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 + ( 3 N 4 F f Q 3 ) c L Q 1 + ( N 4 F f Q 1 ) c = 59.5 + ( 3 4 ( 80 ) 59 10 ) 10 29.5 + ( 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 = ( 16 x y ) 10 6 y = 160 10 x 3 y = 80 5 x ( 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).

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