| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Calculate probabilities using sample mean distribution |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem to find P(X̄ < 3) for n=40 rolls of a fair die. Students need to calculate the mean and variance of a discrete uniform distribution, apply the CLT to get the sampling distribution, then use normal tables. It's a standard textbook exercise with clear steps and no conceptual surprises, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Label full time staff 1–6000, part time staff 1–4000 | M1 | One set of correct numbers |
| Use random numbers to select | M1 | |
| Simple random sample of 120 full time and 80 part time staff | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Enables estimation of statistics/errors for each strata, or "reduce variability", or "more representative", or "reflects population structure" | B1 | NOT "more accurate" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_f = \mu_p,\quad H_1: \mu_f \neq \mu_p\) (accept \(\mu_1, \mu_2\)) | B1 | |
| \(\text{s.e.} = \sqrt{\frac{21}{80} + \frac{19}{80}}\) | M1 | For attempt at s.e.; condone one number wrong; correct s.e. \(= \sqrt{\frac{1}{2}}\) |
| \(z = \frac{52 - 50}{\sqrt{\frac{21}{80} + \frac{19}{80}}} = (2\sqrt{2})\) | M1 | Must be \(\frac{\pm(52-50)}{\sqrt{\frac{p}{q}+\frac{r}{s}}}\) |
| \(= 2.828...\) awrt 2.83 | A1 | |
| Two tailed cv: \(z = 2.5758\) (or prob awrt 0.002 or 0.004) | B1 | |
| \([2.828 > 2.5758]\) significant, reject \(H_0\) | dM1 | Dependent on 2nd M1 |
| Evidence of difference in policy awareness between full time and part time staff | A1ft | Must mention "scores"/"policy awareness" and types of staff; A0 for one-tailed comment |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Can use mean full time and mean part time | B1 | Mention of mean(s) or use of \(\bar{X}\) (clearly referring to full-time or part-time) |
| \(\sim\) Normal | B1 | Distribution can be assumed normal |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Have assumed \(s^2 = \sigma^2\) or variance of sample \(=\) variance of population | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.53 < 2.5758\), not significant or do not reject \(H_0\) | M1 | |
| Insufficient evidence of a difference in mean awareness | A1ft | Accept "no difference in mean scores"; allow ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Training course has closed the gap between full time and part time staff's mean awareness of company policy | B1 | Must imply training was effective; supported by (c) and (f); condone one-tailed comment here |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Label full time staff 1–6000, part time staff 1–4000 | M1 | One set of correct numbers |
| Use random numbers to select | M1 | |
| Simple random sample of 120 full time and 80 part time staff | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Enables estimation of statistics/errors for each strata, or "reduce variability", or "more representative", or "reflects population structure" | B1 | NOT "more accurate" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_f = \mu_p,\quad H_1: \mu_f \neq \mu_p$ (accept $\mu_1, \mu_2$) | B1 | |
| $\text{s.e.} = \sqrt{\frac{21}{80} + \frac{19}{80}}$ | M1 | For attempt at s.e.; condone one number wrong; correct s.e. $= \sqrt{\frac{1}{2}}$ |
| $z = \frac{52 - 50}{\sqrt{\frac{21}{80} + \frac{19}{80}}} = (2\sqrt{2})$ | M1 | Must be $\frac{\pm(52-50)}{\sqrt{\frac{p}{q}+\frac{r}{s}}}$ |
| $= 2.828...$ awrt **2.83** | A1 | |
| Two tailed cv: $z = 2.5758$ (or prob awrt 0.002 or 0.004) | B1 | |
| $[2.828 > 2.5758]$ significant, reject $H_0$ | dM1 | Dependent on 2nd M1 |
| Evidence of difference in policy awareness between full time and part time staff | A1ft | Must mention "scores"/"policy awareness" and types of staff; A0 for one-tailed comment |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Can use mean full time and mean part time | B1 | Mention of mean(s) or use of $\bar{X}$ (clearly referring to full-time or part-time) |
| $\sim$ Normal | B1 | Distribution can be assumed normal |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Have assumed $s^2 = \sigma^2$ or variance of sample $=$ variance of population | B1 | |
## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.53 < 2.5758$, not significant or do not reject $H_0$ | M1 | |
| Insufficient evidence of a difference in mean awareness | A1ft | Accept "no difference in mean scores"; allow ft |
## Part (g):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Training course has closed the gap between full time and part time staff's mean awareness of company policy | B1 | Must imply training was effective; supported by (c) and (f); condone one-tailed comment here |
\begin{enumerate}
\item A fair six-sided die is labelled with the numbers $1,2,3,4,5$ and 6\\
(b) Find an approximation for the probability that the mean of the 40 scores is less than 3\\
\includegraphics[max width=\textwidth, alt={}, center]{0434a6c1-686a-449d-ba16-dbb8e60288e8-24_204_714_237_251}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2018 Q7 [5]}}