| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find standard deviation from probability |
| Difficulty | Challenging +1.2 Part (a) requires understanding that 99% within limits means finding the standard deviation where the mean ±2.576σ equals the boundaries—a straightforward inverse normal calculation. Part (b) involves forming a linear combination of normal distributions (X - 3Y) and finding a probability, which is a standard Further Maths technique but requires careful setup of the combined distribution's mean and variance. The question tests solid understanding of normal distribution properties rather than novel insight. |
| Spec | 2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| For maximum acceptable standard deviation distribution must be symmetrical about Mean \(= 159.45\) | B1 | si |
| \(159.45 + 2.5758\sigma = 163\) OR \(159.45 - 2.5758\sigma = 155.9\) | M1 | or 2.576 from tables |
| \(\sigma = 1.378...g\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Let the random variable \(X\) be the weights of cricket balls. Let the random variable \(Y\) be the weights of the tennis balls. Consider \(W = 3Y - X\) | M1 A1 M1 A1 | — |
| \(E(W) = 16\) | — | — |
| \(\text{Var}(W) = 3^2\text{Var}(Y) + \text{Var}(X) = 16.65\) | — | — |
| \(P(W < 0) = 0.00004\) | M1 A1 | — |
## (a)
For maximum acceptable standard deviation distribution must be symmetrical about Mean $= 159.45$ | B1 | si
$159.45 + 2.5758\sigma = 163$ OR $159.45 - 2.5758\sigma = 155.9$ | M1 | or 2.576 from tables
$\sigma = 1.378...g$ | A1 | —
## (b)
Let the random variable $X$ be the weights of cricket balls. Let the random variable $Y$ be the weights of the tennis balls. Consider $W = 3Y - X$ | M1 A1 M1 A1 | —
$E(W) = 16$ | — | —
$\text{Var}(W) = 3^2\text{Var}(Y) + \text{Var}(X) = 16.65$ | — | —
$P(W < 0) = 0.00004$ | M1 A1 | —
**Total: [9]**
---The rules for the weight of a cricket ball state:
``The ball, when new, shall weigh not less than 155.9 g, nor more than 163 g.''
A company produces cricket balls whose weights are normally distributed. It wants 99\% of the balls it produces to be an acceptable weight.
\begin{enumerate}[label=(\alph*)]
\item What is the largest acceptable standard deviation? [3]
\end{enumerate}
The weights of the cricket balls are in fact normally distributed with mean 159.5 grams and standard deviation 1.2 grams. The company also produces tennis balls. The weights of the tennis balls are normally distributed with mean 58.5 grams and standard deviation 1.3 grams.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the probability that the weight of a randomly chosen cricket ball is more than three times the weight of a randomly chosen tennis ball. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2019 Q3 [9]}}