| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Multiple probability calculations only |
| Difficulty | Standard +0.3 This is a straightforward S1 normal distribution question requiring standard z-score calculations and understanding of independent events. Part (a) and (b) are routine standardization exercises, while part (c) simply applies the independence rule P(A∩B∩C) = P(A)³. No novel insight or complex multi-step reasoning required—slightly easier than average due to its mechanical nature. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(Z < \frac{28-25}{\sqrt{16}}) = P(Z < 0.75) = 0.7734\) | M2 A1 | |
| (b) \(P(-5 < T - 25 < 5) = P(\frac{20-25}{4} < Z < \frac{30-25}{4})\) | M2 | |
| \(= P(-1.25 < Z < 1.25) = 0.8944 - 0.1056 = 0.7888\) | M1 A1 | |
| (c) \(P(T < 23) = P(Z < 0.5) = 0.6915\) | M1 A1 | |
| \(P(\text{3bikes, each} < 23 \text{ mins}) = (0.6915)^3 = 0.3307\) | M1 A1 | (11 marks total) |
(a) $P(Z < \frac{28-25}{\sqrt{16}}) = P(Z < 0.75) = 0.7734$ | M2 A1 |
(b) $P(-5 < T - 25 < 5) = P(\frac{20-25}{4} < Z < \frac{30-25}{4})$ | M2 |
$= P(-1.25 < Z < 1.25) = 0.8944 - 0.1056 = 0.7888$ | M1 A1 |
(c) $P(T < 23) = P(Z < 0.5) = 0.6915$ | M1 A1 |
$P(\text{3bikes, each} < 23 \text{ mins}) = (0.6915)^3 = 0.3307$ | M1 A1 | (11 marks total)5. The time taken in minutes, $T$, for a mechanic to service a bicycle follows a normal distribution with a mean of 25 minutes and a variance of 16 minutes $^ { 2 }$.
Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( T < 28 )$,
\item $\quad \mathrm { P } ( | T - 25 | < 5 )$.
One afternoon the mechanic has 3 bicycles to service.
\item Find the probability that he will take less than 23 minutes on each of the three bicycles.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [11]}}