| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution techniques covering standardization, use of tables, sampling distributions, and inverse normal problems. All parts follow standard S1 procedures with no novel problem-solving required, making it slightly easier than average but still requiring competent execution across multiple related concepts. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 5) = P\left(Z < \frac{5-5.08}{0.05}\right) = P(Z < -1.6) = 1 - P(Z < 1.6) = 1 - 0.9452 = 0.0545\) to \(0.055\) | M1 (m1) (A1) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(5 < X < 5.10) = P(X < 5.10) - (i) = P(Z < 0.4) - (i) = 0.65542 - 0.0548 = 0.6\) to \(0.601\) | M1 (A1) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Variance of \(\bar{X}_1 = 0.05^2/4 = 0.000625\); SD of \(\bar{X}_1 = 0.05/2 = 0.025\) | B1 | |
| \(P(\bar{X}_1 > 5.05) = P\left(Z > \frac{5.05-5.08}{0.025}\right) = P(Z > -1.2) = P(Z < 1.2) = 0.884\) to \(0.886\) | M1 (m1) (A1) | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Zero | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1\% (0.01) ⟹ z = -2.33\) to \(-2.32\) \(z = \frac{5-\mu}{0.05} = -2.3263 = -2.3263\) \(\mu = 5.11\) to \(5.12\) | B1 (M1) (A1) (A1) | |
| Note: \(\frac{5-\mu}{0.05} = 2.3263 ⟹ 5.116 ⟹ B1 M1 A1 A0\) |
**3(a)(i)**
| $P(X < 5) = P\left(Z < \frac{5-5.08}{0.05}\right) = P(Z < -1.6) = 1 - P(Z < 1.6) = 1 - 0.9452 = 0.0545$ to $0.055$ | M1 (m1) (A1) | 3 | Standardising (4.5, 4.95, 5, 5.05 or 5.5) with 5.08 and ($\sqrt{0.05}$, 0.05 or 0.05²) and/or (5.08 – x); (1 – answer) ⟹ M1 max; Or equivalent; must be clear correct method if answer incorrect and answer > 0 |
**3(a)(ii)**
| $P(5 < X < 5.10) = P(X < 5.10) - (i) = P(Z < 0.4) - (i) = 0.65542 - 0.0548 = 0.6$ to $0.601$ | M1 (A1) | 2 | AWFW (0.60062) |
**3(b)(i)**
| Variance of $\bar{X}_1 = 0.05^2/4 = 0.000625$; SD of $\bar{X}_1 = 0.05/2 = 0.025$ | B1 | | CAO; stated or used |
| $P(\bar{X}_1 > 5.05) = P\left(Z > \frac{5.05-5.08}{0.025}\right) = P(Z > -1.2) = P(Z < 1.2) = 0.884$ to $0.886$ | M1 (m1) (A1) | 4 | Standardising 5.05 with 5.08 and 0.025; allow (5.08 – 5.05); Area change; may be implied; AWFW (1 – answer) ⟹ B1 M1 max |
**3(b)(ii)**
| Zero | B1 | 1 | CAO; or equivalent (ignore any working) |
**3(c)**
| $1\% (0.01) ⟹ z = -2.33$ to $-2.32$ $z = \frac{5-\mu}{0.05} = -2.3263 = -2.3263$ $\mu = 5.11$ to $5.12$ | B1 (M1) (A1) (A1) | | AWFW; ignore sign (–2.3263); Standardising 5 with μ and 0.05 or 0.025; allow (μ – 5); Only allow: ±2.05 to ±2.06; ±2.32 to ±2.33; ±2.57 to ±2.58; AWFW (5.1163) |
| Note: $\frac{5-\mu}{0.05} = 2.3263 ⟹ 5.116 ⟹ B1 M1 A1 A0$ | | | Or equivalent inconsistent signs |
**Total for Q3: 14 marks**
---3 UPVC facia board is supplied in lengths labelled as 5 metres. The actual length, $X$ metres, of a board may be modelled by a normal distribution with a mean of 5.08 and a standard deviation of 0.05 .
\begin{enumerate}[label=(\alph*)]
\item Determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 5 )$;
\item $\mathrm { P } ( 5 < X < 5.10 )$.
\end{enumerate}\item Determine the probability that the mean length of a random sample of 4 boards:
\begin{enumerate}[label=(\roman*)]
\item exceeds 5.05 metres;
\item is exactly 5 metres.
\end{enumerate}\item Assuming that the value of the standard deviation remains unchanged, determine the mean length necessary to ensure that only 1 per cent of boards have lengths less than 5 metres.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2009 Q3 [14]}}