| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Custom discrete distribution sample mean |
| Difficulty | Standard +0.8 This is a standard Central Limit Theorem application requiring calculation of population mean and variance from a discrete distribution, then using normal approximation for the sample mean. It's methodical but involves multiple computational steps (E(X), Var(X), standardization) and is a Further Maths topic, placing it moderately above average difficulty. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
| Number on spinner | 1 | 2 | 3 | 4 | 5 |
| Probability | 0.3 | 0.1 | 0.2 | 0.1 | 0.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu = 3\) | B1 | For stating or using mean \(= 3\) |
| \(E(X^2) = 0.3 + 4\times0.1 + 9\times0.2 + 16\times0.1 + 25\times0.3 = 11.6\) or \(\frac{58}{5}\) | M1 | For using the given model to attempt \(E(X^2)\) with at least 3 correct products seen |
| \(\sigma^2 = 11.6 - 3^2 = \mathbf{2.6}\) | A1 | For \(\text{Var}(X) = 2.6\) or \(\sigma = \sqrt{2.6} = 1.6124\ldots\) (awrt \(1.61\)) |
| \(\bar{X} \approx N\!\left(3,\ \dfrac{2.6}{80}\right)\) | M1 | For use of CLT – must use \(\bar{X}\) and normal, or sight of \(N\!\left(3, \sqrt{\dfrac{2.6}{80}}\right)\) with any letter |
| A1ft | For a correct mean and variance, ft their 3 and their 2.6 | |
| \(P(\bar{X} > 3.25) = P\!\left(Z > 1.3867\ldots\right) = 0.0827589\ldots\) awrt \(\mathbf{0.0828}\) | A1 | For using the normal model to find probability awrt \(0.0828\) |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu = 3$ | B1 | For stating or using mean $= 3$ |
| $E(X^2) = 0.3 + 4\times0.1 + 9\times0.2 + 16\times0.1 + 25\times0.3 = 11.6$ or $\frac{58}{5}$ | M1 | For using the given model to attempt $E(X^2)$ with at least 3 correct products seen |
| $\sigma^2 = 11.6 - 3^2 = \mathbf{2.6}$ | A1 | For $\text{Var}(X) = 2.6$ or $\sigma = \sqrt{2.6} = 1.6124\ldots$ (awrt $1.61$) |
| $\bar{X} \approx N\!\left(3,\ \dfrac{2.6}{80}\right)$ | M1 | For use of CLT – must use $\bar{X}$ and normal, or sight of $N\!\left(3, \sqrt{\dfrac{2.6}{80}}\right)$ with any letter |
| | A1ft | For a correct mean and variance, ft their 3 and their 2.6 |
| $P(\bar{X} > 3.25) = P\!\left(Z > 1.3867\ldots\right) = 0.0827589\ldots$ awrt $\mathbf{0.0828}$ | A1 | For using the normal model to find probability awrt $0.0828$ |\begin{enumerate}
\item A biased spinner can land on the numbers $1,2,3,4$ or 5 with the following probabilities.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Number on spinner & 1 & 2 & 3 & 4 & 5 \\
\hline
Probability & 0.3 & 0.1 & 0.2 & 0.1 & 0.3 \\
\hline
\end{tabular}
\end{center}
The spinner will be spun 80 times and the mean of the numbers it lands on will be calculated. Find an estimate of the probability that this mean will be greater than 3.25\\
(6)
\hfill \mbox{\textit{Edexcel FS1 2019 Q3 [6]}}