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Analysis
of data
Clement Radcliffe, Contributor
We will continue the review of statistics
with the solution to last week's homework.
Homework
(1)
Express the following scores in a
frequency table and plot the histogram.
22,
15, 0, 22, 11, 9, 0 14, 20, 9, 16,
5, 11, 24, 16, 5, 11, 24, 5, 5, 22,
15, 9, 9, 11
Solution
Since
the values range from 0-24, it would
be inappropriate to construct a histogram
with 25 bars. Using grouped data as
follows: 0 - 3, 4 - 7, 8 - 11 etc.,
we construct the table by first doing
the tally.
| SCORES |
0
- 3
|
4
- 7
|
8
- 11
|
12
- 15
|
16
- 19
|
20
- 23
|
24
- 27
|
| TALLY |
11
|
1111
|
1111
111
|
111
|
11
|
1111
|
11
|
| FREQUENCY |
2
|
4
|
8
|
3
|
2
|
4
|
2
|
(2)
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 5 inches
on the X axis, and 1 cm to represent
1 day on the Y axis, construct the
histogram to represent the data.
Solution
Please
note the following with respect to
question two:
- Values
that fall between classes, for example
4.5, would pose a problem. In this
case, 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:
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
seedlings on a farm.
| Height(cm)
|
1
- 3
|
4
- 6
|
7
- 9
|
10
- 12
|
13
- 15
|
| Frequency |
4
|
14
|
20
|
9
|
3
|
(a)
Express the above with respect to
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 tendency or average.
These
are the values which best represent
the data, namely mean, median or mode.
Example
The
scores obtained by a class of ten
students in a test were:
3,
3, 4, 4, 4, 6, 6, 7, 11, 12.
Calculate
(i)
the modal mark
(ii)
the median mark
(iii)
the mean mark.
Solution
(i)
The modal mark or mode is the most
frequently occurring mark. In this
case, it is 4.
Answer:
4
(ii)
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
(iii)
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,
the 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
(i)
the modal mark
(ii)
the median mark
(iii)
the mean mark.
Next
week, we will continue to look at
other ways to arrive at informed decisions
from data.
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Greater
Portmore High school students
display their medals during
the Multi-Care Sports Foundation
Athletic Meet at the G.C. Foster
College in Spanish Town, on
Tuesday, May 6.
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Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
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