Edexcel S1 2003 June — Question 6 16 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2003
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeOutliers from raw data
DifficultyModerate -0.8 This is a straightforward S1 question testing standard descriptive statistics procedures: calculating mean, drawing stem-and-leaf, finding quartiles, applying the 1.5×IQR outlier rule, drawing a box plot, and commenting on skewness. All steps are routine recall and application of formulas with no problem-solving or insight required. Easier than average A-level due to being purely procedural.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers

6. The number of bags of potato crisps sold per day in a bar was recorded over a two-week period. The results are shown below. $$20,15,10,30,33,40,5,11,13,20,25,42,31,17$$
  1. Calculate the mean of these data.
  2. Draw a stem and leaf diagram to represent these data.
  3. Find the median and the quartiles of these data. An outlier is an observation that falls either \(1.5 \times\) (interquartile range) above the upper quartile or \(1.5 \times\) (interquartile range) below the lower quartile.
  4. Determine whether or not any items of data are outliers.
  5. On graph paper draw a box plot to represent these data. Show your scale clearly.
  6. Comment on the skewness of the distribution of bags of crisps sold per day. Justify your answer.

(a)
AnswerMarks Guidance
\(\bar{x} = \frac{20+15+\ldots+17}{14} = \frac{312}{14} = 22.2857\ldots\)M1 A1 (awrt 22.3) (2 marks)
(b)
AnswerMarks Guidance
Bags of crisps1 0 means 10
05
10 1 3 5 7
20 0 5
30 1 3
40 2
B1 B1 B1(3 marks)
(c)
AnswerMarks Guidance
\(Q_2 = 20; \quad Q_1 = 13; \quad Q_3 = 31\)B1; B1; B1 (3 marks)
(d)
AnswerMarks Guidance
\(1.5 \times \text{IQR} = 1.5 \times (31-13) = 27\)B1
\(31 + 27 = 58; \quad 13-27 = -14\)M1 both
No outliersA1 B1 (3 marks)
(e)
AnswerMarks
scale and labelB1
\(Q_1 = 13, \quad Q_2 = 20, \quad Q_3 = 31\)B1 ft
Whiskers 5, 42;B1
(f)
AnswerMarks Guidance
\(Q_2 - Q_1 = 7; \quad Q_3 - Q_2 = 11; \quad Q_3 - Q_2 > Q_2 - Q_1\)M1
Positive skewA1 (2 marks)
(13 marks)
**(a)**
| $\bar{x} = \frac{20+15+\ldots+17}{14} = \frac{312}{14} = 22.2857\ldots$ | M1 A1 | (awrt 22.3) (2 marks) |

**(b)**
| Bags of crisps | 1 | 0 means 10 | Totals | | |
|---|---|---|---|---|---|
| 0 | 5 | | | (1) | Label and key |
| 1 | 0 | 1 3 5 7 | | (5) | 2 correct rows |
| 2 | 0 | 0 5 | | (3) | All correct |
| 3 | 0 | 1 3 | | (3) | |
| 4 | 0 | 2 | | (2) | |
| | | | | B1 B1 B1 | (3 marks) |

**(c)**
| $Q_2 = 20; \quad Q_1 = 13; \quad Q_3 = 31$ | B1; B1; B1 | (3 marks) |

**(d)**
| $1.5 \times \text{IQR} = 1.5 \times (31-13) = 27$ | B1 | |
| $31 + 27 = 58; \quad 13-27 = -14$ | M1 | both |
| No outliers | A1 B1 | (3 marks) |

**(e)**
| scale and label | B1 | |
| $Q_1 = 13, \quad Q_2 = 20, \quad Q_3 = 31$ | B1 ft | |
| Whiskers 5, 42; | B1 | |

**(f)**
| $Q_2 - Q_1 = 7; \quad Q_3 - Q_2 = 11; \quad Q_3 - Q_2 > Q_2 - Q_1$ | M1 | |
| Positive skew | A1 | (2 marks) |
| | | (13 marks) |

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6. The number of bags of potato crisps sold per day in a bar was recorded over a two-week period. The results are shown below.

$$20,15,10,30,33,40,5,11,13,20,25,42,31,17$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean of these data.
\item Draw a stem and leaf diagram to represent these data.
\item Find the median and the quartiles of these data.

An outlier is an observation that falls either $1.5 \times$ (interquartile range) above the upper quartile or $1.5 \times$ (interquartile range) below the lower quartile.
\item Determine whether or not any items of data are outliers.
\item On graph paper draw a box plot to represent these data. Show your scale clearly.
\item Comment on the skewness of the distribution of bags of crisps sold per day. Justify your answer.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2003 Q6 [16]}}