| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Standard +0.3 Part (i) is a routine normal distribution probability calculation requiring standardization and table lookup (z-scores). Part (ii) requires working backwards from percentiles to find parameters, involving two simultaneous equations with inverse normal values—a standard but slightly more demanding technique. Overall, this is a straightforward application of normal distribution theory with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(D \sim N(8.3, 0.20^2)\) | ||
| \(P(8.1 < D < 8.3) = P\left(\frac{8.1-8.3}{0.20} < Z < \frac{8.5-8.3}{0.20}\right)\) | M1 | Standardising, either term. |
| \(= \Phi(1.0) - \Phi(-1.0) = 0.8413 - (1 - 0.8413)\) | M1 B1 | Relevant difference of 2 terms s.o.i. Correct table look-up: 0.8413 seen. 1 – ... to deal with negative z value. |
| \(= 0.6826\) | M1 A1 | [5] |
| (ii) Now \(D \sim N(\mu, \sigma^2)\) | ||
| \(P(D < 8.5) = 0.88 \Rightarrow \frac{8.5 - \mu}{\sigma} = 1.175\) | M1 B1 | Set up at least 1 equation for \(\mu\) and \(\sigma\). 1.175 and/or (–)1.282 seen. Both equations correct. |
| \(P(D < 8.1) = 0.10 \Rightarrow \frac{8.1 - \mu}{\sigma} = -1.282\) | A1 | |
| \(\therefore \mu + 1.175\sigma = 8.5\) and \(\mu - 1.282\sigma = 8.1\) | M1 | Attempt to eliminate either \(\mu\) or \(\sigma\). One of \(\sigma\) or \(\mu\) found. c.a.o. |
| \(\therefore 2.457\sigma = 0.4\) | A1 | The other found. c.a.o. |
| \(\therefore \sigma = 0.1628(0...)\) | A1 | Allow 0.163 used and a.w.r.t. 8.31 |
| \(\therefore \mu = 8.5 - 1.175 \times 0.1628 = 8.3087\) | ||
| or \(\mu = 8.1 + 1.282 \times 0.1628\) | A1 | [6] |
**(i)** $D \sim N(8.3, 0.20^2)$ |
$P(8.1 < D < 8.3) = P\left(\frac{8.1-8.3}{0.20} < Z < \frac{8.5-8.3}{0.20}\right)$ | M1 | Standardising, either term.
$= \Phi(1.0) - \Phi(-1.0) = 0.8413 - (1 - 0.8413)$ | M1 B1 | Relevant difference of 2 terms s.o.i. Correct table look-up: 0.8413 seen. 1 – ... to deal with negative z value.
$= 0.6826$ | M1 A1 | [5]
**(ii)** Now $D \sim N(\mu, \sigma^2)$ |
$P(D < 8.5) = 0.88 \Rightarrow \frac{8.5 - \mu}{\sigma} = 1.175$ | M1 B1 | Set up at least 1 equation for $\mu$ and $\sigma$. 1.175 and/or (–)1.282 seen. Both equations correct.
$P(D < 8.1) = 0.10 \Rightarrow \frac{8.1 - \mu}{\sigma} = -1.282$ | A1 |
$\therefore \mu + 1.175\sigma = 8.5$ and $\mu - 1.282\sigma = 8.1$ | M1 | Attempt to eliminate either $\mu$ or $\sigma$. One of $\sigma$ or $\mu$ found. c.a.o.
$\therefore 2.457\sigma = 0.4$ | A1 | The other found. c.a.o.
$\therefore \sigma = 0.1628(0...)$ | A1 | Allow 0.163 used and a.w.r.t. 8.31
$\therefore \mu = 8.5 - 1.175 \times 0.1628 = 8.3087$ |
or $\mu = 8.1 + 1.282 \times 0.1628$ | A1 | [6]A machine is being used to manufacture ball bearings. The diameters of the ball bearings are normally distributed with mean 8.3 mm and standard deviation 0.20 mm.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the diameter of a randomly chosen ball bearing lies between 8.1 mm and 8.5 mm. [5]
\item Following an overhaul of the machine, it is now found that the diameters of 88% of ball bearings are less than 8.5 mm while 10% are less than 8.1 mm. Estimate the new mean and standard deviation of the diameters. [6]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2014 Q6 [11]}}