| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| 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 multi-part normal distribution question requiring standard z-score calculations, understanding of conditional distributions (potatoes sold vs grown), and a straightforward multinomial probability calculation. While it has several parts and requires careful reading to distinguish between 'grown' and 'sold' distributions, all techniques are routine S1 material with no novel problem-solving required. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(W<92) = P\left(Z < \frac{92-140}{40}\right) = [P(Z<-1.2)]\) | M1 | For standardising attempt with 92 or 188, 140 and 40 (o.e.). Accept \(\pm\), ignore inequality |
| \(= 1-0.8849\) | dM1 | Dependent on 1st M1, for attempting \(1-p\) where \(0.5 |
| \(= \text{awrt } \mathbf{11.5}\ (\%) \text{ or } \mathbf{0.115}\) | A1 | For awrt 11.5 (%) or 0.115 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([P(W>q_3) = P(W>92)\times P(W>q_3 | W>92) =]\ (1-\text{(a)})\times0.25 = 0.8849\times0.25\) | M1 |
| \(= 0.221225 = \text{awrt } \mathbf{0.221}\) | A1 | For awrt 0.221 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
\(P(W| W>92) = 0.25\) or \(P(W>q_1 |
W>92) = 0.75\) |
|
| \(P(92 | M1 | For either correct probability statement and \(0.25\) or \(0.75\times(1-\text{their (a)})\) |
| \(P(W | A1 | For \(P(W |
| \(\frac{q_1-140}{40} = -0.42\) (calculator gives \(-0.422513 \sim -0.423404\)) | M1 | For standardising with \(q_1\), 140 and 40 and setting equal to \(z\) where \(0.40< |
| so \(q_1 = 123.2 = \text{awrt } \mathbf{123}\ \text{(g)}\) | A1 | For awrt 123 (condone minor slips in working if correct answer obtained) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{4}\times\frac{1}{4}\times\frac{1}{2}\times3!\) | M1M1 | For \(0.25\times0.25\times0.5\) (o.e.) e.g. \(\frac{1}{32}\) may be seen as decimals or fractions. For \(\times3!\) or \(\times6\) or adding all 6 cases. Must be multiplying probabilities |
| \(= \frac{3}{16}\) or \(0.1875\) | A1 | For \(\frac{3}{16}\) or any exact equivalent |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W<92) = P\left(Z < \frac{92-140}{40}\right) = [P(Z<-1.2)]$ | M1 | For standardising attempt with 92 or 188, 140 and 40 (o.e.). Accept $\pm$, ignore inequality |
| $= 1-0.8849$ | dM1 | Dependent on 1st M1, for attempting $1-p$ where $0.5<p<1$ |
| $= \text{awrt } \mathbf{11.5}\ (\%) \text{ or } \mathbf{0.115}$ | A1 | For awrt 11.5 (%) or 0.115 |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[P(W>q_3) = P(W>92)\times P(W>q_3|W>92) =]\ (1-\text{(a)})\times0.25 = 0.8849\times0.25$ | M1 | For $(1-\text{their (a)})\times0.25$ or $1-[(1-\text{(a)})\times0.75+\text{(a)}] = 1-[0.8849\times0.75+0.1151]$ |
| $= 0.221225 = \text{awrt } \mathbf{0.221}$ | A1 | For awrt 0.221 |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W<q_1|W>92) = 0.25$ or $P(W>q_1|W>92) = 0.75$ | M1 | For a correct conditional prob. statement with $q_1$, 92 and 0.25 or 0.75 |
| $P(92<W<q_1) = 0.25\times0.8849 = \text{``}0.221...\text{''}$ or $P(W>q_1) = 0.75\times0.8849 = 0.663675$ | M1 | For either correct probability statement and $0.25$ or $0.75\times(1-\text{their (a)})$ |
| $P(W<q_1) = 0.221225+0.115 = \text{awrt } \mathbf{0.336}$ or $P(W>q_1) = 0.663675 = \text{awrt } \mathbf{0.664}$ | A1 | For $P(W<q_1) = $ awrt 0.336 or $P(W>q_1) = $ awrt 0.664. NB May be standardised. Award M1M1A1 for either probability clearly stated or marked on a correct sketch |
| $\frac{q_1-140}{40} = -0.42$ (calculator gives $-0.422513 \sim -0.423404$) | M1 | For standardising with $q_1$, 140 and 40 and setting equal to $z$ where $0.40<|z|<0.45$ |
| so $q_1 = 123.2 = \text{awrt } \mathbf{123}\ \text{(g)}$ | A1 | For awrt 123 (condone minor slips in working if correct answer obtained) |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{4}\times\frac{1}{4}\times\frac{1}{2}\times3!$ | M1M1 | For $0.25\times0.25\times0.5$ (o.e.) e.g. $\frac{1}{32}$ may be seen as decimals or fractions. For $\times3!$ or $\times6$ or adding all 6 cases. Must be multiplying probabilities |
| $= \frac{3}{16}$ or $0.1875$ | A1 | For $\frac{3}{16}$ or any exact equivalent |
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If you have additional pages from the mark scheme, please share those and I'll be happy to extract and format the content for you.7. Farmer Adam grows potatoes. The weights of potatoes, in grams, grown by Adam are normally distributed with a mean of 140 g and a standard deviation of 40 g .
Adam cannot sell potatoes with a weight of less than 92 g .
\begin{enumerate}[label=(\alph*)]
\item Find the percentage of potatoes that Adam grows but cannot sell.
The upper quartile of the weight of potatoes sold by Adam is $q _ { 3 }$
\item Find the probability that the weight of a randomly selected potato grown by Adam is more than $q _ { 3 }$
\item Find the lower quartile, $q _ { 1 }$, of the weight of potatoes sold by Adam.
Betty selects a random sample of 3 potatoes sold by Adam.
\item Find the probability that one weighs less than $q _ { 1 }$, one weighs more than $q _ { 3 }$ and one has a weight between $q _ { 1 }$ and $q _ { 3 }$\\
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\hfill \mbox{\textit{Edexcel S1 2018 Q7 [13]}}