| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution with standard procedures: (i) requires standardizing and using tables for P(a < X < b), (ii) involves inverse normal calculation with a given percentage, and (iii) applies binomial distribution using a probability from part (i). All three parts are routine textbook exercises requiring no problem-solving insight, though the multi-part structure and binomial extension in (iii) elevate it slightly above the most basic normal distribution questions. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(46 < X < 53) = P\!\left(\frac{46-49.2}{2.8} < Z < \frac{53-49.2}{2.8}\right)\) | M1 | Using \(\pm\) standardisation formula for either 46 or 53, no continuity correction, \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(P(-1.143 < Z < 1.357)\) | A1 | Both standardisations correct unsimplified |
| \(\Phi(1.357) + \Phi(1.143) - 1 = 0.9126 + 0.8735 - 1\) | M1 | Correct final area |
| \(0.786\) | A1 | Final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{t - 49.2}{2.8} = -1.406\) | B1 | \(\pm 1.406\) seen |
| M1 | An equation using \(\pm\) standardisation formula with a \(z\)-value, condone \(\sigma^2\) or \(\sqrt{\sigma}\) | |
| \(45.3\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 46) = 0.1265\) | M1 | Calculated or ft from (i) |
| \(P(2\text{PB} < 46) = 3(1 - 0.1265)(0.1265)^2\) | M1 | \(3(1-p)p^2,\ 0 < p < 1\) |
| \(0.0419\) | A1 | |
| 3 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(46 < X < 53) = P\!\left(\frac{46-49.2}{2.8} < Z < \frac{53-49.2}{2.8}\right)$ | M1 | Using $\pm$ standardisation formula for either 46 or 53, no continuity correction, $\sigma^2$ or $\sqrt{\sigma}$ |
| $P(-1.143 < Z < 1.357)$ | A1 | Both standardisations correct unsimplified |
| $\Phi(1.357) + \Phi(1.143) - 1 = 0.9126 + 0.8735 - 1$ | M1 | Correct final area |
| $0.786$ | A1 | Final answer |
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{t - 49.2}{2.8} = -1.406$ | **B1** | $\pm 1.406$ seen |
| | **M1** | An equation using $\pm$ standardisation formula with a $z$-value, condone $\sigma^2$ or $\sqrt{\sigma}$ |
| $45.3$ | **A1** | |
| | **3** | |
---
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 46) = 0.1265$ | **M1** | Calculated or ft from **(i)** |
| $P(2\text{PB} < 46) = 3(1 - 0.1265)(0.1265)^2$ | **M1** | $3(1-p)p^2,\ 0 < p < 1$ |
| $0.0419$ | **A1** | |
| | **3** | |7 The shortest time recorded by an athlete in a 400 m race is called their personal best (PB). The PBs of the athletes in a large athletics club are normally distributed with mean 49.2 seconds and standard deviation 2.8 seconds.\\
(i) Find the probability that a randomly chosen athlete from this club has a PB between 46 and 53 seconds.\\
(ii) It is found that $92 \%$ of athletes from this club have PBs of more than $t$ seconds. Find the value of $t$.\\
Three athletes from the club are chosen at random.\\
(iii) Find the probability that exactly 2 have PBs of less than 46 seconds.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S1 2019 Q7 [10]}}