| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard techniques: z-score calculation, inverse normal lookup, conditional probability, and binomial probability. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part structure and part (c)'s conditional probability element. |
| Spec | 2.03d 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/Working | Marks | Guidance |
| \(P(W<18) = P\left(Z < \frac{10-18}{5.4}\right)\) or \(P(Z < -1.481...)\) | M1 | Standardising 10 with 18 and 5.4 (allow \(\pm\)) |
| \(= 1 - 0.9306\) (calc: 0.069239...) | M1 | For \(1-p\) where \(0.91 < p < 0.95\) |
| \(= 0.0694\) | A1 | Answer in range \(0.0692 \leqslant \text{ans} \leqslant 0.0694\). Ans only 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([P(W>j)=0.15 \implies] \frac{j-18}{5.4} = 1.0364\) | M1B1 | M1 for standardising \(j\) with 18 and 5.4, setting equal to \(z\) value \(1<\ |
| \(j = 23.596...\) awrt 23.6 | A1 | awrt 23.6 (calc 23.596740...). [awrt 23.60 scores 3/3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
\([P(W>18 \mid W<\text{"23.59..."})=] \frac{P(18| M1 |
Correct ratio of probability expressions ft their answer to (b) |
|
| \(= \frac{0.5-0.15}{0.85}\) or \(\frac{0.85-0.5}{0.85} = \frac{0.35}{0.85}\) | M1;A1 | 2nd M1 for ratio of probs of form \(\frac{q}{0.85}\) where \(0.15 |
| \(= \frac{35}{85} = \frac{7}{17}\) or allow awrt 0.412 | A1 | 2nd A1 for \(\frac{7}{17}\) or exact equivalent or allow awrt 0.412 (0.4117647...) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.85^2 \times 0.15^2 \times 6\) | M1dM1 | 1st M1 for \(p^2\times(1-p)^2\times k\) for any positive integer \(k\). 2nd dM1 dep on 1st M1 for \(k=6\) or \(3!\) or \(3\times2\) or 4C2 |
| \(= 0.0975375\) awrt 0.0975 | A1 | Allow exact fraction \(\frac{7803}{80000}\). Ans only 3/3 |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W<18) = P\left(Z < \frac{10-18}{5.4}\right)$ or $P(Z < -1.481...)$ | M1 | Standardising 10 with 18 and 5.4 (allow $\pm$) |
| $= 1 - 0.9306$ (calc: 0.069239...) | M1 | For $1-p$ where $0.91 < p < 0.95$ |
| $= 0.0694$ | A1 | Answer in range $0.0692 \leqslant \text{ans} \leqslant 0.0694$. **Ans only 3/3** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[P(W>j)=0.15 \implies] \frac{j-18}{5.4} = 1.0364$ | M1B1 | M1 for standardising $j$ with 18 and 5.4, setting equal to $z$ value $1<\|z\|<2$. B1 for $z = \pm 1.0364$ or better |
| $j = 23.596...$ awrt **23.6** | A1 | awrt 23.6 (calc 23.596740...). [awrt 23.60 scores 3/3] |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[P(W>18 \mid W<\text{"23.59..."})=] \frac{P(18<W<\text{"23.6"})}{P(W<\text{"23.6"})}$ | M1 | Correct ratio of probability expressions ft their answer to (b) |
| $= \frac{0.5-0.15}{0.85}$ or $\frac{0.85-0.5}{0.85} = \frac{0.35}{0.85}$ | M1;A1 | 2nd M1 for ratio of probs of form $\frac{q}{0.85}$ where $0.15<q<0.5$. 1st A1 for correct ratio using $q=0.35$ |
| $= \frac{35}{85} = \frac{7}{17}$ or allow awrt **0.412** | A1 | 2nd A1 for $\frac{7}{17}$ or exact equivalent or allow awrt 0.412 (0.4117647...) |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.85^2 \times 0.15^2 \times 6$ | M1dM1 | 1st M1 for $p^2\times(1-p)^2\times k$ for any positive integer $k$. 2nd dM1 dep on 1st M1 for $k=6$ or $3!$ or $3\times2$ or 4C2 |
| $= 0.0975375$ awrt **0.0975** | A1 | Allow exact fraction $\frac{7803}{80000}$. **Ans only 3/3** |
---3. The weights of packages that arrive at a factory are normally distributed with a mean of 18 kg and a standard deviation of 5.4 kg
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly selected package weighs less than 10 kg
The heaviest 15\% of packages are moved around the factory by Jemima using a forklift truck.
\item Find the weight, in kg , of the lightest of these packages that Jemima will move.
One of the packages not moved by Jemima is selected at random.
\item Find the probability that it weighs more than 18 kg
A delivery of 4 packages is made to the factory. The weights of the packages are independent.
\item Find the probability that exactly 2 of them will be moved by Jemima.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2021 Q3 [13]}}