| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Percentages or proportions given |
| Difficulty | Standard +0.3 This is a straightforward normal distribution problem requiring inverse normal calculation to find μ from a given proportion, followed by a basic binomial probability calculation. Both parts use standard techniques with no conceptual challenges, making it slightly easier than average for A-level statistics. |
| 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/Working | Marks | Guidance |
| \(z = 0.674\) | B1 | \(\pm 0.674\) or rounding to, seen, e.g. 0.6743 |
| \(\frac{1002 - \mu}{8} = 0.674\) | M1 | Standardising and attempting to solve for \(\mu\), must use recognisable \(z\)-value, no cc, no sq rt, no sq |
| \(\mu = 997\) | A1 [3] | Correct answer rounding to 997 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(2) = 3 \times \frac{225}{900} \times \frac{224}{899} \times \frac{675}{898}\) | M1 | \(900 \times 899 \times 898\) or \(^{900}C_3\) seen in denom |
| \(= 0.140\) | A1 [2] | Correct answer not 0.141 or 0.14 |
| OR \(\frac{^{225}C_2 \times ^{675}C_1}{^{900}C_3}\) |
## Question 1:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = 0.674$ | B1 | $\pm 0.674$ or rounding to, seen, e.g. 0.6743 |
| $\frac{1002 - \mu}{8} = 0.674$ | M1 | Standardising and attempting to solve for $\mu$, must use recognisable $z$-value, no cc, no sq rt, no sq |
| $\mu = 997$ | A1 **[3]** | Correct answer rounding to 997 |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(2) = 3 \times \frac{225}{900} \times \frac{224}{899} \times \frac{675}{898}$ | M1 | $900 \times 899 \times 898$ or $^{900}C_3$ seen in denom |
| $= 0.140$ | A1 **[2]** | Correct answer not 0.141 or 0.14 |
| OR $\frac{^{225}C_2 \times ^{675}C_1}{^{900}C_3}$ | | |
---1 The volume of milk in millilitres in cartons is normally distributed with mean $\mu$ and standard deviation 8. Measurements were taken of the volume in 900 of these cartons and it was found that 225 of them contained more than 1002 millilitres.\\
(i) Calculate the value of $\mu$.\\
(ii) Three of these 900 cartons are chosen at random. Calculate the probability that exactly 2 of them contain more than 1002 millilitres.
\hfill \mbox{\textit{CAIE S1 2009 Q1 [5]}}