Edexcel AS Paper 2 2018 June — Question 4 8 marks

Exam BoardEdexcel
ModuleAS Paper 2 (AS Paper 2)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate statistics from discrete frequency table
DifficultyModerate -0.8 This is a straightforward AS-level statistics question requiring basic calculations from a frequency table (mean and standard deviation using standard formulas) and interpretation of summary statistics/box plots in context of the large data set. All parts use direct recall of methods with minimal problem-solving, making it easier than average A-level questions.
Spec2.01a Population and sample: terminology2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers
  1. Helen is studying the daily mean wind speed for Camborne using the large data set from 1987. The data for one month are summarised in Table 1 below.
\begin{table}[h]
Windspeed\(\mathrm { n } / \mathrm { a }\)67891112131416
Frequency13232231212
\captionsetup{labelformat=empty} \caption{Table 1}
\end{table}
  1. Calculate the mean for these data.
  2. Calculate the standard deviation for these data and state the units. The means and standard deviations of the daily mean wind speed for the other months from the large data set for Camborne in 1987 are given in Table 2 below. The data are not in month order. \begin{table}[h]
    Month\(A\)\(B\)\(C\)\(D\)\(E\)
    Mean7.588.268.578.5711.57
    Standard Deviation2.933.893.463.874.64
    \captionsetup{labelformat=empty} \caption{Table 2}
    \end{table}
  3. Using your knowledge of the large data set, suggest, giving a reason, which month had a mean of 11.57 The data for these months are summarised in the box plots on the opposite page. They are not in month order or the same order as in Table 2.
    1. State the meaning of the * symbol on some of the box plots.
    2. Suggest, giving your reasons, which of the months in Table 2 is most likely to be summarised in the box plot marked \(Y\). \includegraphics[max width=\textwidth, alt={}, center]{2edcf965-9c93-4a9b-9395-2d3c023801af-11_1177_1216_324_427}
Question 4:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\bar{x}=10.2\,(2222...)\) awrt 10.2B1 Allow exact fraction \(\frac{184}{18}\)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sigma_x=3.17\,(20227...)\) awrt 3.17B1ft Allow 3.2 from correct expression; accept \(s=3.26(3984...)\); \(\sigma_x=\sqrt{\frac{2062}{18}-\bar{x}^2}\); treating n/a as 0 gives \(\sigma_x=5.59(34...)\) scores 1st B1
Sight of "knots" or "kn" (condone knots/s etc)B1 Accept kn; accept in (a) or (b)
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
OctoberB1 Accept September; 1st B1 choosing October
It is windier in the autumn or month of the hurricane or latest month in the yearB1 For stating Camborne is windier in autumn/winter months; "Sep \(\leqslant\) month \(\leqslant\) Mar" scores B1B1 for "month" = Sep or Oct; B0B1 for other months in range
Part (d)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
They represent outliersB1 Allow "outlier" or idea of extreme value; allow "anomaly"
Part (d)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(Y\) has low median so expect lowish mean (but outlier so \(>7\)) and \(Y\) has big range/IQR or spread so expect larger st.devM1 For comment on location mentioning both median and mean and comment on spread mentioning both range/IQR and standard deviation; choosing \(A\) or \(E\) is M0
Suggests \(B\)A1 Accept \(D\) or \(B\) or \(D\); M1 must be scored; ALT: outlier is (mean\(+3\sigma\)): \(B=19.9\), \(C=18.95\), \(D=20.2\); must see at least one value compared to \(Y\)'s outlier 
## Question 4:

### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x}=10.2\,(2222...)$ awrt **10.2** | B1 | Allow exact fraction $\frac{184}{18}$ |

### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sigma_x=3.17\,(20227...)$ awrt **3.17** | B1ft | Allow 3.2 from correct expression; accept $s=3.26(3984...)$; $\sigma_x=\sqrt{\frac{2062}{18}-\bar{x}^2}$; treating n/a as 0 gives $\sigma_x=5.59(34...)$ scores 1st B1 |
| Sight of "knots" or "kn" (condone knots/s etc) | B1 | Accept kn; accept in (a) or (b) |

### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| October | B1 | Accept September; 1st B1 choosing October |
| It is windier in the autumn or month of the hurricane or latest month in the year | B1 | For stating Camborne is windier in autumn/winter months; "Sep $\leqslant$ month $\leqslant$ Mar" scores B1B1 for "month" = Sep or Oct; B0B1 for other months in range |

### Part (d)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| They represent outliers | B1 | Allow "outlier" or idea of extreme value; allow "anomaly" |

### Part (d)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y$ has low median so expect lowish mean (but outlier so $>7$) **and** $Y$ has big range/IQR or spread so expect larger st.dev | M1 | For comment on location mentioning both median and mean **and** comment on spread mentioning both range/IQR and standard deviation; choosing $A$ or $E$ is M0 |
| Suggests $B$ | A1 | Accept $D$ or $B$ or $D$; M1 must be scored; ALT: outlier is (mean$+3\sigma$): $B=19.9$, $C=18.95$, $D=20.2$; must see at least one value compared to $Y$'s outlier |
\begin{enumerate}
  \item Helen is studying the daily mean wind speed for Camborne using the large data set from 1987. The data for one month are summarised in Table 1 below.
\end{enumerate}

\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Windspeed & $\mathrm { n } / \mathrm { a }$ & 6 & 7 & 8 & 9 & 11 & 12 & 13 & 14 & 16 \\
\hline
Frequency & 13 & 2 & 3 & 2 & 2 & 3 & 1 & 2 & 1 & 2 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}

(a) Calculate the mean for these data.\\
(b) Calculate the standard deviation for these data and state the units.

The means and standard deviations of the daily mean wind speed for the other months from the large data set for Camborne in 1987 are given in Table 2 below. The data are not in month order.

\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Month & $A$ & $B$ & $C$ & $D$ & $E$ \\
\hline
Mean & 7.58 & 8.26 & 8.57 & 8.57 & 11.57 \\
\hline
Standard Deviation & 2.93 & 3.89 & 3.46 & 3.87 & 4.64 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}

(c) Using your knowledge of the large data set, suggest, giving a reason, which month had a mean of 11.57

The data for these months are summarised in the box plots on the opposite page. They are not in month order or the same order as in Table 2.\\
(d) (i) State the meaning of the * symbol on some of the box plots.\\
(ii) Suggest, giving your reasons, which of the months in Table 2 is most likely to be summarised in the box plot marked $Y$.

\includegraphics[max width=\textwidth, alt={}, center]{2edcf965-9c93-4a9b-9395-2d3c023801af-11_1177_1216_324_427}\\

\hfill \mbox{\textit{Edexcel AS Paper 2 2018 Q4 [8]}}
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