| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| 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 standard S1 normal distribution question requiring routine z-score calculations, inverse normal for boundaries, and a straightforward multinomial probability. All techniques are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(W < 120) = P\!\left(Z < \frac{120-165}{35}\right)\) | M1 | For standardising with 120 (allow 210), 165 and 35. Accept \(\pm\) |
| \(= P(Z < -1.2857\ldots) = 1 - 0.9015\) or \(1 - 0.9007285\ldots\) | M1 | For attempting \(1-p\) where \(0.85 < p < 0.95\) |
| \(= 0.09927\ldots =\) awrt \(\mathbf{0.0985{\sim}0.0994}\) | A1 | Correct answer only 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| eg \(P(W > x) = \frac{1}{3}\) gives \(\frac{x-165}{35} = \pm 0.43\) (calculator \(0.430727\ldots\)) | M1B1 | M1 for standardising with \(x\), 165 and 35 and setting equal to a \(z\) value, \(0.4 < |
| Limits \(149.9245\ldots\) to \(180.0754\ldots\) awrt \(\mathbf{150}\) to \(\mathbf{180}\) | A1, A1 | 1st A1 for lower limit awrt 150; 2nd A1 for upper limit awrt 180. SC A0A1 for two limits symmetrically placed about 165 provided M1 scored. Correct answers with no working can score M1B0A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(W < 200\ | \ W > \text{"180"})\) or \(\frac{P(\text{"180"} < W < 200)}{P(W > \text{"180"})}\) or \(\frac{1}{3}\) | M1 |
| \(= \frac{0.8413(44739\ldots) - \frac{2}{3}}{\frac{1}{3}}\) | A1 (num) | For a correct numerator (awrt 0.175) |
| \(= 0.52403\ldots\ (\mathbf{0.523{\sim}0.5264})\) | A1 | For an answer in the range awrt \(0.523{\sim}0.5264\) (use of 180 gives \(0.5263869\ldots\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}\); \(\times 3!\) | M1; M1 | 1st M1 for \(\left(\frac{1}{3}\right)^3\) or equivalent. 2nd M1 for \(p\times 3!\) where \(0 < p < \frac{1}{6}\) |
| \(= \frac{2}{9}\) | A1 | For \(\frac{2}{9}\) or any exact equivalent |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(W < 120) = P\!\left(Z < \frac{120-165}{35}\right)$ | M1 | For standardising with 120 (allow 210), 165 and 35. Accept $\pm$ |
| $= P(Z < -1.2857\ldots) = 1 - 0.9015$ or $1 - 0.9007285\ldots$ | M1 | For attempting $1-p$ where $0.85 < p < 0.95$ |
| $= 0.09927\ldots =$ awrt $\mathbf{0.0985{\sim}0.0994}$ | A1 | Correct answer only 3/3 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| eg $P(W > x) = \frac{1}{3}$ gives $\frac{x-165}{35} = \pm 0.43$ (calculator $0.430727\ldots$) | M1B1 | M1 for standardising with $x$, 165 and 35 and setting equal to a $z$ value, $0.4 < |z| < 0.5$. B1 for use of $z = 0.43$ or better — must see 0.43 or better |
| Limits $149.9245\ldots$ to $180.0754\ldots$ awrt $\mathbf{150}$ to $\mathbf{180}$ | A1, A1 | 1st A1 for lower limit awrt 150; 2nd A1 for upper limit awrt 180. **SC** A0A1 for two limits symmetrically placed about 165 provided M1 scored. Correct answers with no working can score M1B0A1A1 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(W < 200\ |\ W > \text{"180"})$ or $\frac{P(\text{"180"} < W < 200)}{P(W > \text{"180"})}$ or $\frac{1}{3}$ | M1 | For a correct probability statement (either form) ft their 180 or a correct ratio |
| $= \frac{0.8413(44739\ldots) - \frac{2}{3}}{\frac{1}{3}}$ | A1 (num) | For a correct numerator (awrt 0.175) |
| $= 0.52403\ldots\ (\mathbf{0.523{\sim}0.5264})$ | A1 | For an answer in the range awrt $0.523{\sim}0.5264$ (use of 180 gives $0.5263869\ldots$) |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}$; $\times 3!$ | M1; M1 | 1st M1 for $\left(\frac{1}{3}\right)^3$ or equivalent. 2nd M1 for $p\times 3!$ where $0 < p < \frac{1}{6}$ |
| $= \frac{2}{9}$ | A1 | For $\frac{2}{9}$ or any exact equivalent |
---\begin{enumerate}
\item Kris works in the mailroom of a large company and is responsible for all the letters sent by the company. The weights of letters sent by the company, $W$ grams, have a normal distribution with mean 165 g and standard deviation 35 g .\\
(a) Estimate the proportion of letters sent by the company that weigh less than 120 g .
\end{enumerate}
Kris splits the letters to be sent into 3 categories: heavy, medium and light, with $\frac { 1 } { 3 }$ of the letters in each category.\\
(b) Find the weight limits that determine medium letters.
A heavy letter is chosen at random.\\
(c) Find the probability that this letter weighs less than 200 g .
Kris chooses a random sample of 3 letters from those in the mailroom one day.\\
(d) Find the probability that there is one letter in each of the 3 categories.\\
\hfill \mbox{\textit{Edexcel S1 2021 Q4 [13]}}