| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This is a straightforward S1 normal distribution question requiring standard table lookups and z-score calculations. All three parts are routine: (a) direct probability lookup, (b) difference of two probabilities, (c) inverse normal lookup. No problem-solving or conceptual insight needed, just mechanical application of the standard normal transformation formula. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(Z < \frac{45-42}{\sqrt{18}}) = P(Z < 0.71) = 0.7611\) | M2 A1 | |
| (b) \(P(\frac{32-42}{\sqrt{18}} < Z < \frac{38-42}{\sqrt{18}}) = P(-2.36 < Z < -0.94) = P(Z < -0.94) - P(Z < -2.36) = 0.1736 - 0.0091 = 0.1645\) | M2 M1 A1 | |
| (c) \(P(Z < \frac{x-42}{\sqrt{18}}) = 0.95\); \(\frac{x-42}{\sqrt{18}} = 1.6449\) and \(x = 42 + (1.6449 \times \sqrt{18}) = 49.0\) | M1 A1 | Total 11 marks |
**(a)** $P(Z < \frac{45-42}{\sqrt{18}}) = P(Z < 0.71) = 0.7611$ | M2 A1 |
**(b)** $P(\frac{32-42}{\sqrt{18}} < Z < \frac{38-42}{\sqrt{18}}) = P(-2.36 < Z < -0.94) = P(Z < -0.94) - P(Z < -2.36) = 0.1736 - 0.0091 = 0.1645$ | M2 M1 A1 |
**(c)** $P(Z < \frac{x-42}{\sqrt{18}}) = 0.95$; $\frac{x-42}{\sqrt{18}} = 1.6449$ and $x = 42 + (1.6449 \times \sqrt{18}) = 49.0$ | M1 A1 | Total 11 marks
---3. The random variable $X$ is normally distributed with a mean of 42 and a variance of 18 .
Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( X \leq 45 )$,
\item $\mathrm { P } ( 32 \leq X \leq 38 )$,
\item the value of $x$ such that $\mathrm { P } ( X \leq x ) = 0.95$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q3 [11]}}