OCR MEI S1 2010 June — Question 3 7 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeDraw histogram then perform other calculations
DifficultyModerate -0.8 This is a straightforward histogram question with unequal class widths requiring frequency density calculation, plus a simple median location using cumulative frequency. Both are standard S1 techniques with no problem-solving or novel insight required—easier than average A-level.
Spec2.02b Histogram: area represents frequency2.02f Measures of average and spread

3 The lifetimes in hours of 90 components are summarised in the table.
Lifetime \(( x\) hours \()\)\(0 < x \leqslant 20\)\(20 < x \leqslant 30\)\(30 < x \leqslant 50\)\(50 < x \leqslant 65\)\(65 < x \leqslant 100\)
Frequency2413142118
  1. Draw a histogram to illustrate these data.
  2. In which class interval does the median lie? Justify your answer.

(i)
AnswerMarks Guidance
Lifetime (x hours)Frequency Width
\(0 < x \leq 20\)24 20
\(20 < x \leq 30\)13 10
\(30 < x \leq 50\)14 20
\(50 < x \leq 65\)21 15
\(65 < x \leq 100\)18 35
M1 for fds, A1 CAO
Accept any suitable unit for fd such as eg freq per 10 hours.
L1 linear scales on both axes and label on vert axis
W1 width of bars, H1 height of bars5 marks
(ii) Median lies in third class interval \((30 < x \leq 50)\)B1 CAO
Median = 45.5th lifetime (which lies beyond 37 but not as far as 51)E1 dep on B1 2 marks
TOTAL: 7 marks
**(i)** 
| Lifetime (x hours) | Frequency | Width | FD |
|---|---|---|---|
| $0 < x \leq 20$ | 24 | 20 | 1.2 |
| $20 < x \leq 30$ | 13 | 10 | 1.3 |
| $30 < x \leq 50$ | 14 | 20 | 0.7 |
| $50 < x \leq 65$ | 21 | 15 | 1.4 |
| $65 < x \leq 100$ | 18 | 35 | 0.51 |

| M1 for fds, A1 CAO |

Accept any suitable unit for fd such as eg freq per 10 hours. |

| L1 linear scales on both axes and label on vert axis |

| W1 width of bars, H1 height of bars | 5 marks

**(ii)** Median lies in third class interval $(30 < x \leq 50)$ | B1 CAO |

Median = 45.5th lifetime (which lies beyond 37 but not as far as 51) | E1 dep on B1 | 2 marks

**TOTAL: 7 marks**

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3 The lifetimes in hours of 90 components are summarised in the table.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Lifetime $( x$ hours $)$ & $0 < x \leqslant 20$ & $20 < x \leqslant 30$ & $30 < x \leqslant 50$ & $50 < x \leqslant 65$ & $65 < x \leqslant 100$ \\
\hline
Frequency & 24 & 13 & 14 & 21 & 18 \\
\hline
\end{tabular}
\end{center}

(i) Draw a histogram to illustrate these data.\\
(ii) In which class interval does the median lie? Justify your answer.

\hfill \mbox{\textit{OCR MEI S1 2010 Q3 [7]}}