) | M1 | Correct method to find \(\text{Var}(D)\); must be subtracting the correct way round but condone missing brackets |
| \(\bar{D} \sim N\!\left(x+2,\ \frac{3}{n}\right)\) | A1 | For a fully correct distribution; either states \(N\!\left(x+2,\frac{3}{n}\right)\) or accept e.g. "normal" with mean \(= x+2\) and variance \(= \frac{3}{n}\) oe; must be seen in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{"x+2"} = 22.101 + \text{"2"}\ (= 24.101)\) or \(\text{"x+2"} = 24.6 \Rightarrow 24.6 - \text{"2"}\ (= 22.6)\) | M1 | For a correct method to find \(d\) using \(x\) as 22.101 in their "\(x+2\)" from (a) or a correct method to find \(x\) by rearranging their "\(x+2\)" to \(x = 24.6 - \text{"2"}\); implied by 24.101 or 22.6 or \(\pm 0.499\) oe; \(x+2 = 24.6\) on its own is M0 |
| \(24.6 - \text{"2.5758"}\sqrt{\frac{\text{"3"}}{n}} = \text{"24.101"}\) oe | B1M1 | B1 for awrt \(\pm 2.5758\); may be implied by an unrounded value of \(n\) of awrt 79.94. M1 for \(24.6 \pm z\sqrt{\frac{\sigma^2}{n}}\) or \(\text{"22.6"}\pm z\sqrt{\frac{\sigma^2}{n}}\) where \(2.55 < |
| \(n = 80\) | A1cao | Dependent on both the previous M marks; for setting up a valid equation (allow 22.101 instead of "24.101" provided it is correctly paired with "22.6"); their numerical variance must be substituted in for this mark; A1 cao dependent on seeing a correct equation but allow use of \(z = 2.576\); M1B0M1dM1A1 is possible; note awrt 79.94 seen can imply B1 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(D) = x+2$ | M1 | $E(D)$ correct |
| $\text{Var}(D) = \frac{((x+5)-(x-1))^2}{12} [=3]$ | M1 | Correct method to find $\text{Var}(D)$; must be subtracting the correct way round but condone missing brackets |
| $\bar{D} \sim N\!\left(x+2,\ \frac{3}{n}\right)$ | A1 | For a fully correct distribution; either states $N\!\left(x+2,\frac{3}{n}\right)$ or accept e.g. "normal" with mean $= x+2$ and variance $= \frac{3}{n}$ oe; must be seen in (a) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{"x+2"} = 22.101 + \text{"2"}\ (= 24.101)$ or $\text{"x+2"} = 24.6 \Rightarrow 24.6 - \text{"2"}\ (= 22.6)$ | M1 | For a correct method to find $d$ using $x$ as 22.101 in their "$x+2$" from (a) or a correct method to find $x$ by rearranging their "$x+2$" to $x = 24.6 - \text{"2"}$; implied by 24.101 or 22.6 or $\pm 0.499$ oe; $x+2 = 24.6$ on its own is M0 |
| $24.6 - \text{"2.5758"}\sqrt{\frac{\text{"3"}}{n}} = \text{"24.101"}$ oe | B1M1 | B1 for awrt $\pm 2.5758$; may be implied by an unrounded value of $n$ of awrt 79.94. M1 for $24.6 \pm z\sqrt{\frac{\sigma^2}{n}}$ or $\text{"22.6"}\pm z\sqrt{\frac{\sigma^2}{n}}$ where $2.55 < |z| < 2.6$ (ft their mean and variance from (a) or may restart); may be part of an equation; **their numerical variance does not have to be substituted in for this mark** |
| $n = 80$ | A1cao | Dependent on both the previous M marks; for setting up a valid equation (allow 22.101 instead of "24.101" provided it is correctly paired with "22.6"); **their numerical variance must be substituted in for this mark**; A1 cao dependent on seeing a correct equation but allow use of $z = 2.576$; M1B0M1dM1A1 is possible; note awrt 79.94 seen can imply B1 |\begin{enumerate}
\item The continuous random variable $D$ is uniformly distributed over the interval $[ x - 1 , x + 5 ]$ where $x$ is a constant.
\end{enumerate}
A random sample of $n$ observations of $D$ is taken, where $n$ is large.\\
(a) Use the Central Limit Theorem to find an approximate distribution for $\bar { D }$ Give your answer in terms of $n$ and $x$ where appropriate.
The $n$ observations of $D$ have a sample mean of 24.6\\
Given that the lower bound of the $99 \%$ confidence interval for $x$ is 22.101 to 3 decimal places,\\
(b) find the value of $n$
Show your working clearly.
\hfill \mbox{\textit{Edexcel S3 2024 Q7 [8]}}