| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard percentage calculations |
| Difficulty | Moderate -0.8 This is a straightforward normal distribution question requiring standard z-score calculations and inverse normal lookups. Part (a) is direct standardization, part (b) involves simple probability manipulation (finding 1/3 of the upper tail), and part (c) is a routine inverse normal calculation. All techniques are standard S1 content with no problem-solving insight required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 25) = P\!\left(z < \frac{25-40}{12}\right) = P(z < -1.25)\) | M1 | Standardising |
| \(1 - 0.8944\) | M1 | |
| \(0.106\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.8944\) divided by \(3\) | M1 | M1 for \(1 - \text{their (a)}\) divided by 3 |
| \(0.298\) | A1 (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.2981\) gives \(z = 0.53\) | B1 | |
| \(\frac{h - 40}{12} = 0.53\) | M1 | |
| \(h = 46.4\) | A1 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 25) = P\!\left(z < \frac{25-40}{12}\right) = P(z < -1.25)$ | M1 | Standardising |
| $1 - 0.8944$ | M1 | |
| $0.106$ | A1 | |
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.8944$ divided by $3$ | M1 | M1 for $1 - \text{their (a)}$ divided by 3 |
| $0.298$ | A1 (AG) | |
---
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.2981$ gives $z = 0.53$ | B1 | |
| $\frac{h - 40}{12} = 0.53$ | M1 | |
| $h = 46.4$ | A1 | |
---4 Trees in the Redian forest are classified as tall, medium or short, according to their height. The heights can be modelled by a normal distribution with mean 40 m and standard deviation 12 m . Trees with a height of less than 25 m are classified as short.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen tree is classified as short.\\
Of the trees that are classified as tall or medium, one third are tall and two thirds are medium.
\item Show that the probability that a randomly chosen tree is classified as tall is 0.298 , correct to 3 decimal places.
\item Find the height above which trees are classified as tall.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q4 [8]}}