| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: z-score calculation and probability lookup (part a), inverse normal to find standard deviation (part b), and applying independence to combine probabilities (part c). All parts use routine methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X<132) = P\!\left(Z < \dfrac{132-125.4}{18.6}\right) = P(Z < 0.3548)\) | M1 | Use of \(\pm\)standardisation formula with 132 and 125.4 substituted, condone continuity correction \(132\pm0.5\) and use of \(18.6^2\), \(\sqrt{18.6}\) |
| \(0.639\) | A1 | \(0.6385 < p \leqslant 0.639\). If M0 scored, SC B1 for \(0.6385 < p \leqslant 0.639\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{108-117}{\sigma} = -1.175\) | B1 | \(1.1749 < z \leqslant 1.175\) or \(-1.175 \leqslant z < -1.1749\) |
| M1 | 108 and 117 substituted in \(\pm\)standardisation formula, no continuity correction, not \(\sigma^2\), \(\sqrt{\sigma}\), equated to a \(z\)-value | |
| \(\sigma = 7.66\) | A1 | \(7.659 \leqslant \sigma \leqslant 7.66\). If M0 scored, SC B1 for \(7.659 \leqslant \sigma \leqslant 7.66\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(-1.5 < Z < 1.5)\) \([\Phi(1.5) - \Phi(-1.5)]\) \([= 2\Phi(1.5)-1]\) | M1 | Both 1.5 and \(-1.5\) seen as \(z\)-values or appropriate use of 1.5 or \(-1.5\), and no other \(z\)-values in part |
| \(= 2\times\textit{their}\ 0.9332 - 1\) or \(\textit{their}\ 0.9332-(1-\textit{their}\ 0.9332)\) or \(2\times(\textit{their}\ 0.9332-0.5)\) | M1 | Calculating the appropriate area from stated \(\Phi\) of \(z\)-values which must be \(\pm\) the same number. Condone *their* 0.0668 as \((1-\textit{their}\ 0.9332)\) |
| \(0.8664\) | A1 | Accept answers wrt 0.866. If A0 scored SC B1 for answers wrt 0.866 |
| \(0.8664^3 = 0.650[36\ldots]\) | B1 FT | FT *their* 4SF (or better) probability, accept final answers to 3SF |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X<132) = P\!\left(Z < \dfrac{132-125.4}{18.6}\right) = P(Z < 0.3548)$ | M1 | Use of $\pm$standardisation formula with 132 and 125.4 substituted, condone continuity correction $132\pm0.5$ and use of $18.6^2$, $\sqrt{18.6}$ |
| $0.639$ | A1 | $0.6385 < p \leqslant 0.639$. If M0 scored, **SC B1** for $0.6385 < p \leqslant 0.639$ |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{108-117}{\sigma} = -1.175$ | B1 | $1.1749 < z \leqslant 1.175$ or $-1.175 \leqslant z < -1.1749$ |
| | M1 | 108 and 117 substituted in $\pm$standardisation formula, no continuity correction, not $\sigma^2$, $\sqrt{\sigma}$, equated to a $z$-value |
| $\sigma = 7.66$ | A1 | $7.659 \leqslant \sigma \leqslant 7.66$. If M0 scored, **SC B1** for $7.659 \leqslant \sigma \leqslant 7.66$ |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(-1.5 < Z < 1.5)$ $[\Phi(1.5) - \Phi(-1.5)]$ $[= 2\Phi(1.5)-1]$ | M1 | Both 1.5 and $-1.5$ seen as $z$-values **or** appropriate use of 1.5 or $-1.5$, **and** no other $z$-values in part |
| $= 2\times\textit{their}\ 0.9332 - 1$ **or** $\textit{their}\ 0.9332-(1-\textit{their}\ 0.9332)$ **or** $2\times(\textit{their}\ 0.9332-0.5)$ | M1 | Calculating the appropriate area from stated $\Phi$ of $z$-values which must be $\pm$ the same number. Condone *their* 0.0668 as $(1-\textit{their}\ 0.9332)$ |
| $0.8664$ | A1 | Accept answers wrt 0.866. If A0 scored **SC B1** for answers wrt 0.866 |
| $0.8664^3 = 0.650[36\ldots]$ | B1 FT | FT *their* 4SF (or better) probability, accept final answers to 3SF |
---4 In a large population, the systolic blood pressure (SBP) of adults is normally distributed with mean 125.4 and standard deviation 18.6.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the SBP of a randomly chosen adult is less than 132.\\
The SBP of 12-year-old children in the same population is normally distributed with mean 117. Of these children 88\% have SBP more than 108.
\item Find the standard deviation of this distribution.\\
Three adults are chosen at random from this population.
\item Find the probability that each of these three adults has SBP within 1.5 standard deviations of the mean.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q4 [9]}}