| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find standard deviation from probability |
| Difficulty | Standard +0.3 This is a straightforward normal distribution problem requiring inverse normal calculation to find standard deviation from a given probability, then standard probability calculations. Part (iii) requires conceptual understanding of how standard deviation affects spread, but no calculation. All techniques are standard S1 material 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 | Mark | Guidance |
| \(z = \dfrac{4.2 - 3.9}{\sigma}\) | M1 | Standardising, not square root of \(\sigma\), not \(\sigma^2\) |
| \(z = 0.916\) or \(0.915\) | B1 | Accept \(0.915 \leqslant \pm z \leqslant 0.916\) seen |
| \(\sigma = 0.328\) | A1 | Correct final answer (allow \(\frac{20}{61}\) or \(\frac{75}{229}\)) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = \frac{4.4 - 3.9}{\text{their } 0.328}\) or \(z = \frac{3.4 - 3.9}{\text{their } 0.328} = 1.5267 \text{ or } -1.5267\) | M1 | Standardising attempt with 3.4 or 4.4 only, allow square root of \(\sigma\) or \(\sigma^2\) |
| \(\Phi = 0.9364\) | A1 | \(0.936 \leq \Phi \leq 0.937\) or \(0.063 \leq \Phi \leq 0.064\) seen |
| \(\text{Prob} = 2\Phi - 1 = 2(0.9364) - 1\) | M1 | Correct area \(2\Phi - 1\) OE i.e. \(\Phi = -(1-\Phi)\), linked to final solution |
| \(= 0.873\) | A1 | Correct final answer from \(0.9363 \leq \Phi \leq 0.9365\) |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Dividing (0.5) by a larger number gives a smaller \(z\)-value or more spread out as sd larger or use of diagrams | *B1 | No calculations or calculated values present e.g. \(\sigma = 0.656\) seen; reference to spread or \(z\) value required |
| Prob is less than that in (ii) | DB1 | Dependent upon first B1 |
| Total: | 2 |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = \dfrac{4.2 - 3.9}{\sigma}$ | M1 | Standardising, not square root of $\sigma$, not $\sigma^2$ |
| $z = 0.916$ or $0.915$ | B1 | Accept $0.915 \leqslant \pm z \leqslant 0.916$ seen |
| $\sigma = 0.328$ | A1 | Correct final answer (allow $\frac{20}{61}$ or $\frac{75}{229}$) |
| **Total: 3** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = \frac{4.4 - 3.9}{\text{their } 0.328}$ or $z = \frac{3.4 - 3.9}{\text{their } 0.328} = 1.5267 \text{ or } -1.5267$ | M1 | Standardising attempt with 3.4 or 4.4 only, allow square root of $\sigma$ or $\sigma^2$ |
| $\Phi = 0.9364$ | A1 | $0.936 \leq \Phi \leq 0.937$ or $0.063 \leq \Phi \leq 0.064$ seen |
| $\text{Prob} = 2\Phi - 1 = 2(0.9364) - 1$ | M1 | Correct area $2\Phi - 1$ OE i.e. $\Phi = -(1-\Phi)$, linked to final solution |
| $= 0.873$ | A1 | Correct final answer from $0.9363 \leq \Phi \leq 0.9365$ |
| **Total:** | **4** | |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Dividing (0.5) by a larger number gives a smaller $z$-value or more spread out as sd larger or use of diagrams | *B1 | No calculations or calculated values present e.g. $\sigma = 0.656$ seen; reference to spread or $z$ value required |
| Prob is less than that in (ii) | DB1 | Dependent upon first B1 |
| **Total:** | **2** | |5 The lengths of videos of a certain popular song have a normal distribution with mean 3.9 minutes. $18 \%$ of these videos last for longer than 4.2 minutes.\\
(i) Find the standard deviation of the lengths of these videos.\\
(ii) Find the probability that the length of a randomly chosen video differs from the mean by less than half a minute.\\
The lengths of videos of another popular song have a normal distribution with the same mean of 3.9 minutes but the standard deviation is twice the standard deviation in part (i). The probability that the length of a randomly chosen video of this song differs from the mean by less than half a minute is denoted by $p$.\\
(iii) Without any further calculation, determine whether $p$ is more than, equal to, or less than your answer to part (ii). You must explain your reasoning.\\
\hfill \mbox{\textit{CAIE S1 2017 Q5 [9]}}