| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Outliers and box plots |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard techniques: using tables/calculator for probabilities (part a), inverse normal for quartiles (part b), arithmetic with the IQR formula (part c), standardisation (part d), and conditional probability (part e). While multi-part with 5 sections, each step uses routine methods with no novel insight required, making it slightly easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.03d Calculate conditional probability: from first principles2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(L < 3.86) = P\!\left(Z < \frac{3.86-4.5}{0.4}\right)\) | M1 | For standardising with 3.86, 4.5 and 0.4 |
| \(= P(Z < -1.6) = 1 - 0.9452\) or \(1-0.945200... = 0.0548\) | M1 A1 | For \(1-p\) where \(0.5 < p < 1\); awrt 0.0548 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
\(P(L < Q_3)=0.75\) gives \(\frac{Q_3-4.5}{0.4}=0.67\) or \(P(L| M1 B1 |
For standardising with \(Q_3\) or \(Q_1\); for use of 0.67, \( |
|
| \(Q_3 = 4.768\) awrt 4.77 or \(Q_1 = 4.232\) awrt 4.23 | A1 | awrt 4.77 or awrt 4.23 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Q_1 = 4.232\) awrt 4.23 or \(Q_3 = 4.768\) awrt 4.77 | B1ft | ft their part (b)(i); check \(Q_1+Q_3=9\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1.5(Q_3 - Q_1) = 0.804\) or \((0.81)\) | M1 | Use of \(1.5(Q_3-Q_1)\) ft their \(Q_3\) and \(Q_1\) |
| Lower limit \(= 3.428\) \((3.42-3.43)\); Upper limit \(= 5.572\) \((5.57-5.58)\) | A1 A1 | awrt 3.42 to 3.43; awrt 5.57 to 5.58 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(3.42 < L < 5.58) = P\!\left(\frac{3.42-4.5}{0.4} < Z < \frac{5.58-4.5}{0.4}\right)\) | M1 A1ft | For correct standardisation of their 3.42 or 5.58 |
| \(= P(-2.7 < Z < 2.7) = 0.9930\)* | A1* | Allow 0.9930 or \(0.9965-0.0035\) oe; fully correct solution required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(5 < L < 5.58) = P\!\left(\frac{5-4.5}{0.4} < Z < \frac{5.58-4.5}{0.4}\right) = 0.1021\) | M1 A1 | For writing or using \(P(5 < L < 5.58)\); awrt 0.102 |
| \(P(L>5 \mid 3.42 < L < 5.58) = \frac{P(5 < L < 5.58)}{P(3.42 < L < 5.58)} = \frac{0.102}{0.993}\) | M1 | For correct conditional probability statement |
| \(= 0.1027\) awrt 0.103 | A1 | awrt 0.103 |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(L < 3.86) = P\!\left(Z < \frac{3.86-4.5}{0.4}\right)$ | M1 | For standardising with 3.86, 4.5 and 0.4 |
| $= P(Z < -1.6) = 1 - 0.9452$ or $1-0.945200... = 0.0548$ | M1 A1 | For $1-p$ where $0.5 < p < 1$; awrt 0.0548 |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(L < Q_3)=0.75$ gives $\frac{Q_3-4.5}{0.4}=0.67$ or $P(L<Q_1)=0.25$ gives $\frac{Q_1-4.5}{0.4}=-0.67$ | M1 B1 | For standardising with $Q_3$ or $Q_1$; for use of 0.67, $|z|$, 0.675 |
| $Q_3 = 4.768$ awrt 4.77 or $Q_1 = 4.232$ awrt 4.23 | A1 | awrt 4.77 or awrt 4.23 |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Q_1 = 4.232$ awrt 4.23 or $Q_3 = 4.768$ awrt 4.77 | B1ft | ft their part (b)(i); check $Q_1+Q_3=9$ |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $1.5(Q_3 - Q_1) = 0.804$ or $(0.81)$ | M1 | Use of $1.5(Q_3-Q_1)$ ft their $Q_3$ and $Q_1$ |
| Lower limit $= 3.428$ $(3.42-3.43)$; Upper limit $= 5.572$ $(5.57-5.58)$ | A1 A1 | awrt 3.42 to 3.43; awrt 5.57 to 5.58 |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(3.42 < L < 5.58) = P\!\left(\frac{3.42-4.5}{0.4} < Z < \frac{5.58-4.5}{0.4}\right)$ | M1 A1ft | For correct standardisation of their 3.42 or 5.58 |
| $= P(-2.7 < Z < 2.7) = 0.9930$* | A1* | Allow 0.9930 or $0.9965-0.0035$ oe; fully correct solution required |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(5 < L < 5.58) = P\!\left(\frac{5-4.5}{0.4} < Z < \frac{5.58-4.5}{0.4}\right) = 0.1021$ | M1 A1 | For writing or using $P(5 < L < 5.58)$; awrt 0.102 |
| $P(L>5 \mid 3.42 < L < 5.58) = \frac{P(5 < L < 5.58)}{P(3.42 < L < 5.58)} = \frac{0.102}{0.993}$ | M1 | For correct conditional probability statement |
| $= 0.1027$ awrt 0.103 | A1 | awrt 0.103 |\begin{enumerate}
\item The lengths, $L \mathrm {~mm}$, of housefly wings are normally distributed with $L \sim \mathrm {~N} \left( 4.5,0.4 ^ { 2 } \right)$\\
(a) Find the probability that a randomly selected housefly has a wing length of less than 3.86 mm .\\
(b) Find\\
(i) the upper quartile ( $Q _ { 3 }$ ) of $L$\\
(ii) the lower quartile ( $Q _ { 1 }$ ) of $L$
\end{enumerate}
A value that is greater than $Q _ { 3 } + 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$ or smaller than $Q _ { 1 } - 1.5 \times \left( Q _ { 3 } - Q _ { 1 } \right)$ is defined as an outlier.\\
(c) Find these two outlier limits.
A housefly is selected at random.\\
(d) Using standardisation, show that the probability that this housefly is not an outlier is 0.993 to 3 decimal places.
Given that this housefly is not an outlier,\\
(e) showing your working, find the probability that the wing length of this housefly is greater than 5 mm .
\hfill \mbox{\textit{Edexcel S1 2023 Q5 [17]}}