| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Single tail probability P(X < a) or P(X > a) |
| Difficulty | Standard +0.3 This is a standard normal distribution question with routine calculations: (a) single z-score and table lookup, (b) stating independence assumption, (c) multiplying probabilities, (d) solving simultaneous equations from two z-scores. All techniques are textbook exercises requiring no novel insight, though part (d) involves slightly more algebraic manipulation than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(H > 1.6) =]\ 0.091211\ldots =\) awrt \(\mathbf{0.0912}\) | B1 | From calculator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need \(H\) and \(T\) to be independent; or events \(\{H > 1.6\}\) and \(\{T < 300\}\) are independent | B1 | For a suitable reason mentioning or implying \(H\) and \(T\) are independent; allow "they"/"each event"/"\(P(H)\) and \(P(T)\)"/"the variables" and "independent"; B0 for "the results"/"the values" are independent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(T < 300) =]\ 0.124(2816\ldots)\) | M1 | For using model for \(T\) to attempt to find \(P(T < 300)\); e.g. sight of 0.124 or better or sight of \(\pm\left(\dfrac{300-330}{26}\right)\) or \(\pm\left(\dfrac{5-5.5}{0.433\ldots}\right)\) or \(Z = \pm 1.15(3\ldots)\) |
| Prob both is: \(\text{"0.0912..."} \times \text{"0.124..."}\) | M1 | For multiplying their two probabilities together ft part (a) and their \(P(T < 300)\); both values are probabilities; NB M0M1 is possible here |
| \(= 0.011335\ldots =\) awrt \(\mathbf{0.0113}\) | A1 | Correct answer with no incorrect working scores 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{16.3 - \mu}{\sigma} = -0.5244(0051\ldots)\ ,\quad \dfrac{29 - \mu}{\sigma} = 1.2816\) (calc: \(1.28155156\ldots\)) | M1M1 | 1st M1: standardising 16.3 and setting equal to \(z\) value where \(0.5 < |
| e.g. \(29 - 16.3 = \sigma(\text{"1.2816"} - \text{"-"} - \text{"0.5244"})\) | M1 | dep on 1st or 2nd M1; for solving their two linear equations – reach an equation in one variable; may be implied by sight of \(\sigma = 7\) or better |
| \(\sigma = 7.032115\ldots =\) awrt \(\mathbf{7.03}\) | A1 | For \(\sigma =\) awrt 7.03 |
| \(\mu = 19.9876\ldots = \mathbf{19.95}\ ,\ \mu\ ,\ \mathbf{20.0}\) | A1 | For \(\mu\) in \([19.95, 20.0]\); allow 20 from equations with suitable \(z\) values |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(H > 1.6) =]\ 0.091211\ldots =$ awrt $\mathbf{0.0912}$ | B1 | From calculator |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Need $H$ and $T$ to be independent; or events $\{H > 1.6\}$ and $\{T < 300\}$ are independent | B1 | For a suitable reason mentioning or implying $H$ and $T$ are independent; allow "they"/"each event"/"$P(H)$ and $P(T)$"/"the variables" and "independent"; B0 for "the results"/"the values" are independent |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(T < 300) =]\ 0.124(2816\ldots)$ | M1 | For using model for $T$ to attempt to find $P(T < 300)$; e.g. sight of 0.124 or better or sight of $\pm\left(\dfrac{300-330}{26}\right)$ or $\pm\left(\dfrac{5-5.5}{0.433\ldots}\right)$ or $Z = \pm 1.15(3\ldots)$ |
| Prob both is: $\text{"0.0912..."} \times \text{"0.124..."}$ | M1 | For multiplying their two probabilities together ft part (a) and their $P(T < 300)$; both values are probabilities; NB M0M1 is possible here |
| $= 0.011335\ldots =$ awrt $\mathbf{0.0113}$ | A1 | Correct answer with no incorrect working scores 3/3 |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{16.3 - \mu}{\sigma} = -0.5244(0051\ldots)\ ,\quad \dfrac{29 - \mu}{\sigma} = 1.2816$ (calc: $1.28155156\ldots$) | M1M1 | 1st M1: standardising 16.3 and setting equal to $z$ value where $0.5 < |z| < 0.6$; 2nd M1: standardising 29 and setting equal to $z$ value where $1 < |z| < 1.5$ |
| e.g. $29 - 16.3 = \sigma(\text{"1.2816"} - \text{"-"} - \text{"0.5244"})$ | M1 | dep on 1st or 2nd M1; for solving their two linear equations – reach an equation in one variable; may be implied by sight of $\sigma = 7$ or better |
| $\sigma = 7.032115\ldots =$ awrt $\mathbf{7.03}$ | A1 | For $\sigma =$ awrt 7.03 |
| $\mu = 19.9876\ldots = \mathbf{19.95}\ ,\ \mu\ ,\ \mathbf{20.0}$ | A1 | For $\mu$ in $[19.95, 20.0]$; allow 20 from equations with suitable $z$ values |
**NB:** Use of $-0.524$ and 1.28 gives $7.0399\ldots$ and $19.988\ldots$ → M3A0A1; use of $-0.524$ and 1.2816 gives $7.033\ldots$ and $19.99\ldots$ → M3A1A1; use of $-0.5244$ and 1.28 gives $7.038\ldots$ and $19.99\ldots$ → M3**A**1A1; both $z$ values correct to 3dp ($-0.524$ and 1.282) gives $7.032$ and $19.984$ → A1A1
---\begin{enumerate}
\item The records for a school athletics club show that the height, $H$ metres, achieved by students in the high jump is normally distributed with mean 1.4 metres and standard deviation 0.15 metres.\\
(a) Find the proportion of these students achieving a height of more than 1.6 metres.
\end{enumerate}
The records also show that the time, $T$ seconds, to run 1500 metres is normally distributed with mean 330 seconds and standard deviation 26 seconds.
The school's Head would like to use these distributions to estimate the proportion of students from the school athletics club who can jump higher than 1.6 metres and can run 1500 metres in less than 5 minutes.\\
(b) State a necessary assumption about $H$ and $T$ for the Head to calculate an estimate of this proportion.\\
(c) Find the Head's estimate of this proportion.
Students in the school athletics club also throw the discus.\\
The random variable $D \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ represents the distance, in metres, that a student can throw the discus.
Given that $\mathrm { P } ( D < 16.3 ) = 0.30$ and $\mathrm { P } ( D > 29.0 ) = 0.10$\\
(d) calculate the value of $\mu$ and the value of $\sigma$
\hfill \mbox{\textit{Edexcel Paper 3 2024 Q5 [10]}}