| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Discrete uniform distribution sample mean |
| Difficulty | Challenging +1.2 This is a standard Further Maths Statistics question applying the Central Limit Theorem to find probabilities and test power. While it requires understanding of CLT, normal approximation, and hypothesis testing concepts, the execution is straightforward: calculate mean/variance of uniform distribution, apply CLT with n=45, use normal tables for parts (a) and (c), and explain conceptually for parts (b) and (d). The multi-part structure and Further Maths content place it above average difficulty, but it follows a predictable template without requiring novel insight. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Realising \(S\) has a discrete uniform distribution over \(\{1,\ldots,6\}\) | M1 | Setting up model for \(S\) |
| \(\text{E}(S) = 1\times\frac{1}{6}+2\times\frac{1}{6}+3\times\frac{1}{6}+4\times\frac{1}{6}+5\times\frac{1}{6}+6\times\frac{1}{6}\) | M1 | Attempt at expression for \(\text{E}(S)\) |
| \(\text{Var}(S) = \frac{6^2-1}{12}\) or \(1^2\times\frac{1}{6}+\ldots+6^2\times\frac{1}{6} - 3.5^2\) | M1 | Attempt at expression for \(\text{Var}(S)\) |
| \(\text{E}(S) = 3.5\) and \(\text{Var}(S) = \frac{35}{12}\) | A1 | Correct mean and variance |
| \(\bar{S} \sim N(3.5, \ldots)\) | M1 | Use of CLT to find distribution for \(\bar{S}\) |
| \(\text{Var}(\bar{S}) = \frac{35/12}{45} = \frac{7}{108}\), \(\bar{S} \sim N(3.5, 0.0648\ldots)\) | A1 | Correct distribution with correct variance; allow \(\sigma^2=\) awrt 0.0648 |
| \(P(\bar{S} < k) = 0.05 \Rightarrow \frac{k-3.5}{\sqrt{7/108}} = -1.6449\) | M1 | Standardising using model and equating to \(z\)-value \(1< |
| \(k = 3.08122\ldots\) awrt 3.08 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CLT applies since the sample size is large | B1 | Correct explanation about appropriateness of CLT given large sample size (allow \(>30\)) |
| CLT states that the sample mean \(\bar{S}\) is approximately normally distributed | B1 | Requires both sample and mean or \(\bar{S}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| True \(\bar{S} \sim N(4, \frac{3}{45})\) | M1 | Writing/using \(\bar{S} \sim N(4, \frac{3}{45})\); allow \(\sigma^2=\) awrt 0.0667 |
| \(P(\bar{S}<3.1)+P(\bar{S}>3.9)\) or \(1-P(3.1<\bar{S}<3.9)\) | dM1 | Dep on 1st M1; correct probability statement for power |
| Power \(=\) awrt 0.651 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Increase in sample size would decrease variance of \(\bar{S}\) [leading to increase in \(P(\bar{S}>3.9)\) and decrease in \(P(\bar{S}<3.1)\) would be negligible] | B1 | Correct reasoning referring to decrease in variance |
| So the power would increase | dB1 | Dep on 1st B1; correct deduction with no incorrect reasoning |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Realising $S$ has a discrete uniform distribution over $\{1,\ldots,6\}$ | M1 | Setting up model for $S$ |
| $\text{E}(S) = 1\times\frac{1}{6}+2\times\frac{1}{6}+3\times\frac{1}{6}+4\times\frac{1}{6}+5\times\frac{1}{6}+6\times\frac{1}{6}$ | M1 | Attempt at expression for $\text{E}(S)$ |
| $\text{Var}(S) = \frac{6^2-1}{12}$ or $1^2\times\frac{1}{6}+\ldots+6^2\times\frac{1}{6} - 3.5^2$ | M1 | Attempt at expression for $\text{Var}(S)$ |
| $\text{E}(S) = 3.5$ and $\text{Var}(S) = \frac{35}{12}$ | A1 | Correct mean and variance |
| $\bar{S} \sim N(3.5, \ldots)$ | M1 | Use of CLT to find distribution for $\bar{S}$ |
| $\text{Var}(\bar{S}) = \frac{35/12}{45} = \frac{7}{108}$, $\bar{S} \sim N(3.5, 0.0648\ldots)$ | A1 | Correct distribution with correct variance; allow $\sigma^2=$ awrt 0.0648 |
| $P(\bar{S} < k) = 0.05 \Rightarrow \frac{k-3.5}{\sqrt{7/108}} = -1.6449$ | M1 | Standardising using model and equating to $z$-value $1<|z|<2$ |
| $k = 3.08122\ldots$ awrt **3.08** | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CLT applies since the sample size is large | B1 | Correct explanation about appropriateness of CLT given large sample size (allow $>30$) |
| CLT states that the sample mean $\bar{S}$ is approximately normally distributed | B1 | Requires both sample and mean or $\bar{S}$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| True $\bar{S} \sim N(4, \frac{3}{45})$ | M1 | Writing/using $\bar{S} \sim N(4, \frac{3}{45})$; allow $\sigma^2=$ awrt 0.0667 |
| $P(\bar{S}<3.1)+P(\bar{S}>3.9)$ or $1-P(3.1<\bar{S}<3.9)$ | dM1 | Dep on 1st M1; correct probability statement for power |
| Power $=$ awrt **0.651** | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Increase in sample size would decrease variance of $\bar{S}$ [leading to increase in $P(\bar{S}>3.9)$ and decrease in $P(\bar{S}<3.1)$ would be negligible] | B1 | Correct reasoning referring to decrease in variance |
| So the power would increase | dB1 | Dep on 1st B1; correct deduction with no incorrect reasoning |\begin{enumerate}
\item A six-sided die has sides labelled $1,2,3,4,5$ and 6
\end{enumerate}
The random variable $S$ represents the score when the die is rolled.\\
Alicia rolls the die 45 times and the mean score, $\bar { S }$, is calculated.\\
Assuming the die is fair and using a suitable approximation,\\
(a) find, to 3 significant figures, the value of $k$ such that $\mathrm { P } ( \bar { S } < k ) = 0.05$\\
(b) Explain the relevance of the Central Limit Theorem in part (a).
Alicia considers the following hypotheses:\\
$\mathrm { H } _ { 0 }$ : The die is fair\\
$\mathrm { H } _ { 1 }$ : The die is not fair\\
If $\bar { S } < 3.1$ or $\bar { S } > 3.9$, then $\mathrm { H } _ { 0 }$ will be rejected.\\
Given that the true distribution of $S$ has mean 4 and variance 3\\
(c) find the power of this test.\\
(d) Describe what would happen to the power of this test if Alicia were to increase the number of rolls of the die.\\
Give a reason for your answer.
\hfill \mbox{\textit{Edexcel FS1 2020 Q7 [15]}}