CAIE S1 2013 November — Question 4 8 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2013
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeEstimate mean and standard deviation from histogram
DifficultyModerate -0.8 This is a straightforward histogram interpretation question requiring standard calculations of mean and standard deviation from grouped data. The question guides students through reading frequency densities, converting to frequencies, and applying standard formulas—all routine A-level statistics procedures with no novel problem-solving required.
Spec2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation

4 The following histogram summarises the times, in minutes, taken by 190 people to complete a race. \includegraphics[max width=\textwidth, alt={}, center]{df246a50-157b-49f7-bba0-f9b86960b8b9-2_1210_1125_1251_513}
  1. Show that 75 people took between 200 and 250 minutes to complete the race.
  2. Calculate estimates of the mean and standard deviation of the times of the 190 people.
  3. Explain why your answers to part (ii) are estimates.

AnswerMarks Guidance
(i) number \(= 1.5 \times 50 = 75\) (AG)B1 [1] Must see \(1.5 \times 50\)
(ii) freqs are 10, 25, 50, 75, 30 (15, 15)M1, A1 Attempt at freqs not fd; Correct freqs
Mean \(= \frac{(10 \times 125 + 25 \times 162.5 + 50 \times 187.5 + 75 \times 225 + 30 \times 300)/190}{= 40562.5/190 = 213 (213.48 \ldots)}\)M1, A1 attempt at mid points not cw or ucb or lcb; correct mean
\(sd^2 = \frac{10 \times 125^2 + 25 \times 162.5^2 + 50 \times 187.5^2 + 75 \times 225^2 + 30 \times 300^2}{190} - (213.48 \ldots)^2\)M1 subst their \(\Sigma fx^2\) in incorrect variance formula
\(sd = 46.5\) or 46.6A1 [6]
(iii) have used the mid-point of each interval and not the raw dataB1 [1]
**(i)** number $= 1.5 \times 50 = 75$ (AG) | B1 [1] | Must see $1.5 \times 50$

**(ii)** freqs are 10, 25, 50, 75, 30 (15, 15) | M1, A1 | Attempt at freqs not fd; Correct freqs

Mean $= \frac{(10 \times 125 + 25 \times 162.5 + 50 \times 187.5 + 75 \times 225 + 30 \times 300)/190}{= 40562.5/190 = 213 (213.48 \ldots)}$ | M1, A1 | attempt at mid points not cw or ucb or lcb; correct mean

$sd^2 = \frac{10 \times 125^2 + 25 \times 162.5^2 + 50 \times 187.5^2 + 75 \times 225^2 + 30 \times 300^2}{190} - (213.48 \ldots)^2$ | M1 | subst their $\Sigma fx^2$ in incorrect variance formula

$sd = 46.5$ or 46.6 | A1 [6] |

**(iii)** have used the mid-point of each interval and not the raw data | B1 [1] |
4 The following histogram summarises the times, in minutes, taken by 190 people to complete a race.\\
\includegraphics[max width=\textwidth, alt={}, center]{df246a50-157b-49f7-bba0-f9b86960b8b9-2_1210_1125_1251_513}\\
(i) Show that 75 people took between 200 and 250 minutes to complete the race.\\
(ii) Calculate estimates of the mean and standard deviation of the times of the 190 people.\\
(iii) Explain why your answers to part (ii) are estimates.

\hfill \mbox{\textit{CAIE S1 2013 Q4 [8]}}