Marks available | 3 |

Reference code | 11M.1.sl.TZ2.6 |

## Question

A scientist has 100 female fish and 100 male fish. She measures their lengths to the nearest cm. These are shown in the following box and whisker diagrams.

Find the range of the lengths of **all** 200 fish.

Four cumulative frequency graphs are shown below.

Which graph is the best representation of the lengths of the **female** fish?

## Markscheme

correct end points **(A1)(A1)**

max = 27 , min = 4

range = 23 **A1 N3**

**[3 marks]**

Graph 3 **A2 N2**

**[2 marks]**

## Examiners report

While there were a large number of candidates who answered both parts of this question correctly, a surprising number did not know how to find the range of all 200 fish in part (a). Common errors included finding the ranges of the male and female fish separately, or averaging the separate ranges of the male and female fish.

Some candidates did not interpret the cumulative frequency graphs correctly, or just seemed to guess which graph was correct. The most common incorrect “guess” was graph 4, likely because this graph had a more familiar cumulative shape.

Marks available | 2 |

Reference code | 12M.1.sl.TZ2.1 |

## Question

The cumulative frequency curve below represents the marks obtained by 100 students.

Find the median mark.

Find the interquartile range.

## Markscheme

evidence of median position **(M1) **

e.g. 50, line on sketch

median is 56 **A1 N2**

**[2 marks] **

lower quartile \(= 40\) , upper quartile \(= 70\) ** (A1)(A1)**

interquartile range \(= 30\) **A1 N3**

**[3 marks] **

## Examiners report

Overall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (the cumulative frequency) on the *y*-axis, but did not find the median mark on the *x*-axis.

Overall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (the cumulative frequency) on the *y*-axis, but did not find the median mark on the *x*-axis. Similar misunderstanding was shown by some candidates in part (b), when attempting to find the interquartile range.

Marks available | 4 |

Reference code | 15N.1.sl.TZ0.1 |

## Question

The following box-and-whisker plot represents the examination scores of a group of students.

Write down the median score.

The range of the scores is 47 marks, and the interquartile range is 22 marks.

Find the value of

(i) \(c\);

(ii) \(d\).

## Markscheme

60 *A1 N1*

(i) valid approach *(M1)*

*eg *\(\max – \min = {\rm{range}},c = 40 + 47\)

\(c = 87\) *A1 N2*

(ii) valid approach *(M1)*

*eg*\(\;\;\;Q3 – Q1 = IQR,{\text{ }}74 – 22\)

\(d = 52\) *A1 N2*

## Examiners report

Marks available | 2 |

Reference code | 16M.1.sl.TZ1.8 |

## Question

A school collects cans for recycling to raise money. Sam’s class has 20 students.

The number of cans collected by each student in Sam’s class is shown in the following stem and leaf diagram.

The following box-and-whisker plot also displays the number of cans collected by students in Sam’s class.

There are 80 students in the school.

The mean number of cans collected is 39.4. The standard deviation is 18.5.

Each student then collects 2 more cans.

Find the median number of cans collected.

(i) Write down the value of \(a\).

(ii) The interquartile range is 14. Find the value of \(b\).

Sam’s class collected 745 cans. They want an average of 40 cans per student.

How many more cans need to be collected to achieve this target?

The students raise $0.10 for each recycled can.

(i) Find the largest amount raised by a student in Sam’s class.

(ii) The following cumulative frequency curve shows the amounts in dollars raised by all the students in the school. Find the percentage of students in the school who raised more money than anyone in Sam’s class.

(i) Write down the new mean.

(ii) Write down the new standard deviation.

## Markscheme

valid approach *(M1)*

*eg*\(\,\,\,\,\,\)between 10th and 11th, \(\frac{{8 + 8}}{2}\)

median \( = 38\) *A1 N2*

*[2 marks]*

(i) \(a = 20\) *A1 N1*

(ii) valid approach *(M1)*

*eg*\(\,\,\,\,\,\)\({Q_3} – {Q_1},{\text{ }}{Q_1} + 14,{\text{ }}b – 30 = 14\)

\(b = 44\) *A1 N2*

*[3 marks]*

valid approach *(M1)*

*eg*x\(\,\,\,\,\,\)\(40 \times 20,{\text{ }}\frac{{x + 745}}{{20}},{\text{ }}40 – \frac{{745}}{{20}}\)

correct working *(A1)*

*eg*\(\,\,\,\,\,\)\(800 – 745,{\text{ }}20 \times 2.75\)

55 (more cans) *A1 N2*

*[3 marks]*

(i) most cans in Sam’s class \( = 50\) *(A1)*

5 ($) *A1 N2*

(ii) correct value of 64 or 16 *A1*

valid approach *(M1)*

*eg*\(\,\,\,\,\,\)\(\frac{{64}}{{80}},{\text{ }}80\% ,{\text{ }}80 – 64,{\text{ }}\frac{{16}}{{80}}\)

20% *A1 N2*

*[5 marks]*

(i) 41.4 (exact) *A1 N1*

(ii) 18.5 *A1 N1*

*[2 marks]*

## Examiners report

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 – 745 = 15\).

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 – 745 = 15\).

