7.5.3 Statistics, Long Question (Paper 2)

Question 5:

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

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

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

Solution:
(a)

(b)

Interquartile range = Third Quartile – First Quartile

Third Quartile class, Q₃ = ¾ × 80 = 60

Therefore third quartile class is the class 60 – 69.

First Quartile class, Q₁= ¼ × 80 = 20

Therefore first quartile class is the class 30 – 39.

\begin{array}{l} Interquartile Range \\ = L_{Q_{3}} + (\frac{\frac{3 N}{4} - F}{f_{_{Q_{3}}}}) c - L_{Q_{1}} + (\frac{\frac{N}{4} - F}{f_{_{Q_{1}}}}) c \\ = 59.5 + (\frac{\frac{3}{4} (80) - 59}{10}) 10 - 29.5 + (\frac{\frac{1}{4} (80) - 7}{18}) 10 \\ = 59.5 + 1 - (29.5 + 7.22) \\ = 23.78 \end{array}

Question 6:

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

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

Hence, state the modal class.

Solution:

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 + (\frac{\frac{N}{2} - F}{f_{m}}) c \\ 35.5 = 29.5 + (\frac{\frac{40}{2} - (4 + x)}{y}) 10 \\ 6 = (\frac{16 - x}{y}) 10 \\ 6 y = 160 - 10 x \\ 3 y = 80 - 5 x - - - - (2) \end{array}

Substitute (1) into (2):

3y = 80 – 5(18 – y)

3y = 80 – 90 + 5y

–2y = –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).