Statistics Long Question Example 5 & 6


Example 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, 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.
Interquartile Range = L Q 3 +( 3N 4 F f Q 3 )c L Q 1 +( N 4 F f Q 1 )c =59.5+( 3 4 ( 80 )59 10 )1029.5+( 1 4 ( 80 )7 18 )10 =59.5+1( 29.5+7.22 ) =23.78



Example 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
m=L+( N 2 F f m )c 35.5=29.5+( 40 2 ( 4+x ) y )10 6=( 16x y )10 6y=16010x 3y=805x(2)
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).