| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: finding proportions using z-scores, inverse normal calculations, and conditional probability. Part (d) involves basic binomial probability. All parts are routine S1 applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(P(D > 4.3) = P\left(Z > \frac{4.3-3.8}{0.9}\right)\) or \(P(Z > 0.555...)\) | M1 | Standardising 4.3 with 3.8 and 0.9 (allow ±) |
| \(= 1 - 0.7123\) (tables) | M1 | For \(1-p\) where \(0.7 < p < 0.8\) |
| \(= 0.2877\) or awrt 0.288 or awrt 0.289 | A1 | Correct answer only 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{d-3.8}{0.9} = -0.8416\) (calc \(-0.84162123...\)) | M1; B1 | M1 standardising with \(d\), 3.8 and 0.9 and setting equal to \(z\) value \(0.8 < |
| \(d = 3.0425...\) awrt 3.04 | A1 | Condone \(d \geq ...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(P(D > g \mid D > 4.3) = \frac{P(D>g)}{P(D > 4.3) \text{ or (a)}} \left[=\frac{1}{3}\right]\) | M1 | For either expression for conditional prob; ft sight of \(\frac{1}{3}\)(a) to 2 sf |
| \(\therefore P(D > g) = \frac{1}{3}\)(a) \(= 0.096419...\) | A1ft (o.e.) | For \(P(D>g) = 0.096\) or better |
| \(\frac{g-3.8}{0.9} = 1.302228...\) | dM1 | Dep on 1st M1; standardising with \(g\), 3.8 and 0.9, put equal to \(z\) where \( |
| \(g = 4.97200...\) awrt 4.97 or awrt 4.98 | A1 | Condone \(g \geq ...\); correct answer with no incorrect working 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(P(\text{no gold medals}) = \left(\frac{2}{3}\right)^3\) | M1 | For correct probability of no gold medals or 2 of: \(3\left(\frac{2}{3}\right)^2 \times \frac{1}{3}\) or \(3\left(\frac{1}{3}\right)^2 \times \frac{2}{3}\) or \(\left(\frac{1}{3}\right)^3\) |
| \(P(\text{at least one gold}) = 1-\left(\frac{2}{3}\right)^3\) | M1 | For \(1-p^3\) or \(3(p)^2(1-p)+3p(1-p)^2+(1-p)^3\) where \(0 |
| \(= \frac{19}{27}\) | A1 | For \(\frac{19}{27}\) (or exact equivalent) only e.g. \(0.7\dot{0}\dot{3}\) |
# Question 3:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $P(D > 4.3) = P\left(Z > \frac{4.3-3.8}{0.9}\right)$ or $P(Z > 0.555...)$ | M1 | Standardising 4.3 with 3.8 and 0.9 (allow ±) |
| $= 1 - 0.7123$ (tables) | M1 | For $1-p$ where $0.7 < p < 0.8$ |
| $= 0.2877$ or awrt **0.288** or awrt **0.289** | A1 | Correct answer only 3/3 |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{d-3.8}{0.9} = -0.8416$ (calc $-0.84162123...$) | M1; B1 | M1 standardising with $d$, 3.8 and 0.9 and setting equal to $z$ value $0.8 < |z| < 0.9$; B1 for $z = \pm 0.8416$ or better |
| $d = 3.0425...$ awrt **3.04** | A1 | Condone $d \geq ...$ |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $P(D > g \mid D > 4.3) = \frac{P(D>g)}{P(D > 4.3) \text{ or (a)}} \left[=\frac{1}{3}\right]$ | M1 | For either expression for conditional prob; ft sight of $\frac{1}{3}$(a) to 2 sf |
| $\therefore P(D > g) = \frac{1}{3}$(a) $= 0.096419...$ | A1ft (o.e.) | For $P(D>g) = 0.096$ or better |
| $\frac{g-3.8}{0.9} = 1.302228...$ | dM1 | Dep on 1st M1; standardising with $g$, 3.8 and 0.9, put equal to $z$ where $|z|>1$ |
| $g = 4.97200...$ awrt **4.97** or awrt **4.98** | A1 | Condone $g \geq ...$; correct answer with no incorrect working 4/4 |
## Part (d):
| Working | Marks | Notes |
|---------|-------|-------|
| $P(\text{no gold medals}) = \left(\frac{2}{3}\right)^3$ | M1 | For correct probability of no gold medals or 2 of: $3\left(\frac{2}{3}\right)^2 \times \frac{1}{3}$ or $3\left(\frac{1}{3}\right)^2 \times \frac{2}{3}$ or $\left(\frac{1}{3}\right)^3$ |
| $P(\text{at least one gold}) = 1-\left(\frac{2}{3}\right)^3$ | M1 | For $1-p^3$ or $3(p)^2(1-p)+3p(1-p)^2+(1-p)^3$ where $0<p<1$ |
| $= \frac{19}{27}$ | A1 | For $\frac{19}{27}$ (or exact equivalent) only e.g. $0.7\dot{0}\dot{3}$ |
---3. The distance achieved in a long jump competition by students at a school is normally
Students who achieve a distance greater than 4.3 metres receive a medal.
\begin{enumerate}[label=(\alph*)]
\item Find the proportion of students who receive medals.
The school wishes to give a certificate of achievement or a medal to the $80 \%$ of students who achieve a distance of at least $d$ metres.
\item Find the value of $d$.
Of those who received medals, the $\frac { 1 } { 3 }$ who jump the furthest will receive gold medals.
\item Find the shortest distance, $g$ metres, that must be achieved to receive a gold medal.
A journalist from the local newspaper interviews a randomly selected group of 3 medal winners.
\item Find the exact probability that there is at least one gold medal winner in the group.\\
\includegraphics[max width=\textwidth, alt={}, center]{81d5e460-9559-4d25-aa08-6440559aec83-08_79_1153_233_251}
Students who achieve a distance greater than 4.3 metres receive a medal.\\
(a) Find the proportion of students who receive medals.
\begin{center}
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel S1 2020 Q3 [13]}}