| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward normal distribution problem requiring standard z-score calculations and inverse normal lookups. Part (a) involves finding P(X ≥ 184) using standardization, and part (b) requires finding the 20th percentile. Both are routine S1 techniques with no conceptual challenges beyond applying the normal distribution formula correctly. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 184) = P\!\left(Z > \dfrac{184-131}{54}\right)\ [= P(Z > 0.9815)]\) | M1 | Use of \(\pm\)standardisation formula with 184, 131 and 54, no continuity correction. Condone use of \(\sigma^2\), \(\sqrt{\sigma}\). |
| \(1 - 0.837\ [= 0.163]\) | M1 | Calculating the appropriate probability area (leading to their final answer). |
| Percentage \([= 0.163 \times 100] = 16.3\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X < w) = P\!\left(Z < \dfrac{w-131}{54}\right) = 0.2]\) | B1 | \(-0.842 \leqslant z < -0.8415\) or \(0.8415 < z \leqslant 0.842\) seen. |
| \(\dfrac{w-131}{54} = -0.842\) | M1 | Use of the \(\pm\)standardisation formula with 131, 54, \(w\) and a \(z\)-value (not 0.2, 0.8, 0.158, 0.508[0], 0.492[0], 0.7881, 0.2119, 0.5593, 0.4407). |
| \(w = 85.5\) | A1 | \(85.5 \leqslant p \leqslant 85.6\). Signs must be consistent to create a positive answer. |
## Question 3(a):
$P(X > 184) = P\!\left(Z > \dfrac{184-131}{54}\right)\ [= P(Z > 0.9815)]$ | M1 | Use of $\pm$standardisation formula with 184, 131 and 54, no continuity correction. Condone use of $\sigma^2$, $\sqrt{\sigma}$.
$1 - 0.837\ [= 0.163]$ | M1 | Calculating the appropriate probability area (leading to their final answer).
Percentage $[= 0.163 \times 100] = 16.3$ | A1 | AWRT
**Total: 3 marks**
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## Question 3(b):
$[P(X < w) = P\!\left(Z < \dfrac{w-131}{54}\right) = 0.2]$ | B1 | $-0.842 \leqslant z < -0.8415$ or $0.8415 < z \leqslant 0.842$ seen.
$\dfrac{w-131}{54} = -0.842$ | M1 | Use of the $\pm$standardisation formula with 131, 54, $w$ and a $z$-value (not 0.2, 0.8, 0.158, 0.508[0], 0.492[0], 0.7881, 0.2119, 0.5593, 0.4407).
$w = 85.5$ | A1 | $85.5 \leqslant p \leqslant 85.6$. Signs must be consistent to create a positive answer.
**Total: 3 marks**
---3 The weights of oranges can be modelled by a normal distribution with mean 131 grams and standard deviation 54 grams. Oranges are classified as small, medium or large. A large orange weighs at least 184 grams and 20\% of oranges are classified as small.
\begin{enumerate}[label=(\alph*)]
\item Find the percentage of oranges that are classified as large.
\item Find the greatest possible weight of a small orange.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q3 [6]}}