| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This is a straightforward normal distribution question testing standard techniques: z-score calculations, inverse normal for percentiles, and sampling distribution of means. All parts follow routine procedures with no novel problem-solving required, though it's slightly more involved than the most basic single-calculation questions due to multiple parts and the sampling distribution component. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(X \sim N(\mu, 4^2)\), \(\mu = 106\) | ||
| \(P(X < 110) = P\left(Z < \frac{110-106}{4}\right)\) | M1 | Standardising (109.5, 110 or 110.5) with 106 and \((\sqrt{4}, 4\) or \(4^2)\) and/or \((106-x)\) |
| \(= P(Z < 1)\) | A1 | CAO; ignore sign |
| \(= 0.841\) | A1 | AWRT (0.84134) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(\text{underweight}) = P(X < 100)\) | M1 | Use of AFWW 99 to 100 |
| \(= P(Z < -1.5) = 1 - \Phi(1.5)\) | m1 | Area change |
| \(= 1 - 0.93319 = 0.0668\) to \(0.067\) | A1 | AWFW (0.06681) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2\% \Rightarrow z = -2.0537\) | B1 | AWFW 2.05 to 2.06; ignore sign |
| \(z = \frac{100 - \mu}{4}\) | M1 | Standardising AWFW 99 to 100 with \(\mu\) and 4 |
| Thus \(\frac{100 - \mu}{4} = -2.0537\) | m1 | Equating \(z\)-term to \(z\)-value; not using 0.02, 0.98 or \( |
| Thus \(\mu = 108.2\) to \(108.3\) | A1 | AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mu = 108.5\) | B1 | CAO |
| Variance \(= \frac{\sigma^2}{n} = \frac{4^2}{10} = 1.6\) | B1 | CAO; OE |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(\bar{X} > 110) = P\left(Z > \frac{110 - 108.5}{\sqrt{1.6}}\right)\) | M1 | Standardising with \(\mu\) from (i) and \(\left[\sqrt{\frac{\sigma^2}{10}}\right.\) or \(\left.\frac{\sigma^2}{10}\right]\) from (i), and/or \((\mu - x)\) |
| \(= P(Z > 1.19) = 1 - \Phi(1.19)\) | m1 | Area change |
| \(= 0.117\) to \(0.119\) | A1 | AWFW (0.11784) |
# Question 4:
## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $X \sim N(\mu, 4^2)$, $\mu = 106$ | | |
| $P(X < 110) = P\left(Z < \frac{110-106}{4}\right)$ | M1 | Standardising (109.5, 110 or 110.5) with 106 and $(\sqrt{4}, 4$ or $4^2)$ and/or $(106-x)$ |
| $= P(Z < 1)$ | A1 | CAO; ignore sign |
| $= 0.841$ | A1 | AWRT (0.84134) |
## Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(\text{underweight}) = P(X < 100)$ | M1 | Use of AFWW 99 to 100 |
| $= P(Z < -1.5) = 1 - \Phi(1.5)$ | m1 | Area change |
| $= 1 - 0.93319 = 0.0668$ to $0.067$ | A1 | AWFW (0.06681) |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $2\% \Rightarrow z = -2.0537$ | B1 | AWFW 2.05 to 2.06; ignore sign |
| $z = \frac{100 - \mu}{4}$ | M1 | Standardising AWFW 99 to 100 with $\mu$ and 4 |
| Thus $\frac{100 - \mu}{4} = -2.0537$ | m1 | Equating $z$-term to $z$-value; not using 0.02, 0.98 or $|1-z|$ |
| Thus $\mu = 108.2$ to $108.3$ | A1 | AWFW |
## Part (c)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mu = 108.5$ | B1 | CAO |
| Variance $= \frac{\sigma^2}{n} = \frac{4^2}{10} = 1.6$ | B1 | CAO; OE |
## Part (c)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(\bar{X} > 110) = P\left(Z > \frac{110 - 108.5}{\sqrt{1.6}}\right)$ | M1 | Standardising with $\mu$ from (i) and $\left[\sqrt{\frac{\sigma^2}{10}}\right.$ or $\left.\frac{\sigma^2}{10}\right]$ from (i), and/or $(\mu - x)$ |
| $= P(Z > 1.19) = 1 - \Phi(1.19)$ | m1 | Area change |
| $= 0.117$ to $0.119$ | A1 | AWFW (0.11784) |
---4 Chopped lettuce is sold in bags nominally containing 100 grams.\\
The weight, $X$ grams, of chopped lettuce, delivered by the machine filling the bags, may be assumed to be normally distributed with mean $\mu$ and standard deviation 4.
\begin{enumerate}[label=(\alph*)]
\item Assuming that $\mu = 106$, determine the probability that a randomly selected bag of chopped lettuce:
\begin{enumerate}[label=(\roman*)]
\item weighs less than 110 grams;
\item is underweight.
\end{enumerate}\item Determine the minimum value of $\mu$ so that at most 2 per cent of bags of chopped lettuce are underweight. Give your answer to one decimal place.
\item Boxes each contain 10 bags of chopped lettuce. The mean weight of a bag of chopped lettuce in a box is denoted by $\bar { X }$.
Given that $\mu = 108.5$ :
\begin{enumerate}[label=(\roman*)]
\item write down values for the mean and variance of $\bar { X }$;
\item determine the probability that $\bar { X }$ exceeds 110 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2005 Q4 [15]}}