Edexcel S1 2011 June — Question 5 11 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2011
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate statistics from grouped frequency table
DifficultyModerate -0.8 This is a routine S1 statistics question testing standard procedures: finding mid-points, linear interpolation for median, calculating mean/standard deviation from grouped frequency data, and identifying skewness from quartiles. All parts follow textbook methods with no problem-solving or novel insight required. The calculations are straightforward given the formula for Σfx², making this easier than average A-level content.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)

A class of students had a sudoku competition. The time taken for each student to complete the sudoku was recorded to the nearest minute and the results are summarised in the table below.
TimeMid-point, \(x\)Frequency, \(f\)
2 - 852
9 - 127
13 - 15145
16 - 18178
19 - 2220.54
23 - 3026.54
(You may use \(\sum fx^2 = 8603.75\))
  1. Write down the mid-point for the 9 - 12 interval. [1]
  2. Use linear interpolation to estimate the median time taken by the students. [2]
  3. Estimate the mean and standard deviation of the times taken by the students. [5]
The teacher suggested that a normal distribution could be used to model the times taken by the students to complete the sudoku.
  1. Give a reason to support the use of a normal distribution in this case. [1]
On another occasion the teacher calculated the quartiles for the times taken by the students to complete a different sudoku and found \(Q_1 = 8.5 \quad Q_2 = 13.0 \quad Q_3 = 21.0\)
  1. Describe, giving a reason, the skewness of the times on this occasion. [2]

(a)
AnswerMarks
10.5B1
(b)
AnswerMarks
\((Q_2 =) (15.5 +) \frac{\frac{1}{2} \times 30 - 14}{8} \times 3\) or \(\frac{\frac{1}{2} \times 31 - 14}{8} \times 3\)M1
= 15.875 or 16.0625A1
(c)
AnswerMarks
\(\bar{x} = \frac{477.5}{30} = 15.9\) (15.91⅔) [Accept \(\frac{191}{12}\) or \(15\frac{11}{12}\)]M1, A1
\(\sigma = \sqrt{\frac{8603.75}{30} - x^2} = 5.78\) (accept s = 5.88)M1A1ft, A1
(d)
AnswerMarks
Since mean and median are similar (or equal or very close) a normal distribution may be suitable. [Allow mean or median close to mode/modal class]B1
(e)
AnswerMarks
\(Q_3 - Q_2 (= 8) > (4.5 =) Q_2 - Q_1\) Therefore positive skewM1, A1
Notes:
- In parts (a) to (c) a correct answer with no working scores full marks for that value.
- (a) B1 for 10.5 which may be in the table
- (b) M1 for a correct ratio and times 3, ignore the lower boundary for this mark
- A1 for awrt 15.9 (if n = 30 used) or awrt 16.1 (if n+1 = 31 is used)
- (c) 1st M1 for attempt at \(\sum fx\) (this may be seen in the table as fx: 10, 73.5, 70, 136, 82, 106 [condone 1 slip] or awrt 500) and use of \(\frac{\sum fx}{\sum f}\) or a correct expression for mean.
- 1st A1 for awrt 15.9
- 2nd M1 for an attempt at σ or σ², can ft their mean, condone mis-labelling \(\sigma^2 = \sqrt{...}\) etc. Allow use of their \(\sum fx^2\) (awrt 9000)
- 2nd A1ft for a correct expression including square root, ft their mean but not their \(\sum fx^2\).
- No label or correct label is OK but wrong label (e.g. \(\sigma^2 = \sqrt{...}\)) is A0
- 3rd A1 for awrt 5.78, allow s = awrt 5.88. SC Allow M1A1A0 for awrt 5.79 if \(\bar{x}\) correct
- (d) B1 for a reason implying or stating symmetry. "Time is continuous" or "evenly distributed" is B0
- (e) M1 for a clear reason or comparison, values not essential but comparison implying they have been found is required.
- A1 for stating "positive skew". Condone just "positive" but "positive correlation" is A0. Do not allow arguments based on mean and median since this part relates to a different set of data.
## (a)
10.5 | B1 |

