Edexcel Paper 3 2022 June — Question 3 7 marks

Exam BoardEdexcel
ModulePaper 3 (Paper 3)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate mean from coded sums
DifficultyEasy -1.2 This is a straightforward statistics question requiring only basic recall and application of standard formulas. Part (a) is simple data set knowledge, (b) uses given summary statistics with direct formula substitution (mean = Σr/n, standard deviation formula), (c) is a simple comparison, and (d) requires stating why binomial assumptions might fail. No problem-solving or novel insight needed—purely routine calculations and standard reasoning about distributions.
Spec2.01a Population and sample: terminology2.02g Calculate mean and standard deviation

  1. Dian uses the large data set to investigate the Daily Total Rainfall, \(r \mathrm {~mm}\), for Camborne.
    1. Write down how a value of \(0 < r \leqslant 0.05\) is recorded in the large data set.
    Dian uses the data for the 31 days of August 2015 for Camborne and calculates the following statistics $$n = 31 \quad \sum r = 174.9 \quad \sum r ^ { 2 } = 3523.283$$
  2. Use these statistics to calculate
    1. the mean of the Daily Total Rainfall in Camborne for August 2015,
    2. the standard deviation of the Daily Total Rainfall in Camborne for August 2015. Dian believes that the mean Daily Total Rainfall in August is less in the South of the UK than in the North of the UK.
      The mean Daily Total Rainfall in Leuchars for August 2015 is 1.72 mm to 2 decimal places.
  3. State, giving a reason, whether this provides evidence to support Dian's belief. Dian uses the large data set to estimate the proportion of days with no rain in Camborne for 1987 to be 0.27 to 2 decimal places.
  4. Explain why the distribution \(\mathrm { B } ( 14,0.27 )\) might not be a reasonable model for the number of days without rain for a 14-day summer event.

Question 3:
Part (a):
AnswerMarks Guidance
AnswerMark Guidance
trB1 Allow Tr or trace or Trace
Part (b)(i):
AnswerMarks Guidance
AnswerMark Guidance
\(\mu = \frac{174.9}{31} = 5.6419...\) awrt 5.64B1 For a correct mean awrt 5.64
Part (b)(ii):
AnswerMarks Guidance
AnswerMark Guidance
\(\sigma_r = \sqrt{\frac{3523.283}{31}-\mu^2}\)M1 For a correct expression for sd including the \(\sqrt{\phantom{x}}\). Ft their mean
\(= 9.04559...\) awrt 9.05A1 awrt 9.05 (Allow \(s=9.1932...\), awrt 9.19). NB awrt to 9.05 or 9.19 with no working is M1 A1
Part (c):
AnswerMarks Guidance
AnswerMark Guidance
Leuchars is in the North and Camborne is in the SouthM1 For stating Leuchars is North of Camborne oe eg Camborne is further south
The mean is smaller for Leuchars than Camborne therefore there is no evidence that Dian's belief is trueA1ft M1 must be awarded. A correct conclusion and correct comment about the means ft their mean in (b). Allow No. SC for No and there are only 2 places used so there is insufficient data. Mark as M0A1 on epen
Part (d):
AnswerMarks Guidance
AnswerMark Guidance
eg \(p=0.27\) is unlikely to be constantB1 A correct reason referring to: independence (needs context, eg consecutive 14 days unlikely to be independent); probability [of rain] not being constant; or a comment that the proportion of days with no rain will be different over the year
# Question 3:

## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| tr | B1 | Allow Tr or trace or Trace |

## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mu = \frac{174.9}{31} = 5.6419...$ awrt 5.64 | B1 | For a correct mean awrt 5.64 |

## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sigma_r = \sqrt{\frac{3523.283}{31}-\mu^2}$ | M1 | For a correct expression for sd including the $\sqrt{\phantom{x}}$. Ft their mean |
| $= 9.04559...$ awrt 9.05 | A1 | awrt 9.05 (Allow $s=9.1932...$, awrt 9.19). NB awrt to 9.05 or 9.19 with no working is M1 A1 |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Leuchars is in the North and Camborne is in the South | M1 | For stating Leuchars is North of Camborne oe eg Camborne is further south |
| The mean is smaller for Leuchars than Camborne therefore there is no evidence that Dian's belief is true | A1ft | M1 must be awarded. A correct conclusion **and** correct comment about the means ft their mean in (b). Allow No. SC for No **and** there are only 2 places used so there is insufficient data. Mark as M0A1 on epen |

## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| eg $p=0.27$ is unlikely to be constant | B1 | A correct reason referring to: independence (needs context, eg consecutive 14 days unlikely to be independent); probability [of rain] not being constant; or a comment that the proportion of days with no rain will be different over the year |

---
\begin{enumerate}
  \item Dian uses the large data set to investigate the Daily Total Rainfall, $r \mathrm {~mm}$, for Camborne.\\
(a) Write down how a value of $0 < r \leqslant 0.05$ is recorded in the large data set.
\end{enumerate}

Dian uses the data for the 31 days of August 2015 for Camborne and calculates the following statistics

$$n = 31 \quad \sum r = 174.9 \quad \sum r ^ { 2 } = 3523.283$$

(b) Use these statistics to calculate\\
(i) the mean of the Daily Total Rainfall in Camborne for August 2015,\\
(ii) the standard deviation of the Daily Total Rainfall in Camborne for August 2015.

Dian believes that the mean Daily Total Rainfall in August is less in the South of the UK than in the North of the UK.\\
The mean Daily Total Rainfall in Leuchars for August 2015 is 1.72 mm to 2 decimal places.\\
(c) State, giving a reason, whether this provides evidence to support Dian's belief.

Dian uses the large data set to estimate the proportion of days with no rain in Camborne for 1987 to be 0.27 to 2 decimal places.\\
(d) Explain why the distribution $\mathrm { B } ( 14,0.27 )$ might not be a reasonable model for the number of days without rain for a 14-day summer event.

\hfill \mbox{\textit{Edexcel Paper 3 2022 Q3 [7]}}