OCR MEI S1 — Question 3 18 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeOutliers from cumulative frequency diagram
DifficultyModerate -0.3 This is a multi-part statistics question requiring standard techniques: reading cumulative frequency diagrams for median/quartiles/percentiles, applying the 1.5×IQR outlier rule, and basic binomial probability calculations. While it has many parts (6 marks worth), each individual step is routine and requires only direct application of learned procedures without novel problem-solving or conceptual depth.
Spec2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02h Recognize outliers2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities

3 The birth weights in grams of a random sample of 1000 babies are displayed in the cumulative frequency diagram below. \includegraphics[max width=\textwidth, alt={}, center]{79f1015b-7c3d-4576-8d5b-e9fc89d8a49e-2_1266_1546_487_335}
  1. Use the diagram to estimate the median and interquartile range of the data.
  2. Use your answers to part (i) to estimate the number of outliers in the sample.
  3. Should these outliers be excluded from any further analysis? Briefly explain your answer.
  4. Any baby whose weight is below the 10th percentile is selected for careful monitoring. Use the diagram to determine the range of weights of the babies who are selected. \(12 \%\) of new-born babies require some form of special care. A maternity unit has 17 new-born babies. You may assume that these 17 babies form an independent random sample.
  5. Find the probability that
    (A) exactly 2 of these 17 babies require special care,
    (B) more than 2 of the 17 babies require special care.
  6. On 100 independent occasions the unit has 17 babies. Find the expected number of occasions on which there would be more than 2 babies who require special care.

Question 3:
Part (i)
AnswerMarks Guidance
Median \(= 3370\); \(Q_1 = 3050\), \(Q_3 = 3700\); IQR \(= 3700 - 3050 = 650\)B1; B1 for \(Q_3\) or \(Q_1\); B1 for IQR [3]
Part (ii)
AnswerMarks Guidance
Lower limit: \(3050 - 1.5 \times 650 = 2075\); Upper limit: \(3700 + 1.5 \times 650 = 4675\); Approx 40 babies below 2075 and 5 above 4675, total 45B1; B1; M1 (for either); A1 [4]
Part (iii)
AnswerMarks Guidance
Decision based on convincing argument e.g. 'no, because there is nothing to suggest that they are not genuine data items and these data may influence health care provision'E2 for convincing argument [2]
Part (iv)
AnswerMarks Guidance
All babies below 2600 grams in weightB2 CAO [2]
Part (v)(A)
AnswerMarks Guidance
\(X \sim B(17, 0.12)\); \(P(X=2) = \binom{17}{2} \times 0.12^2 \times 0.88^{15} = 0.2878\)M1: \(\binom{17}{2} \times p^2 \times q^{15}\); M1 indep: \(0.12^2 \times 0.88^{15}\); A1 CAO [3]
Part (v)(B)
AnswerMarks Guidance
\(P(X>2) = 1 - (0.2878 + \binom{17}{1} \times 0.12 \times 0.88^{16} + 0.88^{17})\) \(= 1-(0.2878 + 0.2638 + 0.1138) = 0.335\)M1 for \(P(X=1)+P(X=0)\); M1 for \(1-P(X \leq 2)\); A1 CAO [3]
Part (vi)
AnswerMarks Guidance
Expected number of occasions is 33.5B1 FT [1]
## Question 3:

### Part (i)
| Median $= 3370$; $Q_1 = 3050$, $Q_3 = 3700$; IQR $= 3700 - 3050 = 650$ | B1; B1 for $Q_3$ or $Q_1$; B1 for IQR | [3] |

### Part (ii)
| Lower limit: $3050 - 1.5 \times 650 = 2075$; Upper limit: $3700 + 1.5 \times 650 = 4675$; Approx 40 babies below 2075 and 5 above 4675, total 45 | B1; B1; M1 (for either); A1 | [4] |

### Part (iii)
| Decision based on convincing argument e.g. 'no, because there is nothing to suggest that they are not genuine data items and these data may influence health care provision' | E2 for convincing argument | [2] |

### Part (iv)
| All babies below 2600 grams in weight | B2 CAO | [2] |

### Part (v)(A)
| $X \sim B(17, 0.12)$; $P(X=2) = \binom{17}{2} \times 0.12^2 \times 0.88^{15} = 0.2878$ | M1: $\binom{17}{2} \times p^2 \times q^{15}$; M1 indep: $0.12^2 \times 0.88^{15}$; A1 CAO | [3] |

### Part (v)(B)
| $P(X>2) = 1 - (0.2878 + \binom{17}{1} \times 0.12 \times 0.88^{16} + 0.88^{17})$ $= 1-(0.2878 + 0.2638 + 0.1138) = 0.335$ | M1 for $P(X=1)+P(X=0)$; M1 for $1-P(X \leq 2)$; A1 CAO | [3] |

### Part (vi)
| Expected number of occasions is 33.5 | B1 FT | [1] |

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3 The birth weights in grams of a random sample of 1000 babies are displayed in the cumulative frequency diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{79f1015b-7c3d-4576-8d5b-e9fc89d8a49e-2_1266_1546_487_335}
\begin{enumerate}[label=(\roman*)]
\item Use the diagram to estimate the median and interquartile range of the data.
\item Use your answers to part (i) to estimate the number of outliers in the sample.
\item Should these outliers be excluded from any further analysis? Briefly explain your answer.
\item Any baby whose weight is below the 10th percentile is selected for careful monitoring. Use the diagram to determine the range of weights of the babies who are selected.\\
$12 \%$ of new-born babies require some form of special care. A maternity unit has 17 new-born babies. You may assume that these 17 babies form an independent random sample.
\item Find the probability that\\
(A) exactly 2 of these 17 babies require special care,\\
(B) more than 2 of the 17 babies require special care.
\item On 100 independent occasions the unit has 17 babies. Find the expected number of occasions on which there would be more than 2 babies who require special care.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S1  Q3 [18]}}
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