| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Independent probability calculations |
| Difficulty | Moderate -0.5 This is a straightforward normal distribution question requiring standard probability calculations and use of tables/inverse normal. Part (a) involves finding expected number of days using P(X > threshold), part (b) uses given probability to find mean, and part (c) applies binomial approximation to normal. All are routine S1 techniques 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 |
| \(P(X > 3900) = P\!\left(Z > \frac{3900 - 4520}{560}\right)\) | M1 | Standardising: no continuity correction, no square root, no square |
| \(P(Z > -1.107) = \Phi(1.107)\) | M1 | Attempt at correct area: \(\Phi > 0.5\), depends on negative \(z\) |
| \(= 0.8657\) | A1 | Probability rounding to 0.866 |
| Number of days \(= 365 \times 0.8657 = 315\) or \(316\ (315.98)\) | B1FT | FT their wrong probability if previous A0, \(p < 1\); FT must be accurate to 3sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 1.165\) | B1 | \(\pm 1.165\) seen |
| \(1.165 = \frac{8000 - m}{560}\) | M1 | Standardising equation, allow square root; must have \(z\)-value (e.g. not 0.122, 0.878, 0.549, 0.810) |
| \(m = 7350\ (7347.6)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0,1) = (0.878)^6 + {}^6C_1(0.122)^1(0.878)^5\) \((= 0.4581 + 0.3819)\) | 2 | M1M1 |
| \(= 0.840\ (\text{accept } 0.84)\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 3900) = P\!\left(Z > \frac{3900 - 4520}{560}\right)$ | M1 | Standardising: no continuity correction, no square root, no square |
| $P(Z > -1.107) = \Phi(1.107)$ | M1 | Attempt at correct area: $\Phi > 0.5$, depends on negative $z$ |
| $= 0.8657$ | A1 | Probability rounding to 0.866 |
| Number of days $= 365 \times 0.8657 = 315$ or $316\ (315.98)$ | B1FT | FT their wrong probability if previous A0, $p < 1$; FT must be accurate to 3sf |
**Total: 4 marks**
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 1.165$ | B1 | $\pm 1.165$ seen |
| $1.165 = \frac{8000 - m}{560}$ | M1 | Standardising equation, allow square root; must have $z$-value (e.g. not 0.122, 0.878, 0.549, 0.810) |
| $m = 7350\ (7347.6)$ | A1 | |
**Total: 3 marks**
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0,1) = (0.878)^6 + {}^6C_1(0.122)^1(0.878)^5$ $(= 0.4581 + 0.3819)$ | 2 | M1M1 | M1 for correct unsimplified expression; M1 for Binomial term ${}^6C_x\, p^x(1-p)^{6-x}$, $0 < p < 1$ seen |
| $= 0.840\ (\text{accept } 0.84)$ | A1 | |
**Total: 3 marks**4 A petrol station finds that its daily sales, in litres, are normally distributed with mean 4520 and standard deviation 560.\\
(a) Find on how many days of the year (365 days) the daily sales can be expected to exceed 3900 litres.\\
The daily sales at another petrol station are $X$ litres, where $X$ is normally distributed with mean $m$ and standard deviation 560. It is given that $\mathrm { P } ( X > 8000 ) = 0.122$.\\
(b) Find the value of $m$.\\
(c) Find the probability that daily sales at this petrol station exceed 8000 litres on fewer than 2 of 6 randomly chosen days.\\
\hfill \mbox{\textit{CAIE S1 2020 Q4 [10]}}