| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard techniques: (a) basic z-score calculation and table lookup, (b) inverse normal for a percentile, and (c) conditional probability with symmetry. All parts are routine textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(T \sim N(240, 40^2)\); \(\text{P}\!\left(Z > \frac{300-240}{40}\right)\) | M1 | Standardising with 300, 240 and 40; allow \(\pm\) |
| \(= 1 - \text{P}(Z < 1.5)\) or \(1 - 0.9332\) | M1 | Correct method for \(\text{P}(Z > \text{"1.5"})\) |
| \(= \text{awrt } \mathbf{0.0668}\) or 6.68% | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{P}(T < n) = 0.20 \Rightarrow \frac{n - 240}{40} = -0.8416\) | M1 B1 | M1 for standardising with 240, 40, \(n\) and setting \(= \pm z\) where \(0.8 < |
| \(n = \text{awrt } \mathbf{206}\) minutes | A1 | For awrt 206 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{P}(W < \mu - 30 \mid W < \mu) = \frac{\text{P}(W < \mu - 30)}{\text{P}(W < \mu)}\) | M1 | Correct ratio expression (not \(\text{P}([W < 30 - \mu] \cap [W < \mu])\) on numerator) |
| \(= \frac{1 - 0.82}{0.50}\) | A1 | Correct numerical ratio |
| \(= \mathbf{0.36}\) | A1cao | For 0.36 or exact equivalent e.g. \(\frac{9}{25}\); must come from exact values (0.36 rounded from 0.3576 is A0) |
# Question 6:
## Part (a)
| $T \sim N(240, 40^2)$; $\text{P}\!\left(Z > \frac{300-240}{40}\right)$ | M1 | Standardising with 300, 240 and 40; allow $\pm$ |
| $= 1 - \text{P}(Z < 1.5)$ or $1 - 0.9332$ | M1 | Correct method for $\text{P}(Z > \text{"1.5"})$ |
| $= \text{awrt } \mathbf{0.0668}$ or 6.68% | A1 | |
## Part (b)
| $\text{P}(T < n) = 0.20 \Rightarrow \frac{n - 240}{40} = -0.8416$ | M1 B1 | M1 for standardising with 240, 40, $n$ and setting $= \pm z$ where $0.8 < |z| < 0.9$; B1 for $z = \pm 0.8416$ used as z-value |
| $n = \text{awrt } \mathbf{206}$ minutes | A1 | For awrt 206 |
## Part (c)
| $\text{P}(W < \mu - 30 \mid W < \mu) = \frac{\text{P}(W < \mu - 30)}{\text{P}(W < \mu)}$ | M1 | Correct ratio expression (not $\text{P}([W < 30 - \mu] \cap [W < \mu])$ on numerator) |
| $= \frac{1 - 0.82}{0.50}$ | A1 | Correct numerical ratio |
| $= \mathbf{0.36}$ | A1cao | For 0.36 or exact equivalent e.g. $\frac{9}{25}$; must come from exact values (0.36 rounded from 0.3576 is A0) |
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Could you please share the correct page(s) that contain the mark scheme content? This appears to be either a blank page or the content failed to load properly in the image.6. The time, in minutes, taken by men to run a marathon is modelled by a normal distribution with mean 240 minutes and standard deviation 40 minutes.
\begin{enumerate}[label=(\alph*)]
\item Find the proportion of men that take longer than 300 minutes to run a marathon.\\
(3)
Nathaniel is preparing to run a marathon. He aims to finish in the first 20\% of male runners.
\item Using the above model estimate the longest time that Nathaniel can take to run the marathon and achieve his aim.\\
(3)
The time, $W$ minutes, taken by women to run a marathon is modelled by a normal distribution with mean $\mu$ minutes.
Given that $\mathrm { P } ( W < \mu + 30 ) = 0.82$
\item find $\mathrm { P } ( W < \mu - 30 \mid W < \mu )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q6 [9]}}