| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Multiple probability calculations only |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution tables requiring only standardization (z-score calculation) and table lookup for parts (a)-(c), with part (d) testing understanding that P(X=k)=0 for continuous distributions. All calculations are routine with no problem-solving or conceptual challenges beyond basic S1 technique. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(V < 110) = P\!\left(Z < \dfrac{110-106}{2.5}\right)\) | M1 | Standardising 110 with 106 and 2.5; allow \((106-110)\) |
| \(= P(Z < 1.6)\) | A1 | CAO; ignore inequality and sign; may be implied by a correct answer |
| \(= 0.945\) | A1 | AWRT (0.94520) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(V > 100) = P(Z > -2.4) = P(Z < +2.4)\) | M1 | Correct area change; may be implied by a correct answer or by an answer \(> 0.5\) |
| \(= 0.991\) to \(0.992\) | A1 | AWFW (0.99180) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= P(Z < a) - (1 - P(Z < a))\) or \(2 \times P(Z < a) - 1\) | M1 | OE; \(a = 0.8\) is not a requirement; may be implied by \(0.788\) seen or by a correct answer |
| \(= 0.788 - (1 - 0.788) = 0.788 - 0.212\) or \(= 2 \times 0.788 - 1\) | A1 | AWRT (0.78814/0.21186); Condone 0.211; may be implied by a correct answer |
| \(= 0.576\) | A1 | AWRT (0.57628) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(V \neq 106) = 1\) or one or unity or 100% | B1 | CAO; accept nothing else but ignore additional words providing they are not contradictory (eg certain so \(= 1\)) |
# Question 2:
## Part (a)
$V \sim N(106,\ 2.5^2)$
$P(V < 110) = P\!\left(Z < \dfrac{110-106}{2.5}\right)$ | M1 | Standardising 110 with 106 and 2.5; allow $(106-110)$
$= P(Z < 1.6)$ | A1 | CAO; ignore inequality and sign; may be implied by a correct answer
$= 0.945$ | A1 | AWRT (0.94520)
## Part (b)
$P(V > 100) = P(Z > -2.4) = P(Z < +2.4)$ | M1 | Correct area change; may be implied by a correct answer or by an answer $> 0.5$
$= 0.991$ **to** $0.992$ | A1 | AWFW (0.99180)
## Part (c)
$P(104 < V < 108) = P(-a < Z < a)$
$= P(Z < a) - (1 - P(Z < a))$ **or** $2 \times P(Z < a) - 1$ | M1 | OE; $a = 0.8$ is **not** a requirement; may be implied by $0.788$ seen or by a correct answer
$= 0.788 - (1 - 0.788) = 0.788 - 0.212$ **or** $= 2 \times 0.788 - 1$ | A1 | AWRT (0.78814/0.21186); Condone 0.211; may be implied by a correct answer
$= 0.576$ | A1 | AWRT (0.57628)
## Part (d)
$P(V \neq 106) = 1$ **or one or unity or 100%** | B1 | CAO; accept nothing else but ignore additional words providing they are not contradictory (eg certain so $= 1$)
---2 The volume of Everwhite toothpaste in a pump-action dispenser may be modelled by a normal distribution with a mean of 106 ml and a standard deviation of 2.5 ml .
Determine the probability that the volume of Everwhite in a randomly selected dispenser is:
\begin{enumerate}[label=(\alph*)]
\item less than 110 ml ;
\item more than 100 ml ;
\item between 104 ml and 108 ml ;
\item not exactly 106 ml .
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q2 [9]}}