| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring routine z-score calculations and inverse normal lookups. Parts (a) and (b) are straightforward applications of tables, while part (c) requires setting up two equations with standardization, which is slightly more challenging but still a common S1 exercise. Overall slightly easier than average A-level due to being from S1 (not core) and following standard patterns. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a)(i) | \(P(X > 505) = P\left(Z > \frac{505 - 503}{1.6}\right)\) | M1 |
| \(= 1 - P(Z < 1.25) = 1 - 0.8944\) | M1 | 2nd M1 for \(1 - P(Z < 1.25)\) i.e. correct method for finding P(Z > 1.25), e.g. \(1 - p\) where \(0.5 < p < 0.99\) |
| \(= 0.1056\) | A1 | A1 awrt 0.106 |
| (a)(ii) | \(P(501 < X < 505) = 1 - 2 \times 0.1056\) or \(0.8944 - 0.1056\) | M1 |
| \(= 0.7888\) | A1 | A1 awrt 0.789 |
| (b) | \(P(X < w) = 0.9713\) or \(P(X > w) = 0.0287\) (may be implied by \(z = \pm 1.9\)) | M1 |
| \(\frac{w - 503}{1.6} = 1.9\) or \(\frac{(1006 - w) - 503}{1.6} = -1.9\) | M1 | 2nd M1 single standardisation with 503, 1.6 and w (or 1006 − w) and set = ± z value (1.8 < |
| \(w = 506.04...\) | A1 | A1 for awrt 506 (Answer only: 506 scores 0/3, but 506.0...with working send to review) |
| (c) | \(\frac{r - 503}{q} = -2.3263\) | M1A1 |
| \(\frac{r + 6 - 503}{q} = 1.6449\) | M1A1 | 2nd M1 \(\frac{r + 6 - 503}{q} = \) z value where \( |
| \(1.6449q - 6 = -2.3263q\) | ddM1 | |
| \(q = 1.51...\) | A1 | 3rd A1 for awrt 1.51 |
| \(r = 499.48......\) | A1 | 4th A1 for awrt 499 (allow 499.5) |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a)(i) | $P(X > 505) = P\left(Z > \frac{505 - 503}{1.6}\right)$ | M1 | 1st M1 standardising with 505, 503 and 1.6. May be implied by use of 1.25 (Allow ÷) |
| | $= 1 - P(Z < 1.25) = 1 - 0.8944$ | M1 | 2nd M1 for $1 - P(Z < 1.25)$ i.e. correct method for finding P(Z > 1.25), e.g. $1 - p$ where $0.5 < p < 0.99$ |
| | $= 0.1056$ | A1 | A1 awrt 0.106 |
| (a)(ii) | $P(501 < X < 505) = 1 - 2 \times 0.1056$ or $0.8944 - 0.1056$ | M1 | M1: $1 - 2 \times$ their(i) |
| | $= 0.7888$ | A1 | A1 awrt 0.789 |
| (b) | $P(X < w) = 0.9713$ or $P(X > w) = 0.0287$ (may be implied by $z = \pm 1.9$) | M1 | 1st M1 for using symmetry to find area of one tail (may be seen in diagram) |
| | $\frac{w - 503}{1.6} = 1.9$ or $\frac{(1006 - w) - 503}{1.6} = -1.9$ | M1 | 2nd M1 single standardisation with 503, 1.6 and w (or 1006 − w) and set = ± z value (1.8 < |z| < 2) |
| | $w = 506.04...$ | A1 | A1 for awrt 506 (**Answer only: 506 scores 0/3, but 506.0...with working send to review**) |
| (c) | $\frac{r - 503}{q} = -2.3263$ | M1A1 | 1st M1 $\frac{r - 503}{q} = $ z value where $|z| \geq 2$. 1st A1 $\frac{r - 503}{q} =$ awrt −2.3263 (signs must be compatible) |
| | $\frac{r + 6 - 503}{q} = 1.6449$ | M1A1 | 2nd M1 $\frac{r + 6 - 503}{q} = $ z value where $|z| > 1$. 2nd A1 $\frac{r + 6 - 503}{q} =$ awrt 1.6449 (signs must be compatible) |
| | $1.6449q - 6 = -2.3263q$ | ddM1 | |
| | $q = 1.51...$ | A1 | 3rd A1 for awrt 1.51 |
| | $r = 499.48......$ | A1 | 4th A1 for awrt 499 (allow 499.5) |
**Total: 15 marks**
Special Case: Less than 4dp z-values: use of awrt 2.32/2.33/2.34 and awrt 1.64/1.65 could score M1 A0 M1 and then A1 provided both equations have compatible signs. 3rd M1 (dep on both Ms) attempt to solve simultaneous equations leading to value for q or r. 3rd A1 for awrt 1.51. 4th A1 for awrt 499 (allow 499.5).7. A machine fills bottles with water. The volume of water delivered by the machine to a bottle is $X \mathrm { ml }$ where $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$
One of these bottles of water is selected at random.
Given that $\mu = 503$ and $\sigma = 1.6$
\begin{enumerate}[label=(\alph*)]
\item find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X > 505 )$
\item $\mathrm { P } ( 501 < X < 505 )$
\end{enumerate}\item 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 $\mathrm { P } ( X > r + 6 ) = 0.05$
\item find the value of $r$ and the value of $q$\\
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\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q7 [15]}}