| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Independent probability calculations |
| Difficulty | Moderate -0.8 This question tests basic normal distribution calculations with straightforward z-score computations and standard probability lookups. Part (a) requires a single z-score calculation and multiplication by sample size; part (b) is even more routine as it directly asks about probabilities within a specified number of standard deviations from the mean. Both parts are standard textbook exercises requiring only recall and direct application of formulas, with no problem-solving or conceptual insight needed. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X<54.8)] = P\left(Z < \frac{54.8-55.6}{1.2}\right)\) | M1 | Use of \(\pm\) standardisation formula with 54.8, 55.6 and 1.2 substituted. Condone \(1.2^2\), \(\sqrt{1.2}\) or continuity correction of 54.75 or 54.85 |
| \([= P(Z < -0.6667)] = 1 - 0.7477\) | M1 | Appropriate area \(\Phi\), from final process, must be probability |
| \(= 0.2523\) | A1 | \(0.252 \leqslant p \leqslant 0.2525\); if A0 scored S CB1 for \(0.252 \leqslant p \leqslant 0.2525\) |
| [Expected number \(=\)] \(400 \times 0.2523 = 100.92\); 100 or 101 | B1 FT | FT *their* 4SF (or better) probability from a normal calculation. Must be a single integer answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = \Phi(\tfrac{1}{2}) - \Phi(-\tfrac{1}{2}) =\right]\) \(2\Phi\left(\frac{1}{2}\right)-1 = 2\times \textit{their}\; 0.6915 - 1\) | M1 | {Both \(\frac{1}{2}\) and \(-\frac{1}{2}\) seen as \(z\)-values or appropriate use of \(+\frac{1}{2}\) or \(-\frac{1}{2}\)} and {no other \(z\)-values in part}. Condone \(\frac{56.2-55.6}{1.2}\) and \(\frac{55[.0]-55.6}{1.2}\) seen as \(z\)-values |
| or \(\textit{their}\; 0.6915 - (1 - \textit{their}\; 0.6915)\) or \(2\times(0.6915 - 0.5)\) | M1 | Calculating the appropriate area from stated phis of \(z\)-values which must be \(\pm\) the same number |
| \(0.383\) | A1 | \(0.3829 \leqslant z \leqslant 0.383\); if A0 scored SC B1 for \(0.3829 \leqslant z \leqslant 0.383\) |
## Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X<54.8)] = P\left(Z < \frac{54.8-55.6}{1.2}\right)$ | M1 | Use of $\pm$ standardisation formula with 54.8, 55.6 and 1.2 substituted. Condone $1.2^2$, $\sqrt{1.2}$ or continuity correction of 54.75 or 54.85 |
| $[= P(Z < -0.6667)] = 1 - 0.7477$ | M1 | Appropriate area $\Phi$, from final process, must be probability |
| $= 0.2523$ | A1 | $0.252 \leqslant p \leqslant 0.2525$; if A0 scored **S CB1** for $0.252 \leqslant p \leqslant 0.2525$ |
| [Expected number $=$] $400 \times 0.2523 = 100.92$; 100 or 101 | B1 FT | FT *their* 4SF (or better) probability from a normal calculation. Must be a single integer answer |
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## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = \Phi(\tfrac{1}{2}) - \Phi(-\tfrac{1}{2}) =\right]$ $2\Phi\left(\frac{1}{2}\right)-1 = 2\times \textit{their}\; 0.6915 - 1$ | M1 | {Both $\frac{1}{2}$ and $-\frac{1}{2}$ seen as $z$-values **or** appropriate use of $+\frac{1}{2}$ or $-\frac{1}{2}$} **and** {no other $z$-values in part}. Condone $\frac{56.2-55.6}{1.2}$ and $\frac{55[.0]-55.6}{1.2}$ seen as $z$-values |
| or $\textit{their}\; 0.6915 - (1 - \textit{their}\; 0.6915)$ or $2\times(0.6915 - 0.5)$ | M1 | Calculating the appropriate area from stated phis of $z$-values which must be $\pm$ the same number |
| $0.383$ | A1 | $0.3829 \leqslant z \leqslant 0.383$; if A0 scored **SC B1** for $0.3829 \leqslant z \leqslant 0.383$ |
---2 The lengths of the rods produced by a company are normally distributed with mean 55.6 mm and standard deviation 1.2 mm .
\begin{enumerate}[label=(\alph*)]
\item In a random sample of 400 of these rods, how many would you expect to have length less than 54.8 mm ?
\item Find the probability that a randomly chosen rod produced by this company has a length that is within half a standard deviation of the mean.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q2 [7]}}