| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find standard deviation from probability |
| Difficulty | Standard +0.3 This is a straightforward inverse normal distribution problem requiring students to use tables/calculator to find the z-score for 20% in the upper tail (z ≈ 0.84), then solve 125 = 90 + 0.84σ for σ. Part (b) is routine application, and part (c) tests conceptual understanding of model validity. Slightly above average due to the inverse normal step, but still a standard S1 question with clear structure and familiar techniques. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.04g Normal distribution properties: empirical rule (68-95-99.7), points of inflection |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(L\) represent length of visit; \(L\sim N(90,\sigma^2)\) | ||
| \(P(L<125)=0.80\) or \(P(L>125)=0.20\) | ||
| \(P\!\left(Z<\frac{125-90}{\sigma}\right)=0.8\) | M1 | Standardising; \(\pm(125-90)\), \(\sigma/\sigma^2/\sqrt{\phantom{x}}/\sigma\) |
| \(\frac{125-90}{\sigma}=0.8416\) | B1 | \(0.8416\) |
| \(\frac{\pm(125-90)}{\sigma}=z\) value | M1 | |
| \(\sigma=\frac{35}{0.8416}=41.587\ldots\) | A1 | AWRT \(41.6\) (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(L<25)=P\!\left(Z<\frac{25-90}{41.587\ldots}\right)\) | M1 | Standardising \(25\), \(90\), their \(+\)ve \(41.587\) |
| \(=P(Z<-1.56)\) | ||
| \(=1-P(Z<1.56)\) | M1 | For use of symmetry or \(\Phi(-z)=1-\Phi(z)\); \(p<0.5\) |
| \(=0.0594\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(90+3\sigma=215\Rightarrow 6.25\text{ pm}\) for latest arrival | B1 | Based on \(2\sigma/3\sigma\) rule |
| \(90+2\sigma=173.\dot{3}\Rightarrow 7.07\text{ pm}\) for latest arrival | B1 | |
| \(\therefore\) This normal distribution is not suitable. | (2 marks) |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $L$ represent length of visit; $L\sim N(90,\sigma^2)$ | | |
| $P(L<125)=0.80$ or $P(L>125)=0.20$ | | |
| $P\!\left(Z<\frac{125-90}{\sigma}\right)=0.8$ | M1 | Standardising; $\pm(125-90)$, $\sigma/\sigma^2/\sqrt{\phantom{x}}/\sigma$ |
| $\frac{125-90}{\sigma}=0.8416$ | B1 | $0.8416$ |
| $\frac{\pm(125-90)}{\sigma}=z$ value | M1 | |
| $\sigma=\frac{35}{0.8416}=41.587\ldots$ | A1 | AWRT $41.6$ **(4 marks)** |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(L<25)=P\!\left(Z<\frac{25-90}{41.587\ldots}\right)$ | M1 | Standardising $25$, $90$, their $+$ve $41.587$ |
| $=P(Z<-1.56)$ | | |
| $=1-P(Z<1.56)$ | M1 | For use of symmetry or $\Phi(-z)=1-\Phi(z)$; $p<0.5$ |
| $=0.0594$ | A1 | **(3 marks)** |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $90+3\sigma=215\Rightarrow 6.25\text{ pm}$ for latest arrival | B1 | Based on $2\sigma/3\sigma$ rule |
| $90+2\sigma=173.\dot{3}\Rightarrow 7.07\text{ pm}$ for latest arrival | B1 | |
| $\therefore$ This normal distribution is **not** suitable. | | **(2 marks)** |
**(9 marks total)**5. A health club lets members use, on each visit, its facilities for as long as they wish. The club's records suggest that the length of a visit can be modelled by a normal distribution with mean 90 minutes. Only $20 \%$ of members stay for more than 125 minutes.
\begin{enumerate}[label=(\alph*)]
\item Find the standard deviation of the normal distribution.
\item Find the probability that a visit lasts less than 25 minutes.
The club introduce a closing time of 10:00 pm. Tara arrives at the club at 8:00 pm.
\item Explain whether or not this normal distribution is still a suitable model for the length of her visit.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q5 [9]}}