| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This is a standard S1 normal distribution question with routine calculations: parts (a) require basic z-score standardization and table lookup, part (b) uses symmetry to find a value, and part (c) involves simultaneous equations with inverse normal. All techniques are textbook exercises 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 |
| VIAV SIHI NI BIIYM ION OO | V34V SIHI NI IIIYM ION OO | V38V SIHI NI JLIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X > 505) = P\!\left(Z > \frac{505-503}{1.6}\right)\) | M1 | Standardising with 505, 503 and 1.6; allow \(\pm\) |
| \(= 1 - P(Z < 1.25) = 1 - 0.8944\) | M1 | Correct method for \(P(Z > 1.25)\), e.g. \(1-p\) where \(0.5 < p < 0.99\) |
| \(= 0.1056\) | A1 | awrt \(\mathbf{0.106}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(501 < X < 505) = 1 - 2 \times 0.1056\) or \(0.8944 - 0.1056\) | M1 | \(1 - 2 \times\) their (i) |
| \(= 0.7888\) | A1 | awrt \(\mathbf{0.789}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X < w) = 0.9713\) or \(P(X > w) = 0.0287\) (implied by \(z = \pm 1.9\)) | M1 | For using symmetry to find area of one tail |
| \(\frac{w - 503}{1.6} = 1.9\) or \(\frac{(1006-w)-503}{1.6} = -1.9\) | M1 | Single standardisation with 503, 1.6 and \(w\) (or \(1006-w\)); set \(= \pm z\) where \(1.8 < \ |
| \(w = 506.04\ldots\) | A1 | awrt \(\mathbf{506}\) (answer only 506 scores 0/3; 506.0... with no working send to review) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{r - 503}{q} = -2.3263\) | M1 A1 | M1: \(\frac{r-503}{q} = z\) where \(\ |
| \(\frac{r + 6 - 503}{q} = 1.6449\) | M1 A1 | M1: \(\frac{r+6-503}{q} = z\) where \(\ |
| \(1.6449q - 6 = -2.3263q\) | ddM1 | Dependent on both M marks; attempt to solve simultaneous equations |
| \(q = 1.51\ldots\) | A1 | awrt \(\mathbf{1.51}\) |
| \(r = 499.48\ldots\) | A1 | awrt \(\mathbf{499}\) (allow 499.5) |
## Question 7:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X > 505) = P\!\left(Z > \frac{505-503}{1.6}\right)$ | M1 | Standardising with 505, 503 and 1.6; allow $\pm$ |
| $= 1 - P(Z < 1.25) = 1 - 0.8944$ | M1 | Correct method for $P(Z > 1.25)$, e.g. $1-p$ where $0.5 < p < 0.99$ |
| $= 0.1056$ | A1 | awrt $\mathbf{0.106}$ |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(501 < X < 505) = 1 - 2 \times 0.1056$ or $0.8944 - 0.1056$ | M1 | $1 - 2 \times$ their (i) |
| $= 0.7888$ | A1 | awrt $\mathbf{0.789}$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X < w) = 0.9713$ or $P(X > w) = 0.0287$ (implied by $z = \pm 1.9$) | M1 | For using symmetry to find area of one tail |
| $\frac{w - 503}{1.6} = 1.9$ or $\frac{(1006-w)-503}{1.6} = -1.9$ | M1 | Single standardisation with 503, 1.6 and $w$ (or $1006-w$); set $= \pm z$ where $1.8 < \|z\| < 2$ |
| $w = 506.04\ldots$ | A1 | awrt $\mathbf{506}$ (answer only 506 scores 0/3; 506.0... with no working send to review) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{r - 503}{q} = -2.3263$ | M1 A1 | M1: $\frac{r-503}{q} = z$ where $\|z\| > 2$; A1: awrt $-2.3263$ (signs must be compatible) |
| $\frac{r + 6 - 503}{q} = 1.6449$ | M1 A1 | M1: $\frac{r+6-503}{q} = z$ where $\|z\| > 1$; A1: awrt $1.6449$ (signs must be compatible) |
| $1.6449q - 6 = -2.3263q$ | ddM1 | Dependent on both M marks; attempt to solve simultaneous equations |
| $q = 1.51\ldots$ | A1 | awrt $\mathbf{1.51}$ |
| $r = 499.48\ldots$ | A1 | awrt $\mathbf{499}$ (allow 499.5) |
The image appears to be a blank page (page 472) from Pearson Edexcel International Advanced Level Mathematics Sample Assessment Materials (SAMs), Issue 3, June 2018. There is no mark scheme content visible on this page to extract.\begin{enumerate}
\item A machine fills bottles with water. The volume of water delivered by the machine to a bottle is $X \mathrm { ml }$ where $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$
\end{enumerate}
One of these bottles of water is selected at random.\\
Given that $\mu = 503$ and $\sigma = 1.6$\\
(a) find\\
(i) $\mathrm { P } ( X > 505 )$\\
(ii) $\mathrm { P } ( 501 < X < 505 )$\\
(b) Find $w$ such that $\mathrm { P } ( 1006 - w < X < w ) = 0.9426$
Following adjustments to the machine, the volume of water delivered by the machine to a bottle is such that $\mu = 503$ and $\sigma = q$
Given that $\mathrm { P } ( X < r ) = 0.01$ and $\mathrm { P } ( X > r + 6 ) = 0.05$\\
(c) find the value of $r$ and the value of $q$\\
$\_\_\_\_$ VAYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JIIIM ION OC\\
VEYV SIHI NI ELIYM ION OC
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIAV SIHI NI BIIYM ION OO & V34V SIHI NI IIIYM ION OO & V38V SIHI NI JLIYM ION OC \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{Edexcel S1 2018 Q7 [15]}}