| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Combined events across distributions |
| Difficulty | Moderate -0.3 This is a straightforward two-part normal distribution question requiring standard techniques: (i) using inverse normal with a given percentile to find the mean, and (ii) calculating a basic probability using the normal CDF. Both parts are routine applications with no problem-solving insight 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 |
| \(-0.674 = \dfrac{7 - \mu}{2.6}\) | B1 | \(\pm 0.674\) seen only |
| M1 | Standardising must have a recognisable z-value, no cc and 2.6 | |
| M1 | For solving their equation with recognisable z-value, \(\mu\) and 2.6 not \(1 - 0.674\) or 0.326, allow cc | |
| \(\mu = 8.75\) | A1 4 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 6.2) = P\!\left(z > \dfrac{6.2 - 6.5}{2.6}\right)\) | M1 | Standardising, no cc on the 6.2 |
| \(= P(z > -0.1154)\) | M1 | prob \(> 0.5\) |
| \(= 0.546\) | A1 3 | Correct answer |
## Question 4:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $-0.674 = \dfrac{7 - \mu}{2.6}$ | B1 | $\pm 0.674$ seen only |
| | M1 | Standardising must have a recognisable z-value, no cc and 2.6 |
| | M1 | For solving their equation with recognisable z-value, $\mu$ and 2.6 not $1 - 0.674$ or 0.326, allow cc |
| $\mu = 8.75$ | A1 **4** | Correct answer |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 6.2) = P\!\left(z > \dfrac{6.2 - 6.5}{2.6}\right)$ | M1 | Standardising, no cc on the 6.2 |
| $= P(z > -0.1154)$ | M1 | prob $> 0.5$ |
| $= 0.546$ | A1 **3** | Correct answer |
---4 In a certain country the time taken for a common infection to clear up is normally distributed with mean $\mu$ days and standard deviation 2.6 days. $25 \%$ of these infections clear up in less than 7 days.\\
(i) Find the value of $\mu$.
In another country the standard deviation of the time taken for the infection to clear up is the same as in part (i), but the mean is 6.5 days. The time taken is normally distributed.\\
(ii) Find the probability that, in a randomly chosen case from this country, the infection takes longer than 6.2 days to clear up.
\hfill \mbox{\textit{CAIE S1 2008 Q4 [7]}}