| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Percentages or proportions given |
| Difficulty | Standard +0.3 This is a straightforward S1 normal distribution question requiring standard techniques: z-score calculation and inverse normal lookup, followed by a simple binomial probability calculation. All parts follow routine procedures with no novel problem-solving 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 |
|---|---|
| (a) \([P(D > 50) =] P\left(Z > \frac{50-32}{12}\right)\) | M1 (3) |
| \(= 1 - P(Z < 1.5)\) or \(1 - 0.9332\) | M1 |
| \(=\) awrt \(0.0668\) or 6.68% | A1cso |
| (b) \(P(D > d) = 0.191 + 0.0668 = 0.2578\) or \(P(D < d) = 0.7422\) | B1 (4) |
| \(\frac{d-32}{12} = 0.65\) (calc gives 0.65014... or 0.65012...) | M1A1 |
| \(d = 39.8\) | A1 |
| (c) \(0.0668 \times 0.191^2 [= 0.0024369...]\) | M1 (3) |
| \([...] \times 3\) | M1 |
| \(= 0.00731079... = \) awrt \(0.0073\) | A1 |
**(a)** $[P(D > 50) =] P\left(Z > \frac{50-32}{12}\right)$ | M1 (3) |
$= 1 - P(Z < 1.5)$ or $1 - 0.9332$ | M1 |
$=$ awrt $0.0668$ or 6.68% | A1cso |
**(b)** $P(D > d) = 0.191 + 0.0668 = 0.2578$ or $P(D < d) = 0.7422$ | B1 (4) |
$\frac{d-32}{12} = 0.65$ (calc gives 0.65014... or 0.65012...) | M1A1 |
$d = 39.8$ | A1 |
**(c)** $0.0668 \times 0.191^2 [= 0.0024369...]$ | M1 (3) |
$[...] \times 3$ | M1 |
$= 0.00731079... = $ awrt $0.0073$ | A1 |
[10]
**Notes:**
(a) 1st M1 for standardising with 50, 32 and 12. Allow +. 2nd M1 for $1 - P(Z < 1.5)$ seen i.e. a correct method for finding $P(Z > 1.5)$ e.g. $1 -$ tables value. A1cso for awrt 0.0668 with both Ms scored and no incorrect working seen. Condone incomplete notation and condone use of different letters for Z.
(b) B1 for awrt 0.2578 (calc = 0.257807...) or awrt 0.7422 (calc = 0.742192...). may be implied by $z =$ awrt 0.65. M1 for standardising with 32 and 12, i.e. $\pm \frac{d-32}{12}$ (equating to a probability is M0). 1st A1 for $z =$ awrt 0.65 and a correct equation in $d$ (with compatible signs). 2nd A1 for awrt 39.8.
(c) 1st M1 for $0.0668 \times 0.191^2$ or sight of awrt 0.0024 (may be seen embedded in part of an expression, e.g. $\eta \times 0.0668 \times 0.191^2$) (condone 6.68% × 19.1% × 19.1% if the final answer given is < 1). 2nd M1 for any expression of the form $3pq^2$ where $p$ and $q$ are both probabilities. A1 for awrt 0.0073. allow awrt 0.73% but 0.73 is A0.7. One event at Pentor sports day is throwing a tennis ball. The distance a child throws a tennis ball is modelled by a normal distribution with mean 32 m and standard deviation 12 m . Any child who throws the tennis ball more than 50 m is awarded a gold certificate.
\begin{enumerate}[label=(\alph*)]
\item Show that, to 3 significant figures, 6.68\% of children are awarded a gold certificate.
A silver certificate is awarded to any child who throws the tennis ball more than $d$ metres but less than 50 m .
Given that 19.1\% of the children are awarded a silver certificate,
\item find the value of $d$.
Three children are selected at random from those who take part in the throwing a tennis ball event.
\item Find the probability that 1 is awarded a gold certificate and 2 are awarded silver certificates. Give your answer to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q7 [10]}}