| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Outliers and box plots |
| Difficulty | Standard +0.8 Part (a) requires solving simultaneous equations using inverse normal distribution to find two unknown parameters (μ and σ), which is non-standard and requires careful manipulation of z-scores. Part (b) combines quartiles with outlier definitions and normal distribution probability, requiring multiple conceptual steps. This goes beyond routine S1 questions which typically give parameters directly. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{162 - \mu}{\sigma} = -1.2816\) (Calculator gives –1.28155…) | M1 A1 | For standardising with \(\mu\) and \(\sigma\) and setting = to a z value with \( |
| or \(\frac{175 - \mu}{\sigma} = 1.04\) (Calculator gives 1.03987…) | ||
| \(\mu - 1.2816\sigma = 162\) | ||
| \(\mu + 1.04\sigma = 175\) | ||
| \(2.3216\sigma = 13\) | dM1 | Dependent on previous M1. For solving their 2 linear simultaneous equations (Can be implied by both correct answers) If answers are incorrect then we need to see evidence of correct working |
| \(\sigma = 5.5995...\) awrt 5.6 | A1 | \(\mu = 169.1764...\) awrt 169 |
| (b) \(Q_1 = 208.26\) or \(Q_3 - Q_1 = 13.48\) | B1 | For \(Q_1 = 208.26\) or \(Q_3 - Q_1 = 13.48\) Do not accept rounded values |
| \(221.74 + 1.5(221.74 - 'Q_1') = '241.96'\) or \('Q_1' - 1.5(221.74 - 'Q_1') = '188.04'\) | M1 | For a correct method for finding 1 outlier limit, ft their \(Q_1\) or their IQR. You will need to check that these are correct if no working shown. |
| Probability of an outlier | 1 – Probability of not an outlier | dM1 |
| \(\left[P\left(B > \text{'241.96'}\right) =\right]\) | \(\left[P\left(\text{'188.04'} < B < \text{'241.96'}\right) =\right]\) | |
| \(P\left(Z > \frac{\text{241.96'} - 215}{10}\right) \left[= P(Z > 2.70)\right]\) | \(P\left(\frac{\text{'188.04'} - 215}{10} < Z < \frac{\text{'241.96'} - 215}{10}\right)\) | |
| or \(\left[P\left(B < \text{'188.04'}\right) =\right]\) | \(\left[= P(-2.70) < Z < P(2.70)\right]\) | |
| \(P\left(Z < \frac{\text{'188.04'} - 215}{10}\right) \left[= P(Z < -2.70)\right]\) | ||
| or | ||
| \(= 0.0035\) (Calculator gives 0.0034883…) | \(= 0.993\) (Calculator gives 0.99298…) | M1 |
| \(P(\text{Outlier}) = 2 \times \text{'0.0035'}\) | \(P(\text{Outlier}) = 1 - \text{'0.993'}\) | A1 |
| \(= 0.007\) (Calculator gives 0.006976…) | \(= 0.007\) (Calculator gives 0.007017…) | If using the RHS of the MS: for 1 – their probability |
| awrt 0.007 | awrt 0.007 |
**(a)** $\frac{162 - \mu}{\sigma} = -1.2816$ (Calculator gives –1.28155…) | M1 A1 | For standardising with $\mu$ and $\sigma$ and setting = to a z value with $|z| > 1$
or $\frac{175 - \mu}{\sigma} = 1.04$ (Calculator gives 1.03987…) | |
$\mu - 1.2816\sigma = 162$ | |
$\mu + 1.04\sigma = 175$ | |
$2.3216\sigma = 13$ | dM1 | Dependent on previous M1. For solving their 2 linear simultaneous equations (Can be implied by both correct answers) If answers are incorrect then we need to see evidence of correct working
$\sigma = 5.5995...$ awrt 5.6 | A1 | $\mu = 169.1764...$ awrt 169
**(b)** $Q_1 = 208.26$ or $Q_3 - Q_1 = 13.48$ | B1 | For $Q_1 = 208.26$ or $Q_3 - Q_1 = 13.48$ Do not accept rounded values
$221.74 + 1.5(221.74 - 'Q_1') = '241.96'$ or $'Q_1' - 1.5(221.74 - 'Q_1') = '188.04'$ | M1 | For a correct method for finding 1 outlier limit, ft their $Q_1$ or their IQR. You will need to check that these are correct if no working shown.
| |
**Probability of an outlier** | **1 – Probability of not an outlier** | dM1 | Dependent on previous M1. For standardising using their limit(s), 215 and 10 (allow ±) May be implied by ± awrt 2.70 or awrt 0.0035 or awrt 0.993 or final answer of awrt 0.007
$\left[P\left(B > \text{'241.96'}\right) =\right]$ | $\left[P\left(\text{'188.04'} < B < \text{'241.96'}\right) =\right]$ | |
$P\left(Z > \frac{\text{241.96'} - 215}{10}\right) \left[= P(Z > 2.70)\right]$ | $P\left(\frac{\text{'188.04'} - 215}{10} < Z < \frac{\text{'241.96'} - 215}{10}\right)$ | |
or $\left[P\left(B < \text{'188.04'}\right) =\right]$ | $\left[= P(-2.70) < Z < P(2.70)\right]$ | |
$P\left(Z < \frac{\text{'188.04'} - 215}{10}\right) \left[= P(Z < -2.70)\right]$ | |
or | |
$= 0.0035$ (Calculator gives 0.0034883…) | $= 0.993$ (Calculator gives 0.99298…) | M1 | If using the LHS of the MS: for multiplying their probability by 2
$P(\text{Outlier}) = 2 \times \text{'0.0035'}$ | $P(\text{Outlier}) = 1 - \text{'0.993'}$ | A1 | or
$= 0.007$ (Calculator gives 0.006976…) | $= 0.007$ (Calculator gives 0.007017…) | If using the RHS of the MS: for 1 – their probability
awrt 0.007 | awrt 0.007 | | May be implied by awrt 0.007
**Total: 10 marks**\begin{enumerate}
\item An orchard produces apples.
\end{enumerate}
The weights, $A$ grams, of its apples are normally distributed with mean $\mu$ grams and standard deviation $\sigma$ grams.
It is known that
$$\mathrm { P } ( A < 162 ) = 0.1 \text { and } \mathrm { P } ( 162 < A < 175 ) = 0.7508$$
(a) Calculate the value of $\mu$ and the value of $\sigma$
A second orchard also produces apples.\\
The weights, $B$ grams, of its apples have distribution $B \sim N \left( 215,10 ^ { 2 } \right)$\\
An outlier is a value that is\\
greater than $\mathrm { Q } _ { 3 } + 1.5 \times \left( \mathrm { Q } _ { 3 } - \mathrm { Q } _ { 1 } \right)$ or smaller than $\mathrm { Q } _ { 1 } - 1.5 \times \left( \mathrm { Q } _ { 3 } - \mathrm { Q } _ { 1 } \right)$\\
An apple is selected at random from this second orchard.\\
Using $\mathrm { Q } _ { 3 } = 221.74$ grams,\\
(b) find the probability that this apple is an outlier.
\hfill \mbox{\textit{Edexcel S1 2024 Q8}}