OCR S1 2006 June — Question 7 13 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2006
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeEstimate mean and standard deviation from frequency table
DifficultyModerate -0.8 This is a routine S1 statistics question requiring standard procedures: calculating mean/SD from grouped frequency data, finding IQR from cumulative frequencies, and conceptual understanding of how changing class boundaries affects summary statistics. All techniques are straightforward textbook applications with no novel problem-solving required.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation

7 In a UK government survey in 2000, smokers were asked to estimate the time between their waking and their having the first cigarette of the day. For heavy smokers, the results were as follows.
Time between waking
and first cigarette
1 to 4
minutes
5 to 14
minutes
15 to 29
minutes
30 to 59
minutes
At least 60
minutes
Percentage of smokers312719149
Times are given correct to the nearest minute.
  1. Assuming that 'At least 60 minutes' means 'At least 60 minutes but less than 240 minutes', calculate estimates for the mean and standard deviation of the time between waking and first cigarette for these smokers.
  2. Find an estimate for the interquartile range of the time between waking and first cigarette for these smokers. Give your answer correct to the nearest minute.
  3. The meaning of 'At least 60 minutes' is now changed to 'At least 60 minutes but less than 480 minutes'. Without further calculation, state whether this would cause an increase, a decrease or no change in the estimated value of
    1. the mean,
    2. the standard deviation,
    3. the interquartile range.

AnswerMarks Guidance
7(i)Midpoints attempted \(\geq 2\) classes \(\Sigma f / 100\) or \(\Sigma f / \Sigma f\) attempted \(\geq 2\) terms or \(x\) within class, not class width Mean \(= 27.2\) (to 3 sfs) (not 27.25) art 27.2 from fully correct wking M1: —; M1: —
\(\Sigma x^2f\) or \(\Sigma(x - \bar{x})^2 f\) \(\geq 2\) terms \(\sqrt{(\Sigma x^2f / 100- \bar{x}^2)} \) or \(\sqrt{(\Sigma(x - \bar{x})^2f / 100)}\) or \(\sqrt{\Sigma f}\) fully corr method, not /neg \(= 40.5\) to 41.1 (3 sfs)A1: 6 27.2; 240702.25; 40.82; 27.25; 242050; 40.96 allow class widths for 2nd M1 only
7(ii)Recog LQ in 1st class & UQ in 3rd class Graph: Attempt 25(25)th value Attempt 75(75)th value Interp: LQ \(= 3.0\) to 4.3 UQ \(= 27\) to 29 Subtract IQR \(= 23\) or 24 or 25 B1: —; M1: —; M1: —; A1: 4
7(iii)(a)Increase B1: 1
(b)Increase B1: 1
(c)No change B1: 1
Total: 13
7(i) | Midpoints attempted $\geq 2$ classes $\Sigma f / 100$ or $\Sigma f / \Sigma f$ attempted $\geq 2$ terms or $x$ within class, not class width Mean $= 27.2$ (to 3 sfs) (not 27.25) art 27.2 from fully correct wking | M1: —; M1: — | Correct (149.5): 2720.5/100 With 150: 2725/100 Allow Ms & poss As |
| | | $\Sigma x^2f$ or $\Sigma(x - \bar{x})^2 f$ $\geq 2$ terms $\sqrt{(\Sigma x^2f / 100- \bar{x}^2)} $ or $\sqrt{(\Sigma(x - \bar{x})^2f / 100)}$ or $\sqrt{\Sigma f}$ fully corr method, not /neg $= 40.5$ to 41.1 (3 sfs) | A1: 6 | 27.2; 240702.25; 40.82; 27.25; 242050; 40.96 allow class widths for 2nd M1 only |

7(ii) | Recog LQ in 1st class & UQ in 3rd class Graph: Attempt 25(25)th value Attempt 75(75)th value Interp: LQ $= 3.0$ to 4.3 UQ $= 27$ to 29 Subtract IQR $= 23$ or 24 or 25 | B1: —; M1: —; M1: —; A1: 4 | both nec'y dep B1or M1 integer, dep M2 |

7(iii)(a) | Increase | B1: 1 | Ignore "probably" etc |
| (b) | Increase | B1: 1 | |
| (c) | No change | B1: 1 | |

**Total: 13**

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7 In a UK government survey in 2000, smokers were asked to estimate the time between their waking and their having the first cigarette of the day. For heavy smokers, the results were as follows.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Time between waking \\
and first cigarette \\
\end{tabular} & \begin{tabular}{ c }
1 to 4 \\
minutes \\
\end{tabular} & \begin{tabular}{ c }
5 to 14 \\
minutes \\
\end{tabular} & \begin{tabular}{ c }
15 to 29 \\
minutes \\
\end{tabular} & \begin{tabular}{ c }
30 to 59 \\
minutes \\
\end{tabular} & \begin{tabular}{ c }
At least 60 \\
minutes \\
\end{tabular} \\
\hline
Percentage of smokers & 31 & 27 & 19 & 14 & 9 \\
\hline
\end{tabular}
\end{center}

Times are given correct to the nearest minute.\\
(i) Assuming that 'At least 60 minutes' means 'At least 60 minutes but less than 240 minutes', calculate estimates for the mean and standard deviation of the time between waking and first cigarette for these smokers.\\
(ii) Find an estimate for the interquartile range of the time between waking and first cigarette for these smokers. Give your answer correct to the nearest minute.\\
(iii) The meaning of 'At least 60 minutes' is now changed to 'At least 60 minutes but less than 480 minutes'. Without further calculation, state whether this would cause an increase, a decrease or no change in the estimated value of
\begin{enumerate}[label=(\alph*)]
\item the mean,
\item the standard deviation,
\item the interquartile range.
\end{enumerate}

\hfill \mbox{\textit{OCR S1 2006 Q7 [13]}}