| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find mean from probability statement |
| Difficulty | Moderate -0.8 Part (i) is a direct normal distribution probability calculation requiring standardization and table lookup. Part (ii) reverses this process—finding a mean from a given probability—but still follows a standard procedure: look up the z-value for 0.8888, then solve μ = 0 - z√40. Both parts are routine applications of normal distribution techniques with no problem-solving insight required, making this 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/Working | Marks | Guidance |
| \(P(X > 0) = 1 - \Phi\left(\frac{0 - (-15.1)}{\sqrt{62}}\right)\) | M1 | Standardising, sq rt, no cc |
| \(= 1 - \Phi(1.918) = 1 - 0.9724\) | M1 | Prob \(< 0.5\) after use of normal tables |
| \(= 0.0276\) or answer rounding to | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = -1.22\) | B1 | \(z = \pm 1.22\) |
| \(-1.22 = \frac{0 - \mu}{\sqrt{40}}\) | M1 | An equation in \(\mu\), recognisable \(z\), \(\sqrt{40}\), no cc |
| \(\mu = 7.72\) c.a.o | A1 [3] | Correct answer c.w.o from same sign on both sides |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X > 0) = 1 - \Phi\left(\frac{0 - (-15.1)}{\sqrt{62}}\right)$ | M1 | Standardising, sq rt, no cc |
| $= 1 - \Phi(1.918) = 1 - 0.9724$ | M1 | Prob $< 0.5$ after use of normal tables |
| $= 0.0276$ or answer rounding to | A1 **[3]** | Correct answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = -1.22$ | B1 | $z = \pm 1.22$ |
| $-1.22 = \frac{0 - \mu}{\sqrt{40}}$ | M1 | An equation in $\mu$, recognisable $z$, $\sqrt{40}$, no cc |
| $\mu = 7.72$ c.a.o | A1 **[3]** | Correct answer c.w.o from same sign on both sides |
---3 (i) The daily minimum temperature in degrees Celsius ( ${ } ^ { \circ } \mathrm { C }$ ) in January in Ottawa is a random variable with distribution $\mathrm { N } ( - 15.1,62.0 )$. Find the probability that a randomly chosen day in January in Ottawa has a minimum temperature above $0 ^ { \circ } \mathrm { C }$.\\
(ii) In another city the daily minimum temperature in ${ } ^ { \circ } \mathrm { C }$ in January is a random variable with distribution $\mathrm { N } ( \mu , 40.0 )$. In this city the probability that a randomly chosen day in January has a minimum temperature above $0 ^ { \circ } \mathrm { C }$ is 0.8888 . Find the value of $\mu$.
\hfill \mbox{\textit{CAIE S1 2008 Q3 [6]}}