| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Topic | Normal Distribution |
| Type | Find k given probability statement |
| Difficulty | Standard +0.8 This is a multi-step normal distribution problem requiring z-score calculations, probability computations, and solving an equation involving the normal CDF. Part (ii) requires setting up and solving a non-standard equation where the threshold value x is unknown, demanding algebraic manipulation with normal distribution functions—significantly more challenging than routine A-level statistics questions. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(z = \frac{114 - 120}{6} = -1\) | B1 | State or imply \(z = \pm 1\) |
| \(P(X > 114) = P(Z > -1)\) | M1 | Attempts \(P(Z > \pm 1)\) |
| \(= 0.8413\) | A1 | Concludes 0.8413 |
| Expected profit = \(15 \times 0.8413\) | M1 | Uses Profit = Number \(\times\) Prob anywhere |
| \(= 12.6195p = £12.62\) | A1 | Obtain £12.62 or equiv 0.8413) seen. No units seen A0 |
| (ii) \(20P(X > x) + 3P(X \le x) = 19.17\) | M1 | State probabilities (may be wrong way round) and make one further step to reduce to a single probability |
| \(20(1 - P(X \le x)) + 3P(X \le x) = 19.17\) | ||
| \(P(X \le x) = 0.04882\) | A1 | Obtain 0.04882 or 0.9512 |
| \((1 - 0.04882) = 0.9512\) | M1 | Use \(\Phi^{-1}(0.9512)\) |
| \(\Phi^{-1}(0.9512) = 1.657\) | ||
| so \(z = -1.657\) | A1 | Allow \(\pm [1.655, 1.660]\) |
| \(\frac{x - 120}{6} = -1.657\) or equiv negative version | M1, B1 | Award for sight of \(\frac{x-120}{6} = \pm\) (their) \(z\) value |
| \(x = 110\) | A1 | [12] |
(i) $z = \frac{114 - 120}{6} = -1$ | B1 | State or imply $z = \pm 1$
$P(X > 114) = P(Z > -1)$ | M1 | Attempts $P(Z > \pm 1)$
$= 0.8413$ | A1 | Concludes 0.8413
Expected profit = $15 \times 0.8413$ | M1 | Uses Profit = Number $\times$ Prob anywhere
$= 12.6195p = £12.62$ | A1 | Obtain £12.62 or equiv 0.8413) seen. No units seen A0
(ii) $20P(X > x) + 3P(X \le x) = 19.17$ | M1 | State probabilities (may be wrong way round) and make one further step to reduce to a single probability
$20(1 - P(X \le x)) + 3P(X \le x) = 19.17$ |
$P(X \le x) = 0.04882$ | A1 | Obtain 0.04882 or 0.9512
$(1 - 0.04882) = 0.9512$ | M1 | Use $\Phi^{-1}(0.9512)$
$\Phi^{-1}(0.9512) = 1.657$ |
so $z = -1.657$ | A1 | Allow $\pm [1.655, 1.660]$
$\frac{x - 120}{6} = -1.657$ or equiv negative version | M1, B1 | Award for sight of $\frac{x-120}{6} = \pm$ (their) $z$ value
$x = 110$ | A1 | [12] | Obtain 110 (= 110.058)A firm produces chocolate bars whose weights are normally distributed with mean 120 g and standard deviation 6 g.
\begin{enumerate}[label=(\roman*)]
\item Bars which weigh more than 114 g are sold at a profit of 15p per bar. The remaining bars are sold at no profit. Show that the expected profit per 100 bars is £12.62. [5]
\item It is subsequently decided that bars which weigh more than $x$ g should be sold at a profit of 20p per bar. Those which weigh $x$ g or less are sold to employees at a profit of 3p per bar. The expected profit per 100 bars is £19.17. Find the value of $x$. [7]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q15 [12]}}