| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring routine standardisation (z-scores), inverse normal calculations, and basic conditional probability. Part (a) is direct z-score lookup, part (b) is inverse normal with percentiles, and part (c) uses symmetry of normal distribution with conditional probability—all textbook techniques with no novel insight required. Slightly above average difficulty due to the conditional probability element in part (c), but still well within standard S1 scope. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X < 18) = P\!\left(Z < \pm\left(\frac{18-15}{2}\right)[= \pm 1.5]\right)\) | M1 | For standardising correctly; may be implied by \(\pm 1.5\) |
| \(= 0.9332\) awrt 0.933 | A1 | awrt 0.933 (do not ISW) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x-15}{2} = 0.2533\) | M1 | For correct standardisation \(=\) a \(z\) value such that \(0.25 < |
| B1 | For use of awrt \(\pm 0.2533\) | |
| \(x = 15.506\ldots\) awrt 15.5 | A1 | awrt 15.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(T > \mu - 10) = 0.975\) | M1 | For correct probability of 0.975; may be seen as denominator of fraction; may be implied by \( |
| \(\frac{(\mu \pm 10) - \mu}{\sigma} = \pm 1.96 \Rightarrow \sigma = \frac{10}{1.96} [= 5.10\ldots]\) | M1 | For \(\frac{\mu+10-\mu}{\sigma} = 1.96\) or \(\frac{\mu-10-\mu}{\sigma} = -1.96\) leading to value for \(\sigma\); may be implied by \(\pm 0.98\) |
| \(P(T > \mu - 5) = P\!\left(Z > \frac{\mu - 5 - \mu}{\text{"5.10..."}}\right)[= -0.98]\big)[= 0.836\ldots]\) | M1 | For correct method to find \(P(T > \mu-5)\) using their \(\sigma\); may be implied by \(-0.98\); if \(P(T < \mu+5)\) calculated may be implied by 0.98 |
| \(P(T > \mu-5 \mid T > \mu-10) = \frac{\text{"0.836..."}}{\text{"0.975"}}\) | M1 | For \(\frac{p}{0.975}\) where \(0.5 < p < 0.975\) (must be probability not \(z\) value); if denominator incorrect only follow through their \(P(T>\mu-10)\) if clearly labelled and \(> 0.95\) |
| \(= 0.8579\ldots\) awrt 0.858 | A1 | awrt 0.858 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X < 18) = P\!\left(Z < \pm\left(\frac{18-15}{2}\right)[= \pm 1.5]\right)$ | M1 | For standardising correctly; may be implied by $\pm 1.5$ |
| $= 0.9332$ awrt 0.933 | A1 | awrt 0.933 (do not ISW) |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x-15}{2} = 0.2533$ | M1 | For correct standardisation $=$ a $z$ value such that $0.25 < |z| < 0.26$ |
| | B1 | For use of awrt $\pm 0.2533$ |
| $x = 15.506\ldots$ awrt 15.5 | A1 | awrt 15.5 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(T > \mu - 10) = 0.975$ | M1 | For correct probability of 0.975; may be seen as denominator of fraction; may be implied by $|z| = 1.96$ or better |
| $\frac{(\mu \pm 10) - \mu}{\sigma} = \pm 1.96 \Rightarrow \sigma = \frac{10}{1.96} [= 5.10\ldots]$ | M1 | For $\frac{\mu+10-\mu}{\sigma} = 1.96$ or $\frac{\mu-10-\mu}{\sigma} = -1.96$ leading to value for $\sigma$; may be implied by $\pm 0.98$ |
| $P(T > \mu - 5) = P\!\left(Z > \frac{\mu - 5 - \mu}{\text{"5.10..."}}\right)[= -0.98]\big)[= 0.836\ldots]$ | M1 | For correct method to find $P(T > \mu-5)$ using their $\sigma$; may be implied by $-0.98$; if $P(T < \mu+5)$ calculated may be implied by 0.98 |
| $P(T > \mu-5 \mid T > \mu-10) = \frac{\text{"0.836..."}}{\text{"0.975"}}$ | M1 | For $\frac{p}{0.975}$ where $0.5 < p < 0.975$ (must be probability not $z$ value); if denominator incorrect only follow through their $P(T>\mu-10)$ if clearly labelled and $> 0.95$ |
| $= 0.8579\ldots$ awrt 0.858 | A1 | awrt 0.858 |
---\begin{enumerate}
\item A competition consists of two rounds.
\end{enumerate}
The time, in minutes, taken by adults to complete round one is modelled by a normal distribution with mean 15 minutes and standard deviation 2 minutes.\\
(a) Use standardisation to find the proportion of adults that take less than 18 minutes to complete round one.
Only the fastest $60 \%$ of adults from round one take part in round two.\\
(b) Use standardisation to find the longest time that an adult can take to complete round one if they are to take part in round two.
The time, $T$ minutes, taken by adults to complete round two is modelled by a normal distribution with mean $\mu$
Given that $\mathrm { P } ( \mu - 10 < T < \mu + 10 ) = 0.95$\\
(c) find $\mathrm { P } ( T > \mu - 5 \mid T > \mu - 10 )$
\hfill \mbox{\textit{Edexcel S1 2024 Q5 [10]}}