| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a standard S1 normal distribution question covering routine techniques: z-score calculations for probabilities, inverse normal for percentiles, and finding standard deviation from a given probability. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average but still requiring competent execution of multiple standard procedures. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X < 250) = P\left(Z < \dfrac{250-205}{25}\right)\) | M1 | Standardising (249.5, 250 or 250.5) with 205 and (\(\sqrt{25}\), 25 or \(25^2\)) and/or \((205-x)\); CAO ignore sign |
| \(= P(Z < 1.8)\) | A1 | |
| \(= 0.964\) | A1 | AWRT (0.96407) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(200 < X < 250) = \text{(i)} - P(X < 200)\) | M1 | Or equivalent |
| \(= \text{(i)} - P(Z < -0.2) = \text{(i)} - [1 - \Phi(0.2)]\) | M1 | Area change |
| \(= 0.96407 - (1 - 0.57926) = 0.543\) | A1 | AWRT (0.54333) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((100-30)\% = 70\% \Rightarrow z = 0.524\) to \(0.525\) | B1 | AWFW; ignore sign \((-0.5244)\) |
| \(\dfrac{s - 205}{25} = -0.5244\) | M1 | Equating \(z\)-term involving 205 and 25 to \(z\) value; not using 0.3, 0.7 or \( |
| \(s = 191.9\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((100-20)\% = 80\% \Rightarrow z = 0.841\) to \(0.842\) | B1 | AWFW; ignore sign (0.8416) |
| \(\dfrac{m - 205}{25} = 0.8416\) | (M1) | Only if not awarded in (i); not using 0.2 or 0.8; allow \((205-m)\) |
| \(m = 226.0\) | A1 | AWRT; accept 226 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(90\% \Rightarrow z = 1.28\) | B1 | AWRT; ignore sign (1.2816) |
| \(z = \dfrac{200 - 175}{\sigma}\) | M1 | Standardising 200 with 175 and \(\sigma\); do not allow \(175 - 200\) |
| \(\dfrac{25}{\sigma} = 1.2816\) | m1 | Equating \(z\)-term to \(z\)-value; not using 0.9 or 0.1 |
| \(\sigma = 19.5\) | A1 | AWRT |
# Question 2:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X < 250) = P\left(Z < \dfrac{250-205}{25}\right)$ | M1 | Standardising (249.5, 250 or 250.5) with 205 and ($\sqrt{25}$, 25 or $25^2$) and/or $(205-x)$; CAO ignore sign |
| $= P(Z < 1.8)$ | A1 | |
| $= 0.964$ | A1 | AWRT (0.96407) |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(200 < X < 250) = \text{(i)} - P(X < 200)$ | M1 | Or equivalent |
| $= \text{(i)} - P(Z < -0.2) = \text{(i)} - [1 - \Phi(0.2)]$ | M1 | Area change |
| $= 0.96407 - (1 - 0.57926) = 0.543$ | A1 | AWRT (0.54333) |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(100-30)\% = 70\% \Rightarrow z = 0.524$ to $0.525$ | B1 | AWFW; ignore sign $(-0.5244)$ |
| $\dfrac{s - 205}{25} = -0.5244$ | M1 | Equating $z$-term involving 205 and 25 to $z$ value; not using 0.3, 0.7 or $|1-z|$; allow $(205-s)$ |
| $s = 191.9$ | A1 | AWRT |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(100-20)\% = 80\% \Rightarrow z = 0.841$ to $0.842$ | B1 | AWFW; ignore sign (0.8416) |
| $\dfrac{m - 205}{25} = 0.8416$ | (M1) | Only if not awarded in (i); not using 0.2 or 0.8; allow $(205-m)$ |
| $m = 226.0$ | A1 | AWRT; accept 226 |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $90\% \Rightarrow z = 1.28$ | B1 | AWRT; ignore sign (1.2816) |
| $z = \dfrac{200 - 175}{\sigma}$ | M1 | Standardising 200 with 175 and $\sigma$; do not allow $175 - 200$ |
| $\dfrac{25}{\sigma} = 1.2816$ | m1 | Equating $z$-term to $z$-value; not using 0.9 or 0.1 |
| $\sigma = 19.5$ | A1 | AWRT |
---2 The weight, $X$ grams, of a particular variety of orange is normally distributed with mean 205 and standard deviation 25.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that the weight of an orange is:
\begin{enumerate}[label=(\roman*)]
\item less than 250 grams;
\item between 200 grams and 250 grams.
\end{enumerate}\item A wholesaler decides to grade such oranges by weight. He decides that the smallest 30 per cent should be graded as small, the largest 20 per cent graded as large, and the remainder graded as medium.
Determine, to one decimal place, the maximum weight of an orange graded as:
\begin{enumerate}[label=(\roman*)]
\item small;
\item medium.
\end{enumerate}\item The weight, $Y$ grams, of a second variety of orange is normally distributed with mean 175.
Given that 90 per cent of these oranges weigh less than 200 grams, calculate the standard deviation of their weights.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2005 Q2 [15]}}