| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find standard deviation from probability |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard inverse normal table lookups and basic probability calculations. Part (a) uses symmetry, part (b) involves finding σ from P(X > 256) = 0.01 using tables (z = 2.326), and part (c) applies independent probability multiplication. All techniques are routine for this specification with no novel problem-solving required. |
| 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) | 98% (Condone 0.98) | B1 |
| (b) | \(z = \pm 2.3263\) (or better: calculator gives 2.326347877...) | B1 |
| \(\frac{256-250}{\sigma} = 2.3263\) | M1 | |
| \(\sigma = 2.579\ldots\) | A1 | awrt 2.58 |
| (3) | ||
| (c) | \([P(X < 246 \cup X > 254) =]\) | M1 |
| \(2 \times P\left(Z > \frac{254-250}{\text{"2.579..."}}\right)\) or \(1 - P\left(\frac{246-250}{\text{"2.579"}} < Z < \frac{254-250}{\text{"2.579"}}\right)\) | ||
| \(= 2 \times P(Z > 1.55)\) or \(1 - P(-1.55 < Z < 1.55) = 0.12(12)\) | A1 | |
| \(P(\text{both bags outside range}) = (0.1212)^2 = 0.01468\ldots\) | dM1, A1 | 1st M1 for attempt to find sum of the area above 254 and below 246 or \(2 \times\) area above 254 or \(2 \times\) area below 246. 1st A1 for awrt 0.12 (NB 1 - 0.1212 = 0.8788 is A0 here and 1st M0 too). 2nd dM1 for \(p^2\) dependent on previous M1. 2nd A1 for awrt 0.0146 (use of calculator value) or 0.0147. |
| \(\underline{\text{awrt } 0.0146/7}\) | ||
| (4) | ||
| [8 marks] | ||
| Notes | ||
| (b) | B1 for \(\pm 2.3263\) or better seen and used, can be with \(\sigma^2\) (may be implied by \(\sigma = \text{awrt } 2.579\)). M1 for standardising with 256 or 244, 250 and \(\sigma\) and equating to a z-value [\ | z\ |
| z = 2.33: Use of \(z = 2.33\) leads to \(\sigma = 2.575\ldots\) can score B0M1A1. | ||
| z = 2.32: Special case: use of \(z = 2.32\) from tables gives 2.586... \(\sigma = \text{awrt } 2.59\) can score B0M1A1. | ||
| Ans only: B1M1A1 can be awarded for sight of at least \(\sigma = \text{awrt } 2.5791\) or awrt 2.5792 | ||
| (c) | 1st M1 for attempt to find sum of the area above 254 and below 246 or \(2 \times\) area above 254 or \(2 \times\) area below 246 (2 \(\times\) needed). Allow ft of their \(\sigma\) (provided \(\sigma > 0\)). | |
| 1st A1 for awrt 0.12 (NB 1 - 0.1212 = 0.8788 is A0 here and 1st M0 too). | ||
| 2nd dM1 for \(p^2\) dependent on previous M1. | ||
| 2nd A1 for awrt 0.0146 (use of calculator value) or 0.0147. | ||
| 'B1' for those who use 1 tail only and get 0.06... but then do \((0.06...)^2\) Score as M0A0M1A0. Do not award for \(2 \times (0.06...)^2\) or \(3 \times (0.06...)^2\). | ||
| SC |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | 98% (Condone 0.98) | B1 | (1) |
| (b) | $z = \pm 2.3263$ (or better: calculator gives 2.326347877...) | B1 | |
| | $\frac{256-250}{\sigma} = 2.3263$ | M1 | |
| | $\sigma = 2.579\ldots$ | A1 | awrt 2.58 |
| | | (3) | |
| (c) | $[P(X < 246 \cup X > 254) =]$ | M1 | |
| | $2 \times P\left(Z > \frac{254-250}{\text{"2.579..."}}\right)$ or $1 - P\left(\frac{246-250}{\text{"2.579"}} < Z < \frac{254-250}{\text{"2.579"}}\right)$ | | |
| | $= 2 \times P(Z > 1.55)$ or $1 - P(-1.55 < Z < 1.55) = 0.12(12)$ | A1 | |
| | $P(\text{both bags outside range}) = (0.1212)^2 = 0.01468\ldots$ | dM1, A1 | 1st M1 for attempt to find sum of the area above 254 and below 246 or $2 \times$ area above 254 or $2 \times$ area below 246. 1st A1 for awrt 0.12 (NB 1 - 0.1212 = 0.8788 is A0 here and 1st M0 too). 2nd dM1 for $p^2$ dependent on previous M1. 2nd A1 for awrt 0.0146 (use of calculator value) or 0.0147. |
| | $\underline{\text{awrt } 0.0146/7}$ | | |
| | | (4) | |
| | | [8 marks] | |
| **Notes** | | | |
| (b) | B1 for $\pm 2.3263$ or better seen and used, can be with $\sigma^2$ (may be implied by $\sigma = \text{awrt } 2.579$). M1 for standardising with 256 or 244, 250 and $\sigma$ and equating to a z-value [\|z\| > 2]. A1 for awrt 2.58 from correct working. | | |
| | z = 2.33: Use of $z = 2.33$ leads to $\sigma = 2.575\ldots$ can score B0M1A1. | | |
| | z = 2.32: Special case: use of $z = 2.32$ from tables gives 2.586... $\sigma = \text{awrt } 2.59$ can score B0M1A1. | | |
| | Ans only: B1M1A1 can be awarded for sight of at least $\sigma = \text{awrt } 2.5791$ or awrt 2.5792 | | |
| (c) | 1st M1 for attempt to find sum of the area above 254 and below 246 or $2 \times$ area above 254 or $2 \times$ area below 246 (2 $\times$ needed). Allow ft of their $\sigma$ (provided $\sigma > 0$). | | |
| | 1st A1 for awrt 0.12 (NB 1 - 0.1212 = 0.8788 is A0 here and 1st M0 too). | | |
| | 2nd dM1 for $p^2$ dependent on previous M1. | | |
| | 2nd A1 for awrt 0.0146 (use of calculator value) or 0.0147. | | |
| | 'B1' for those who use 1 tail only and get 0.06... but then do $(0.06...)^2$ Score as M0A0M1A0. Do not award for $2 \times (0.06...)^2$ or $3 \times (0.06...)^2$. | | |
| **SC** | | | |\begin{enumerate}
\item In a factory, a machine is used to fill bags of rice. The weights of bags of rice are modelled using a normal distribution with mean 250 g .
\end{enumerate}
Only $1 \%$ of the bags of rice weigh more than 256 g .\\
(a) Write down the percentage of bags of rice with weights between 244 g and 256 g .\\
(b) Find the standard deviation of the weights of the bags of rice.
An inspection consists of selecting a bag of rice at random and checking if its weight is within 4 g of the mean. If the weight is more than 4 g away from the mean, then a second bag of rice is selected at random and checked. If the weight of each of the 2 bags of rice is more than 4 g away from the mean, then the machine is shut down.\\
(c) Find the probability that the machine is shut down following an inspection.
\hfill \mbox{\textit{Edexcel S1 2017 Q6 [8]}}