**Question 5**:

The table shows the cumulative frequency distribution for the distance travelled by 80 children in a competition.

Distance (cm) |
<20 |
<30 |
<40 |
<50 |
<60 |
<70 |
<80 |
<90 |

Number of children |
1 |
7 |
25 |
34 |
59 |
69 |
79 |
80 |

**(a)**Based on the table above, copy and complete the table below.

Distance (cm) |
10-19 |
|||||||

Frequency |

**(b)**Without drawing an ogive, estimate the interquartile range of this data.

*Solution:***(a)**

Distance (cm) |
10-19 |
20-29 |
30-39 |
40-49 |
50-59 |
60-69 |
70-79 |
80-89 |

Frequency |
1 |
6 |
18 |
9 |
25 |
10 |
10 |
1 |

Cumulative Frequency |
1 |
7 |
25 |
34 |
59 |
69 |
79 |
80 |

**(b)**

Interquartile range = Third Quartile – First Quartile

Third Quartile class,

*Q*_{3}= ¾ × 80 = 60Therefore third quartile class is the class 60 – 69.

First Quartile class,

*Q*_{1}= ¼ × 80 = 20Therefore first quartile class is the class 30 – 39.

$\begin{array}{l}\text{InterquartileRange}\\ ={L}_{{Q}_{3}}+\left(\frac{\frac{3N}{4}-F}{{f}_{{}_{{Q}_{3}}}}\right)c-{L}_{{Q}_{1}}+\left(\frac{\frac{N}{4}-F}{{f}_{{}_{{Q}_{1}}}}\right)c\\ =59.5+\left(\frac{\frac{3}{4}\left(80\right)-59}{10}\right)10-29.5+\left(\frac{\frac{1}{4}\left(80\right)-7}{18}\right)10\\ =59.5+1-\left(29.5+7.22\right)\\ =23.78\end{array}$
**Question 6**:

Table shows the daily salary obtained by 40 workers in a construction site.

Daily Salary (RM) |
Number of workers |

10 – 19 |
4 |

20 – 29 |
x |

30 – 39 |
y |

40 – 49 |
10 |

50 – 59 |
8 |

Given that the median daily salary is RM35.5, find the value of

*x*and of*y*.Hence, state the modal class.

*Solution:*Daily Salary (RM) |
Frequency |
Cumulative frequency |

10 – 19 |
4 |
4 |

20 – 29 |
x |
4 + x |

30 – 39 |
y |
4 + x + y |

40 – 49 |
10 |
14 + x + y |

50 – 59 |
8 |
22 + x + y |

Total workers = 40

22 +

*x*+*y*= 40*x*= 18 –

*y*——(1)

Median daily salary = 35.5

Median class is 30 – 39

$\begin{array}{l}m=L+\left(\frac{\frac{N}{2}-F}{{f}_{m}}\right)c\\ 35.5=29.5+\left(\frac{\frac{40}{2}-\left(4+x\right)}{y}\right)10\\ 6=\left(\frac{16-x}{y}\right)10\\ 6y=160-10x\\ 3y=80-5x----(2)\end{array}$

Substitute (1) into (2):

3

*y*= 80 – 5(18 –*y*)3

*y*= 80 – 90 + 5*y*–2

*y*= –10*y*= 5

Substitute

*y*= 5 into (1)*x*= 18 – 5 = 13

**Thus**

*x*= 13 and*y*= 5.**The modal class is 20 – 29 daily salary (RM).**

what class lesson is this?