| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Independent probability calculations |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard z-score calculations and table lookups. Part (a) involves three routine probability calculations with given parameters. Part (b) requires solving simultaneous equations from inverse normal lookups, which is slightly more demanding but still a standard textbook exercise with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X < 1.90) = P\left(Z < \frac{1.90 - 1.81}{0.08}\right) = P(Z < 1.125)\) | M1 | Standardising with \(\mu = 1.81\), \(\sigma = 0.08\) |
| \(= \Phi(1.125) = 0.8697\) | A1 | Accept 0.8694 to 0.8700 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 1.85) = 1 - P\left(Z < \frac{1.85 - 1.81}{0.08}\right) = 1 - \Phi(0.5)\) | M1 | Standardising correctly |
| \(= 1 - 0.6915 = 0.3085\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(1.81 < X < 1.85) = \Phi(0.5) - \Phi(0)\) | M1 | Correct method |
| \(= 0.6915 - 0.5 = 0.1915\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(Y < 9.25) = 0.88 \Rightarrow z = \Phi^{-1}(0.88)\) | M1 | Using inverse normal |
| \(z = 1.175\) | A1 | Accept 1.17 to 1.18 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(Y > 8.75) = 0.975 \Rightarrow P(Y < 8.75) = 0.025\) | M1 | Correct use of symmetry |
| \(\frac{8.75 - \mu}{\sigma} = -1.96\) | A1 | |
| Two simultaneous equations: \(9.25 - \mu = 1.175\sigma\) and \(8.75 - \mu = -1.96\sigma\) | DM1 | Solving simultaneously |
| \(\sigma = \frac{9.25 - 8.75}{1.175 + 1.96} = \frac{0.5}{3.135} = 0.1595\) | A1 | \(\sigma \approx 0.160\) |
| \(\mu = 9.25 - 1.175 \times 0.1595 \approx 9.062\) | A1 | \(\mu \approx 9.06\) |
# Question 5:
## Part 5(a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X < 1.90) = P\left(Z < \frac{1.90 - 1.81}{0.08}\right) = P(Z < 1.125)$ | M1 | Standardising with $\mu = 1.81$, $\sigma = 0.08$ |
| $= \Phi(1.125) = 0.8697$ | A1 | Accept 0.8694 to 0.8700 |
## Part 5(a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 1.85) = 1 - P\left(Z < \frac{1.85 - 1.81}{0.08}\right) = 1 - \Phi(0.5)$ | M1 | Standardising correctly |
| $= 1 - 0.6915 = 0.3085$ | A1 | |
## Part 5(a)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(1.81 < X < 1.85) = \Phi(0.5) - \Phi(0)$ | M1 | Correct method |
| $= 0.6915 - 0.5 = 0.1915$ | A1 | |
## Part 5(b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(Y < 9.25) = 0.88 \Rightarrow z = \Phi^{-1}(0.88)$ | M1 | Using inverse normal |
| $z = 1.175$ | A1 | Accept 1.17 to 1.18 |
## Part 5(b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(Y > 8.75) = 0.975 \Rightarrow P(Y < 8.75) = 0.025$ | M1 | Correct use of symmetry |
| $\frac{8.75 - \mu}{\sigma} = -1.96$ | A1 | |
| Two simultaneous equations: $9.25 - \mu = 1.175\sigma$ and $8.75 - \mu = -1.96\sigma$ | DM1 | Solving simultaneously |
| $\sigma = \frac{9.25 - 8.75}{1.175 + 1.96} = \frac{0.5}{3.135} = 0.1595$ | A1 | $\sigma \approx 0.160$ |
| $\mu = 9.25 - 1.175 \times 0.1595 \approx 9.062$ | A1 | $\mu \approx 9.06$ |5
\begin{enumerate}[label=(\alph*)]
\item Wooden lawn edging is supplied in 1.8 m length rolls. The actual length, $X$ metres, of a roll may be modelled by a normal distribution with mean 1.81 and standard deviation 0.08 .
Determine the probability that a randomly selected roll has length:
\begin{enumerate}[label=(\roman*)]
\item less than 1.90 m ;
\item greater than 1.85 m ;
\item between 1.81 m and 1.85 m .
\end{enumerate}\item Plastic lawn edging is supplied in 9 m length rolls. The actual length, $Y$ metres, of a roll may be modelled by a normal distribution with mean $\mu$ and standard deviation $\sigma$.
An analysis of a batch of rolls, selected at random, showed that
$$\mathrm { P } ( Y < 9.25 ) = 0.88$$
\begin{enumerate}[label=(\roman*)]
\item Use this probability to find the value of $z$ such that
$$9.25 - \mu = z \times \sigma$$
where $z$ is a value of $Z \sim \mathrm {~N} ( 0,1 )$.
\item Given also that
$$\mathrm { P } ( Y > 8.75 ) = 0.975$$
find values for $\mu$ and $\sigma$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q5 [12]}}