Generally, candidates were very successful with this question, appearing to move easily between the three different representations of data. The main conceptual errors appeared in part d) where a percentage of 100 was found (instead of 80) and in part e) where the new standard deviation was often given as 20.5. Arithmetic errors seemed to be the other factor, with a surprising number of candidates finding in part c) that \(800 – 745 = 15\).

Marks available | 2 |

Reference code | 16M.1.sl.TZ2.2 |

## Question

There are 10 items in a data set. The sum of the items is 60.

The variance of this data set is 3. Each value in the set is multiplied by 4.

Find the mean.

(i) Write down the value of the new mean.

(ii) Find the value of the new variance.

## Markscheme

correct approach *(A1)*

*eg*\(\,\,\,\,\,\)\(\frac{{60}}{{10}}\)

\({\text{mean}} = 6\) *A1 N2*

(i) new mean \( = 24\) *A1 N1*

(ii) valid approach *(M1)*

*eg*\(\,\,\,\,\,\)\({\text{variance}} \times {(4)^2},{\text{ }}3 \times 16\), new standard deviation \( = 4\sqrt 3 \)

new variance \( = 48\) *A1 N2*

*[3 marks]*

## Examiners report

While most candidates were able to answer part (a) of this question correctly, they were not as successful in part (b). It seems that the item on the Maths SL syllabus dealing with the “effect of constant changes to the original data” was skipped over in many schools.

While most candidates were able to answer part (a) of this question correctly, they were not as successful in part (b). It seems that the item on the Maths SL syllabus dealing with the “effect of constant changes to the original data” was skipped over in many schools.

Marks available | 1 |

Reference code | 17M.1.sl.TZ2.8 |

## Question

A city hired 160 employees to work at a festival. The following cumulative frequency curve shows the number of hours employees worked during the festival.

The city paid each of the employees £8 per hour for the first 40 hours worked, and £10 per hour for each hour they worked after the first 40 hours.

Find the median number of hours worked by the employees.

Write down the number of employees who worked 50 hours or less.

Find the amount of money an employee earned for working 40 hours;

Find the amount of money an employee earned for working 43 hours.

Find the number of employees who earned £200 or less.

Only 10 employees earned more than £\(k\). Find the value of \(k\).

## Markscheme

evidence of median position *(M1)*

*eg*\(\,\,\,\,\,\)80th employee

40 hours *A1**N2*

*[2 marks]*

130 employees *A1**N1*

*[1 mark]*

£320 *A1**N1*

*[1 mark]*

splitting into 40 and 3 *(M1)*

*eg*\(\,\,\,\,\,\)3 hours more, \(3 \times 10\)

correct working *(A1)*

*eg*\(\,\,\,\,\,\)\(320 + 3 \times 10\)

£350 *A1**N3*

*[3 marks]*

valid approach *(M1)*

*eg*\(\,\,\,\,\,\)200 is less than 320 so 8 pounds/hour, \(200 \div 8,{\text{ }}25,{\text{ }}\frac{{200}}{{320}} = \frac{x}{{40}}\),

18 employees *A2**N3*

*[3 marks]*

valid approach *(M1)*

*eg*\(\,\,\,\,\,\)\(160 – 10\)

60 hours worked *(A1)*

correct working *(A1)*

*eg*\(\,\,\,\,\,\)\(40(8) + 20(10),{\text{ }}320 + 200\)

\(k = 520\) *A1**N3*

*[4 marks]*

## Examiners report

Marks available | 2 |

Reference code | 18M.1.sl.TZ1.2 |

## Question

The following box-and-whisker plot shows the number of text messages sent by students in a school on a particular day.

Find the value of the interquartile range.

One student sent* k* text messages, where *k* > 11 . Given that *k* is an outlier, find the least value of *k*.

## Markscheme

recognizing *Q*_{1} or *Q*_{3} (seen anywhere) **(M1)**

eg 4,11 , indicated on diagram

IQR = 7 **A1 N2**

**[2 marks]**

recognizing the need to find 1.5 IQR **(M1)**

*eg * 1.5 × IQR, 1.5 × 7

valid approach to find *k * ** (M1)**

*eg *10.5 + 11, 1.5 × IQR + *Q*_{3}

21.5 **(A1) **

*k* = 22 **A1 N3**

**Note:** If no working shown, award * N2* for an answer of 21.5.

**[4 marks]**

## Examiners report

Marks available | 2 |

Reference code | 18M.1.sl.TZ2.3 |

## Question

A data set has *n* items. The sum of the items is 800 and the mean is 20.

The standard deviation of this data set is 3. Each value in the set is multiplied by 10.

Find *n*.

Write down the value of the new mean.

Find the value of the new variance.

## Markscheme

correct approach **(A1)**

*eg* \(\frac{{800}}{n} = 20\)

40 **A1 N2**

**[2 marks]**

200 **A1 N1**

**[1 mark]**

**METHOD 1**

recognizing variance = *σ *^{2} **(M1)**

*eg* 3^{2} = 9

correct working to find new variance **(A1)**

*eg σ *^{2 }× 10^{2}, 9 × 100

900 ** A1 N3**

**METHOD 2**

new standard deviation is 30 ** (A1)**

recognizing variance = *σ *^{2} **(M1)**

*e*g 3^{2} = 9, 30^{2}

900 ** A1 N3**

**[3 marks]**