| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution tables requiring standardization to z-scores and basic probability multiplication for independent events. Part (a) involves routine table lookups, and part (b) simply requires raising a probability to the 6th power. No problem-solving insight needed, just mechanical application of standard techniques taught early in S1. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 10.5) = P\left(Z < \frac{10.5-10.2}{0.15}\right) = P(Z < 2) = 0.977\) | M1, A1, A1 | Standardising (10.45, 10.5 or 10.55) with 10.2 and \(\sqrt{0.15}, 0.15 \text{ or } 0.15^2\) and/or \((10.2 - x)\); CAO; ignore inequality and sign. May be implied by a correct answer. AWRT (0.97725) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(10.0 < X < 10.5) = [C's (a)(i)] - P(X < 10.0) = (a)(i) - P(Z < -1.33) = (a)(i) - (1 - p) = 0.97725 - (1 - 0.90824) = 0.885 \text{ to } 0.887\) | M1, m1, A1 | Or equivalent; must be clear correct method if answer incorrect and answer > 0. Method correct using \(-1.3\) gives 0.88 to 0.881 \(\Rightarrow\) M1 m1 A0. Area change. May be implied by a correct answer or answer > 0.5. AWFW (0.88604). M1 m1 A1 for 0.90824 \(-\) [1 \(-\) (a)(i)] = 0.0886. M1 m0 A0 for (a)(i) \(-\) 0.90824 = 0.0685. M0 m0 A0 for answer < 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.56 \text{ to } 0.565\) | B1F, M1, A1 | Correct value or F on value used or implied in (a)(ii) providing > 0.5. Use of \(-1.3\) gives 0.9032. Accept any probability to power 6. AWFW |
## Part (a)(i)
$X \sim N(10.2, 0.15^2)$
$P(X < 10.5) = P\left(Z < \frac{10.5-10.2}{0.15}\right) = P(Z < 2) = 0.977$ | M1, A1, A1 | Standardising (10.45, 10.5 or 10.55) with 10.2 and $\sqrt{0.15}, 0.15 \text{ or } 0.15^2$ and/or $(10.2 - x)$; CAO; ignore inequality and sign. May be implied by a correct answer. AWRT (0.97725)
## Part (a)(ii)
$P(10.0 < X < 10.5) = [C's (a)(i)] - P(X < 10.0) = (a)(i) - P(Z < -1.33) = (a)(i) - (1 - p) = 0.97725 - (1 - 0.90824) = 0.885 \text{ to } 0.887$ | M1, m1, A1 | Or equivalent; must be clear correct method if answer incorrect and answer > 0. Method correct using $-1.3$ gives 0.88 to 0.881 $\Rightarrow$ M1 m1 A0. Area change. May be implied by a correct answer or answer > 0.5. AWFW (0.88604). M1 m1 A1 for 0.90824 $-$ [1 $-$ (a)(i)] = 0.0886. M1 m0 A0 for (a)(i) $-$ 0.90824 = 0.0685. M0 m0 A0 for answer < 0
## Part (b)
$P(X > 10) = p[$from (a)(ii)$] = 0.908 \text{ to } 0.909$
$P(6 \text{ rolls} > 10) = 0.90824^6$
$0.56 \text{ to } 0.565$ | B1F, M1, A1 | Correct value or F on value used or implied in (a)(ii) providing > 0.5. Use of $-1.3$ gives 0.9032. Accept any probability to power 6. AWFW
Note: B0F M1 A0 is possible
---1 Draught excluder for doors and windows is sold in rolls of nominal length 10 metres.\\
The actual length, $X$ metres, of draught excluder on a roll may be modelled by a normal distribution with mean 10.2 and standard deviation 0.15 .
\begin{enumerate}[label=(\alph*)]
\item Determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 10.5 )$;
\item $\mathrm { P } ( 10.0 < X < 10.5 )$.
\end{enumerate}\item A customer randomly selects six 10 -metre rolls of the draught excluder.
Calculate the probability that all six rolls selected contain more than 10 metres of draught excluder.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2010 Q1 [9]}}