| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | October |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Finding unknown boundaries |
| Difficulty | Standard +0.3 This is a multi-part normal distribution question requiring standardization, inverse normal calculations, and ratio work. Part (a) is routine verification, (b) requires finding a percentile (standard S1 skill), (c) involves working with ratios to find boundaries within a conditional distribution (slightly more demanding), and (d) is straightforward arithmetic. While it has multiple parts and requires careful organization, all techniques are standard S1 material with no novel problem-solving required. Slightly easier than average due to the guided structure and routine nature of each component. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(H \sim N(25.1, 5.5^2)\); \(P(H > 30.8) = P\!\left(Z > \frac{30.8 - 25.1}{5.5}\right)\) or \(P(Z > 1.03636...)\) | M1 | Standardising 30.8 with 25.1 and 5.5 (allow \(\pm\)); allow \(z = 1.04\) |
| \(= 1 - 0.8508\) | M1 | For \(1 - p\) where \(0.84 < p < 0.86\) |
| \(= 0.1492\) or better (calc: 0.1500...) | A1cso | Answer of 0.1492 or better with evidence of both M's |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{y - 25.1}{5.5} = -1.6449\) | M1B1 | Standardising \(y\) with 25.1 and 5.5; use of \(z = \pm 1.6449\) or better |
| \(y = 16.053...\), so range is awrt 16.1 ~ 30.8 | A1 | For awrt 16.1 (ISW) or range [16.1, 30.8] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(H < d) = 0.05 + 0.2 + 0.3 [= 0.55]\) | M1 | Correct method to calculate \(P(H < d)\) |
| \(\frac{d - 25.1}{5.5} = 0.13\) (Calc 0.12566) | M1 | Correct standardisation \(= z\) where \(0.125 \leq |
| \(d = 0.13 \times 5.5 + 25.1 = 25.815\) (25.791... calc) | A1cso | Both method marks awarded, no errors seen and awrt 25.82 or awrt 25.79 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(H < m) = 0.05 + 0.2 [= 0.25]\) | M1 | Correct method for \(P(H < m)\); allow \(0.05 +\) awrt 0.200 |
| \(\frac{m - 25.1}{5.5} = -0.67\) (Calc 0.674489) | M1M1 | Standardising \(m\) with 25.1 and 5.5 and setting equal to awrt \(-0.67\) |
| \(m =\) awrt 21.4 | A1 | For \(m =\) awrt 21.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Height \(= 2 \times \text{"m"} + 3 \times 25.8 + 3 \times 30.8\,[+8]\) | M1 | For \(2 \times \text{"m"} + 3 \times 25.8 + 3 \times 30.8\,[+n]\) where \(n\) is integer \(\geq 0\) |
| \(= 220.6\) awrt 221 (cm) | A1 | For awrt 221 (cm) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(H > 25.8 \mid "16.1" < H < 30.8)\) or \(\dfrac{P(25.8 < H < 30.8)}{1-(0.15+0.05)}\) | M1 | 1st M1: for a correct conditional probability statement ft their answer to (b) i.e. their \(y\) |
| \(= \dfrac{0.8508 - 0.5517}{0.8}\) (tables) or \(\dfrac{0.299345...}{0.8}\) (calc) \(\approx \dfrac{3}{8}\) | M1 A1cso | 2nd M1: for a ratio of probs of the form \(\dfrac{q}{0.8}\) where \(q = 0.3\) to 1sf |
| (3) | A1: for probability of approx \(\dfrac{3}{8}\) |
