| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Find parameter from normal approximation |
| Difficulty | Standard +0.3 This is a standard S2 question combining normal distribution with binomial probability. Part (a) requires inverse normal calculation, part (b) is straightforward binomial probability, and part (c) involves normal approximation to binomial with continuity correction. All techniques are routine for S2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(M < 10) = P\!\left(Z < \frac{12-14}{\sigma}\right) = 0.1\) | ||
| \(\Rightarrow \frac{12-14}{\sigma} = -1.2816\) | M1 | Standardising (\(\pm\)) with 12, 14 and \(\sigma\); setting equal to a \(z\) value where \( |
| B1 | \(\pm 1.2816\) or better | |
| \(\sigma = 1.5605\ldots \approx\) awrt 1.56 minutes | A1 | awrt 1.56; do not allow answer written as an exact fraction (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T \sim B(15, 0.1)\) | B1 | Writing or using \(B(15, 0.1)\) |
| \(P(T \leq 1)\) | M1 | Writing \(P(T \leq 1)\) or \(P(T < 2)\); any letter may be used |
| \(= 0.549\) | A1 | awrt 0.549; NB 0.549 gets B1 M1 A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T \sim B(n, 0.1)\); \(T\) can be approximated by \(N(0.1n,\ 0.09n)\) | B1 | mean \(= 0.1n\) and Var \(= 0.09n\) oe; may be seen in an attempt at standardisation |
| \(P\!\left(Z < \frac{8.5 - 0.1n}{\sqrt{0.09n}}\right) = 0.3085\) | M1 | Using a continuity correction either 8.5 or 7.5 in an attempt at standardised form; allow 0.09 for sd |
| B1 | A \(z\) value of awrt \(\pm\, 0.5\) | |
| \(\frac{8.5 - 0.1n}{\sqrt{0.09n}} = -0.5\) or \(\frac{8.5 - 0.1x^2}{0.3x} = -0.5\) | M1 | Standardising using their mean and sd (if not given, must be correct here) and one of 7.5, 8, 8.5, 9 or 9.5 and equal to a \(z\) value where \( |
| A1 | Correct equation in any form; ISW; do not allow if they have \(0.3n\) rather than \(0.3\sqrt{n}\) | |
| \(0.1n - 0.15\sqrt{n} - 8.5 = 0\); \(\sqrt{n} = 10\) | M1A1 | M1 using quadratic formula, completing the square, factorising or any correct method to solve their 3-term quadratic; if they write the formula down allow one slip; if formula written down it must be correct for their equation; may be implied by seeing 10 or 8.5; must show working if equation used is not correct; 2nd A1 awrt 10.0 – do not need to see \(n\) or \(\sqrt{n}\); allow \(n = 10\); may be implied by 100 |
| \(n = 100\) | A1cso | 100; if they have a second answer of 72.25 they must reject it to get this final mark (8 marks) |
## Question 5:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(M < 10) = P\!\left(Z < \frac{12-14}{\sigma}\right) = 0.1$ | | |
| $\Rightarrow \frac{12-14}{\sigma} = -1.2816$ | M1 | Standardising ($\pm$) with 12, 14 and $\sigma$; setting equal to a $z$ value where $|z| > 1$ |
| | B1 | $\pm 1.2816$ or better |
| $\sigma = 1.5605\ldots \approx$ awrt 1.56 minutes | A1 | awrt 1.56; do **not** allow answer written as an exact fraction **(3 marks)** |
### Part (b)
$T$ represents number less than 12 minutes; $T \sim B(15, 0.1)$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \sim B(15, 0.1)$ | B1 | Writing or using $B(15, 0.1)$ |
| $P(T \leq 1)$ | M1 | Writing $P(T \leq 1)$ or $P(T < 2)$; any letter may be used |
| $= 0.549$ | A1 | awrt 0.549; **NB** 0.549 gets B1 M1 A1 **(3 marks)** |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T \sim B(n, 0.1)$; $T$ can be approximated by $N(0.1n,\ 0.09n)$ | B1 | mean $= 0.1n$ and Var $= 0.09n$ oe; may be seen in an attempt at standardisation |
| $P\!\left(Z < \frac{8.5 - 0.1n}{\sqrt{0.09n}}\right) = 0.3085$ | M1 | Using a continuity correction either 8.5 or 7.5 in an attempt at standardised form; allow 0.09 for sd |
| | B1 | A $z$ value of awrt $\pm\, 0.5$ |
| $\frac{8.5 - 0.1n}{\sqrt{0.09n}} = -0.5$ or $\frac{8.5 - 0.1x^2}{0.3x} = -0.5$ | M1 | Standardising using their mean and sd (if not given, must be correct here) **and** one of 7.5, 8, 8.5, 9 or 9.5 **and** equal to a $z$ value where $|z| > 0.4$; allow any form |
| | A1 | Correct equation in **any form**; ISW; do **not** allow if they have $0.3n$ rather than $0.3\sqrt{n}$ |
| $0.1n - 0.15\sqrt{n} - 8.5 = 0$; $\sqrt{n} = 10$ | M1A1 | M1 using quadratic formula, completing the square, factorising or any correct method to solve their 3-term quadratic; if they write the formula down allow one slip; if formula written down it must be correct for their equation; may be implied by seeing 10 or 8.5; must show working if equation used is not correct; **2nd A1** awrt 10.0 – do not need to see $n$ or $\sqrt{n}$; allow $n = 10$; may be implied by 100 |
| $n = 100$ | A1cso | 100; if they have a second answer of 72.25 they must reject it to get this final mark **(8 marks)** |
---5. The time taken for a randomly selected person to complete a test is $M$ minutes, where $M \sim \mathrm {~N} \left( 14 , \sigma ^ { 2 } \right)$
Given that $10 \%$ of people take less than 12 minutes to complete the test,
\begin{enumerate}[label=(\alph*)]
\item find the value of $\sigma$
Graham selects 15 people at random.
\item Find the probability that fewer than 2 of these people will take less than 12 minutes to complete the test.
Jovanna takes a random sample of $n$ people.
Using a normal approximation, the probability that fewer than 9 of these $n$ people will take less than 12 minutes to complete the test is 0.3085 to 4 decimal places.
\item Find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q5 [14]}}