|
Tables
Clement Radcliffe,Contributor
We
will begin this week's lesson with
the solution to last week's Homework.
Homework
The
number of words in each of the first
100 sentences of a book is recorded
in the table below.
Number
of
words |
0
-10 |
11-20 |
21-30 |
31
-40 |
41
-50 |
51-60 |
61
-70 |
71
- 80 |
81-90 |
91-100 |
| Frequency |
21 |
16 |
15 |
14 |
11 |
9 |
8 |
3 |
3 |
3 |
Using
a suitable scale, represent the data
on both a histogram and a frequency
polygon.
We
will now continue the lesson with
another example.
The
table below shows the number of inches
of rainfall which fell over a period
of time.
| Inches
of Rainfall |
0
- 4 |
5
- 9 |
10
- 14 |
15
- 19 |
20
- 24 |
25
- 29 |
| Number
of days |
5 |
8 |
3 |
1 |
2 |
1 |
Using
a scale of 2 cm to represent five
inches on the x axis, and 1 cm to
represent 1 day on the y axis, construct
the histogram to represent the data.
Please
note the following with respect to
the question above.
- Values
that fall between classes would
pose a problem. In this case, for
example, we would have to bring
4.5 to the nearest whole number.
- Class
boundaries are recommended when
the values are continuous variables.
- Class
Intervals are converted to class
boundaries as follows:
| Class
Intervals |
0
- 4 |
5
- 9 |
10
- 14 |
15
- 19 |
20
- 24 |
25
- 29 |
| Class
Boundaries |
0.5-4.5 |
4.5-9.5 |
9.5-14.5 |
14.5-19.5 |
19.45-24.5 |
24.5-29.5 |
You
will notice the following:
- Any
value between 0 and 29 can be assigned
to a class without difficulty.
- Of
necessity, the bars will touch.
- 0.5
is added and subtracted from the
class Intervals to obtain the class
boundaries.
- The
frequency polygon is constructed
by joining the midpoint of the top
of each bar.
Using
the above, please attempt the following:
The
table below shows the height of orange
seedling in a farm.
| Class
Intervals |
0
- 4 |
5
- 9 |
10
- 14 |
15
- 19 |
20
- 24 |
25
- 29 |
| Class
Boundaries |
0.5-4.5 |
4.5-9.5 |
9.5-14.5 |
14.5-19.5 |
19.45-24.5 |
24.5-29.5 |
(a)
Express the above with respect to
the class boundaries.
(b)
Draw the histogram and frequency polygon
to represent the data.
Solution
(a)
Height
(cm)
Class Boundaries |
1
- 3
0.5-3.5
|
4
- 6
3.5-6.5 |
7
- 9
6.5-9.5 |
10
- 12
9.5-12.5 |
13
- 15
12.5 -15.5 |
| Frequency |
4
|
14
|
20
|
9
|
3
|
(b)
We
will now proceed with analysis of
data.
ANALYSIS
OF DATA.
The
aim is to arrive at informed decisions
from the data. The following is one
way in which this may be done;
(a)
Measures of central tendencies or
average.
These
are the values which best represent
the data namely: mean, median and
mode.
Example
The
scores obtained by a class of 10 students
in a test were:
3,
3, 4, 4, 4, 6, 6, 7, 11, 12.
Calculate
(1)
The modal score.
(2)
The median mark.
(3)
The mean mark.
Solution
(1)
The modal value is the most frequently
occurring mark. In this case, it is
4.
Answer:
4
(2)
The median mark is the middle value
when scores are arranged in order
of size. When there is an odd number
of scores, it is the single middle
value. However, it is the average
of the two middle scores when the
number of scores is even.
From
the values given, the 5th mark is
4 and the 6th mark is 6
...
The median mark is the average of
the 5th and 6th values as there is
an even number of values(10).
=
(4 + 6)/2 = 5
(3)
The mean mark = Sum of scores/Number
of scores
=
(3 + 3 + 4 + 4+ 4+ 6 + 6 + 7 + 11
+12)/10 = 60/10 = 6
As
the average is the value which best
represents the group, you should be
able to determine when it is appropriate
to use any of the three: the mean,
median or the mode.
Now,
please work this example for homework.
Six
students earned the following marks
in a test: 5, 3, 4, 6, 5, 7
Calculate
(1)
The modal value
(2)
The median mark
(3)
The mean mark.
Next
week we will continue to look at the
other ways to arrive at informed decisions
from data.
Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
|