# Question 6:
## Part (a)
| $H \sim N(25.1, 5.5^2)$; $P(H > 30.8) = P\!\left(Z > \frac{30.8 - 25.1}{5.5}\right)$ or $P(Z > 1.03636...)$ | M1 | Standardising 30.8 with 25.1 and 5.5 (allow $\pm$); allow $z = 1.04$ |
| $= 1 - 0.8508$ | M1 | For $1 - p$ where $0.84 < p < 0.86$ |
| $= 0.1492$ or better (calc: 0.1500...) | A1cso | Answer of 0.1492 or better with evidence of both M's |
## Part (b)
| $\frac{y - 25.1}{5.5} = -1.6449$ | M1B1 | Standardising $y$ with 25.1 and 5.5; use of $z = \pm 1.6449$ or better |
| $y = 16.053...$, so range is awrt **16.1 ~ 30.8** | A1 | For awrt 16.1 (ISW) or range [16.1, 30.8] |
## Part (c)(i)
| $P(H < d) = 0.05 + 0.2 + 0.3 [= 0.55]$ | M1 | Correct method to calculate $P(H < d)$ |
| $\frac{d - 25.1}{5.5} = 0.13$ (Calc 0.12566) | M1 | Correct standardisation $= z$ where $0.125 \leq |z| \leq 0.13$ |
| $d = 0.13 \times 5.5 + 25.1 = 25.815$ (25.791... calc) | A1cso | Both method marks awarded, no errors seen **and** awrt 25.82 or awrt 25.79 |
## Part (c)(ii)
| $P(H < m) = 0.05 + 0.2 [= 0.25]$ | M1 | Correct method for $P(H < m)$; allow $0.05 +$ awrt 0.200 |
| $\frac{m - 25.1}{5.5} = -0.67$ (Calc 0.674489) | M1M1 | Standardising $m$ with 25.1 and 5.5 and setting equal to awrt $-0.67$ |
| $m =$ awrt **21.4** | A1 | For $m =$ awrt 21.4 |
## Part (d)
| Height $= 2 \times \text{"m"} + 3 \times 25.8 + 3 \times 30.8\,[+8]$ | M1 | For $2 \times \text{"m"} + 3 \times 25.8 + 3 \times 30.8\,[+n]$ where $n$ is integer $\geq 0$ |
| $= 220.6$ awrt **221** (cm) | A1 | For awrt 221 (cm) |
# Question ALT 1 (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(H > 25.8 \mid "16.1" < H < 30.8)$ **or** $\dfrac{P(25.8 < H < 30.8)}{1-(0.15+0.05)}$ | M1 | 1st M1: for a correct conditional probability statement ft their answer to (b) i.e. their $y$ |
| $= \dfrac{0.8508 - 0.5517}{0.8}$ (tables) **or** $\dfrac{0.299345...}{0.8}$ (calc) $\approx \dfrac{3}{8}$ | M1 A1cso | 2nd M1: for a ratio of probs of the form $\dfrac{q}{0.8}$ where $q = 0.3$ to 1sf |
| | (3) | A1: for probability of approx $\dfrac{3}{8}$ |\begin{enumerate}
\item Xiang is designing shelves for a bookshop. The height, $H \mathrm {~cm}$, of books is modelled by the normal distribution with mean 25.1 cm and standard deviation 5.5 cm\\
(a) Show that $\mathrm { P } ( H > 30.8 ) = 0.15$
\end{enumerate}
Xiang decided that the smallest $5 \%$ of books and books taller than 30.8 cm would not be placed on the shelves. All the other books will be placed on the shelves.\\
(b) Find the range of heights of books that will be placed on the shelves.\\
(3)
The books that will be placed on the shelves have heights classified as small, medium or large.\\
The numbers of small, medium and large books are in the ratios $2 : 3 : 3$\\
(c) The medium books have heights $x \mathrm {~cm}$ where $m < x < d$\\
(i) Show that $d = 25.8$ to 1 decimal place.\\
(ii) Find the value of $m$
Xiang wants 2 shelves for small books, 3 shelves for medium books and 3 shelves for large books.\\
These shelves will be placed one above another and made of wood that is 1 cm thick.\\
(d) Work out the minimum total height needed.
\hfill \mbox{\textit{Edexcel S1 2021 Q6 [15]}}