## (b)
$(Q_2 =) (15.5 +) \frac{\frac{1}{2} \times 30 - 14}{8} \times 3$ or $\frac{\frac{1}{2} \times 31 - 14}{8} \times 3$ | M1 |
= 15.875 or 16.0625 | A1 |

## (c)
$\bar{x} = \frac{477.5}{30} = 15.9$ (15.91⅔) [Accept $\frac{191}{12}$ or $15\frac{11}{12}$] | M1, A1 |
$\sigma = \sqrt{\frac{8603.75}{30} - x^2} = 5.78$ (accept s = 5.88) | M1A1ft, A1 |

## (d)
Since mean and median are similar (or equal or very close) a normal distribution may be suitable. [Allow mean or median close to mode/modal class] | B1 |

## (e)
$Q_3 - Q_2 (= 8) > (4.5 =) Q_2 - Q_1$ Therefore positive skew | M1, A1 |

**Notes:**
- **In parts (a) to (c)** a correct answer with no working scores full marks for that value.
- **(a)** B1 for 10.5 which may be in the table
- **(b)** M1 for a correct ratio and times 3, ignore the lower boundary for this mark
  - A1 for awrt 15.9 (if n = 30 used) or awrt 16.1 (if n+1 = 31 is used)
- **(c)** 1st M1 for attempt at $\sum fx$ (this may be seen in the table as fx: 10, 73.5, 70, 136, 82, 106 [condone 1 slip] or awrt 500) and use of $\frac{\sum fx}{\sum f}$ or a correct expression for mean.
  - 1st A1 for awrt 15.9
  - 2nd M1 for an attempt at σ or σ², can ft their mean, condone mis-labelling $\sigma^2 = \sqrt{...}$ etc. Allow use of their $\sum fx^2$ (awrt 9000)
  - 2nd A1ft for a correct expression including square root, ft their mean but not their $\sum fx^2$.
  - No label or correct label is OK but wrong label (e.g. $\sigma^2 = \sqrt{...}$) is A0
  - 3rd A1 for awrt 5.78, allow s = awrt 5.88. **SC Allow M1A1A0 for awrt 5.79 if $\bar{x}$ correct**
- **(d)** B1 for a reason implying or stating symmetry. "Time is continuous" or "evenly distributed" is B0
- **(e)** M1 for a clear reason or comparison, values not essential but comparison implying they have been found is required.
  - A1 for stating "positive skew". Condone just "positive" but "positive correlation" is A0. **Do not allow arguments based on mean and median since this part relates to a different set of data.**

---
A class of students had a sudoku competition. The time taken for each student to complete the sudoku was recorded to the nearest minute and the results are summarised in the table below.

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Time & Mid-point, $x$ & Frequency, $f$ \\
\hline
2 - 8 & 5 & 2 \\
\hline
9 - 12 & & 7 \\
\hline
13 - 15 & 14 & 5 \\
\hline
16 - 18 & 17 & 8 \\
\hline
19 - 22 & 20.5 & 4 \\
\hline
23 - 30 & 26.5 & 4 \\
\hline
\end{tabular}
\end{center}

(You may use $\sum fx^2 = 8603.75$)

\begin{enumerate}[label=(\alph*)]
\item Write down the mid-point for the 9 - 12 interval. [1]

\item Use linear interpolation to estimate the median time taken by the students. [2]

\item Estimate the mean and standard deviation of the times taken by the students. [5]
\end{enumerate}

The teacher suggested that a normal distribution could be used to model the times taken by the students to complete the sudoku.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Give a reason to support the use of a normal distribution in this case. [1]
\end{enumerate}

On another occasion the teacher calculated the quartiles for the times taken by the students to complete a different sudoku and found

$Q_1 = 8.5 \quad Q_2 = 13.0 \quad Q_3 = 21.0$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Describe, giving a reason, the skewness of the times on this occasion. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2011 Q5 [11]}}