| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Independent probability calculations |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: (a) basic z-score calculation and table lookup, (b) solving simultaneous equations from two given probabilities (routine but slightly more involved), and (c) applying the empirical rule. All parts use well-practiced 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 |
| \(P(X < 6) = P(Z < \frac{6-5.2}{1.5}) = P(Z < 0.5333)\) | M1 | 6, 5.2, 1.5 substituted into \(\pm\) standardisation formula, condone \(1.5^2\), continuity correction \(\pm 0.5\) |
| \(0.703\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z_1 = \frac{3-\mu}{\sigma} = -1.329\) | B1 | \(1.328 < z_1 \leqslant 1.329\) or \(-1.329 \leqslant z_1 < -1.328\) |
| \(z_2 = \frac{8-\mu}{\sigma} = 0.878\) | B1 | \(0.877 < z_2 \leqslant 0.878\) or \(-0.878 \leqslant z_2 < -0.877\) |
| Solve to find at least one unknown: \(\frac{3-\mu}{\sigma} = -1.329\), \(\frac{8-\mu}{\sigma} = 0.878\) | M1 | Use of \(\pm\) standardisation formula once with \(\mu\), \(\sigma\), a z-value (not 0.8179, 0.7910, 0.5367, 0.5753, 0.19, 0.092 etc.) and 3 or 8, condone continuity correction but not \(\sigma^2\) or \(\sqrt{\sigma}\) |
| M1 | Use either elimination or substitution method to solve two equations in \(\mu\) and \(\sigma\) | |
| \(\sigma = 2.27\), \(\mu = 6.01\) | A1 | \(2.26 \leqslant \sigma \leqslant 2.27\), \(6.01 \leqslant \mu \leqslant 6.02\) |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(Z<-1)+P(Z>1)]\ \Phi(1)-\Phi(-1) = 2-2\Phi(1) = 2-2\times 0.8413\) | M1 | Identify 1 and \(-1\) as the appropriate z-values |
| M1 | Calculating appropriate area from stated phis of z-values which must be \(\pm\) the same number | |
| \(0.3174\) | A1 | Accept AWRT 0.317 |
| Number of leaves: \(2000 \times 0.3174 = 634.8\) so \(634\) or \(635\) | B1 FT | FT *their* 4 s.f. (or better) probability, final answer must be positive integer, no approximation or rounding stated |
| 4 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 6) = P(Z < \frac{6-5.2}{1.5}) = P(Z < 0.5333)$ | M1 | 6, 5.2, 1.5 substituted into $\pm$ standardisation formula, condone $1.5^2$, continuity correction $\pm 0.5$ |
| $0.703$ | A1 | |
| | **2** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z_1 = \frac{3-\mu}{\sigma} = -1.329$ | B1 | $1.328 < z_1 \leqslant 1.329$ or $-1.329 \leqslant z_1 < -1.328$ |
| $z_2 = \frac{8-\mu}{\sigma} = 0.878$ | B1 | $0.877 < z_2 \leqslant 0.878$ or $-0.878 \leqslant z_2 < -0.877$ |
| Solve to find at least one unknown: $\frac{3-\mu}{\sigma} = -1.329$, $\frac{8-\mu}{\sigma} = 0.878$ | M1 | Use of $\pm$ standardisation formula once with $\mu$, $\sigma$, a z-value (not 0.8179, 0.7910, 0.5367, 0.5753, 0.19, 0.092 etc.) and 3 or 8, condone continuity correction but not $\sigma^2$ or $\sqrt{\sigma}$ |
| | M1 | Use either elimination or substitution method to solve two equations in $\mu$ and $\sigma$ |
| $\sigma = 2.27$, $\mu = 6.01$ | A1 | $2.26 \leqslant \sigma \leqslant 2.27$, $6.01 \leqslant \mu \leqslant 6.02$ |
| | **5** | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(Z<-1)+P(Z>1)]\ \Phi(1)-\Phi(-1) = 2-2\Phi(1) = 2-2\times 0.8413$ | M1 | Identify 1 and $-1$ as the appropriate z-values |
| | M1 | Calculating appropriate area from stated phis of z-values which must be $\pm$ the same number |
| $0.3174$ | A1 | Accept AWRT 0.317 |
| Number of leaves: $2000 \times 0.3174 = 634.8$ so $634$ or $635$ | B1 FT | FT *their* 4 s.f. (or better) probability, final answer must be positive integer, no approximation or rounding stated |
| | **4** | |5 The lengths, in cm, of the leaves of a particular type are modelled by the distribution $\mathrm { N } \left( 5.2,1.5 ^ { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen leaf of this type has length less than 6 cm .\\
The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than 3 cm long and 95 are more than 8 cm long.
\item Find estimates for the mean and standard deviation of the lengths of leaves of this type.
\item In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean?
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q5 [11]}}