Edexcel S3 2018 Specimen — Question 7 5 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Year2018
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeCalculate probabilities using sample mean distribution
DifficultyStandard +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.
Spec5.05a Sample mean distribution: central limit theorem

  1. 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}

Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Label full time staff 1–6000, part time staff 1–4000M1 One set of correct numbers
Use random numbers to selectM1
Simple random sample of 120 full time and 80 part time staffA1
Part (b):
AnswerMarks Guidance
Answer/WorkingMark 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):
AnswerMarks Guidance
Answer/WorkingMark 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 2.83A1
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 staffA1ft Must mention "scores"/"policy awareness" and types of staff; A0 for one-tailed comment
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Can use mean full time and mean part timeB1 Mention of mean(s) or use of \(\bar{X}\) (clearly referring to full-time or part-time)
\(\sim\) NormalB1 Distribution can be assumed normal
Part (e):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Have assumed \(s^2 = \sigma^2\) or variance of sample \(=\) variance of populationB1
Part (f):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2.53 < 2.5758\), not significant or do not reject \(H_0\)M1
Insufficient evidence of a difference in mean awarenessA1ft Accept "no difference in mean scores"; allow ft
Part (g):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Training course has closed the gap between full time and part time staff's mean awareness of company policyB1 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]}}