| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Symmetric probability given |
| Difficulty | Moderate -0.8 This is a straightforward inverse normal distribution problem requiring standard table/calculator use to find μ from a given probability, followed by a trivial symmetry observation. Part (a) is routine S1 material with no conceptual challenges, and part (b) tests basic understanding of the normal distribution's symmetry property. Easier than average A-level questions. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{23 - \mu}{5} = "1.40"\) (o.e) giving \(\mu = 16\) (or awrt 16.0) | B1, M1A1ft, A1 | For awrt ± 1.40 or better seen anywhere. Condone 1.4 instead of 1.40. For attempting to standardise with 23 and 5 and μ, accept ±. e.g. \(\frac{23 - \mu}{25} = 1.40\) can score B1M0 (since using 25 not 5 for standardising); \(\frac{23 - \mu}{5} = 0.9192\) can score B0M1 (since have correct standardisation). Can accept equivalent equations e.g. \(23 - \mu = 5 \times "1.40"\). 1st A1ft for standardised expression = to a z value ( |
| Answer | Marks | Guidance |
|---|---|---|
| 0.4192 | B1 | For 0.4192 (but accept 3sf accuracy if 0.9192 − 0.5 is seen) |
## (a)
$\frac{23 - \mu}{5} = "1.40"$ (o.e) giving $\mu = 16$ (or awrt 16.0) | B1, M1A1ft, A1 | For awrt ± 1.40 or better seen anywhere. Condone 1.4 instead of 1.40. For attempting to standardise with 23 and 5 and μ, accept ±. e.g. $\frac{23 - \mu}{25} = 1.40$ can score B1M0 (since using 25 not 5 for standardising); $\frac{23 - \mu}{5} = 0.9192$ can score B0M1 (since have correct standardisation). Can accept equivalent equations e.g. $23 - \mu = 5 \times "1.40"$. **1st A1ft** for standardised expression = to a z value (|z| > 1). Signs must be compatible. Follow through their z. e.g. $\frac{23 - \mu}{5} = $ their z where z > 1 or $\frac{\mu - 23}{5} = $ their z where z < −1. **2nd A1** for 16 or awrt 16.0 if they are using a more accurate z. **Correct answer only scores 4/4 but if any working is seen apply scheme**
## (b)
0.4192 | B1 | For 0.4192 (but accept 3sf accuracy if 0.9192 − 0.5 is seen)
---The random variable $X \sim \text{N}(\mu, 5^2)$ and $\text{P}(X < 23) = 0.9192$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\mu$. [4]
\item Write down the value of $\text{P}(\mu < X < 23)$. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2011 Q2 [5]}}