| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution problem requiring students to find μ and σ from two given probabilities using inverse normal tables, then calculate a third probability. While it involves multiple steps and careful manipulation of z-scores, it follows a well-practiced routine with no conceptual surprises—slightly easier than average due to its predictable structure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Bell shaped curve with 4 values marked | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P\left(Z \leq \frac{66-\mu}{\sigma}\right) = 0.0359 \Rightarrow 66 - \mu = -1.80\sigma\) | B1 | -1.80 seen |
| Clear attempt including standardization either way | M1, A1 | or equivalent |
| \(81 - \mu = 1.20\sigma\) | B1A1 | 1.20 or equivalent |
| Subtracting: \(15 = 1.20\sigma + 1.80\sigma \Rightarrow \sigma = 5\) | given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mu = 66 + 1.8 \times 5 = 75\) | M1A1, B1 | Clear attempt to solve, cso; 75 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(69 \leq X \leq 83) = P\left(\frac{69-75}{5} \leq Z \leq \frac{83-75}{5}\right)\) | M1 | standardize both either way |
| \(= P(-1.20 \leq Z \leq 1.60)\) | A1 | -1.20, 1.60 seen |
| \(= 0.8301\) | A1 | 4 dp |
## Question 3:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Bell shaped curve with 4 values marked | B1 | |
### Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $P\left(Z \leq \frac{66-\mu}{\sigma}\right) = 0.0359 \Rightarrow 66 - \mu = -1.80\sigma$ | B1 | -1.80 seen |
| Clear attempt including standardization either way | M1, A1 | or equivalent |
| $81 - \mu = 1.20\sigma$ | B1A1 | 1.20 or equivalent |
| Subtracting: $15 = 1.20\sigma + 1.80\sigma \Rightarrow \sigma = 5$ | | **given answer** |
### Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mu = 66 + 1.8 \times 5 = 75$ | M1A1, B1 | Clear attempt to solve, cso; 75 |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(69 \leq X \leq 83) = P\left(\frac{69-75}{5} \leq Z \leq \frac{83-75}{5}\right)$ | M1 | standardize both either way |
| $= P(-1.20 \leq Z \leq 1.60)$ | A1 | -1.20, 1.60 seen |
| $= 0.8301$ | A1 | 4 dp |
---3. The random variable $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$.
It is known that
$$\mathrm { P } ( X \leq 66 ) = 0.0359 \text { and } \mathrm { P } ( X \geq 81 ) = 0.1151 .$$
\begin{enumerate}[label=(\alph*)]
\item In the space below, give a clearly labelled sketch to represent these probabilities on a Normal curve.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the value of $\sigma$ is 5 .
\item Find the value of $\mu$.
\end{enumerate}\item Find $\mathrm { P } ( 69 \leq X \leq 83 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q3 [12]}}