| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution with standard bookwork: (i) requires a single z-score calculation and table lookup for P(X < 1.100), and (ii) requires finding the inverse normal value for the 90th percentile. Both parts are routine A-level statistics questions with no problem-solving or conceptual challenges beyond direct formula application. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(W \sim \mathrm{N}(1.349, 0.236^2)\) | M1 | Standardising. |
| \(\mathrm{P}(W < 1.100) = \mathrm{P}\left(Z < \frac{1.100 - 1.349}{0.236}\right) = \Phi(-1.055) = 1 - 0.8543 = 0.1457\) | A1, M1, B1, A1 | c.a.o.; \(1 - \ldots\) to deal with negative \(z\) value; Correct table look-up, including use of difference columns: e.g. 0.8543 seen, it abs(c's \(z\)-value). If c's negative \(z\)-value. Note: \(z = -1.05 \Rightarrow p = 1 - 0.8531 = 0.1469\); \(z = -1.06 \Rightarrow p = 1 - 0.8554 = 0.1446\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Phi^{-1}(0.9) = 1.282\) | B1 | Correct table look-up. |
| \(w = 1.349 + 1.282 \times 0.236 = 1.6515\ldots \approx 1.65\) (kg) | M1, A1 | c.a.o. |
### (i)
$W \sim \mathrm{N}(1.349, 0.236^2)$ | M1 | Standardising.
$\mathrm{P}(W < 1.100) = \mathrm{P}\left(Z < \frac{1.100 - 1.349}{0.236}\right) = \Phi(-1.055) = 1 - 0.8543 = 0.1457$ | A1, M1, B1, A1 | c.a.o.; $1 - \ldots$ to deal with negative $z$ value; Correct table look-up, including use of difference columns: e.g. 0.8543 seen, it abs(c's $z$-value). If c's negative $z$-value. Note: $z = -1.05 \Rightarrow p = 1 - 0.8531 = 0.1469$; $z = -1.06 \Rightarrow p = 1 - 0.8554 = 0.1446$ | [5]
### (ii)
$\Phi^{-1}(0.9) = 1.282$ | B1 | Correct table look-up.
$w = 1.349 + 1.282 \times 0.236 = 1.6515\ldots \approx 1.65$ (kg) | M1, A1 | c.a.o. | [3]
---The weights of pineapples on sale at a wholesaler are normally distributed with mean $1.349$ kg and standard deviation $0.236$ kg. Before going on sale the pineapples are classified as 'Small', 'Medium', 'Large' and 'Extra Large'.
\begin{enumerate}[label=(\roman*)]
\item A pineapple is classified as 'Small' if it weighs less than $1.100$ kg. Find the probability that a randomly chosen pineapple will be classified as 'Small'. [5]
\item $10\%$ of pineapples are classified as 'Extra Large'. Find the minimum weight required for a pineapple to be classified as 'Extra Large'. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q2 [8]}}