Mathematics
Assalamualaikum to you all read my blog.I am the first time in blog
Mathematics
Assalamualaikum to you all read my blog.I am the first time in blog
Saturday, 30 May 2020
Tuesday, 12 July 2016
MEASURES OF CENTRAL TENDENCY
DEFINITION MEASURES OF CENTRAL TENDENCY;
A measure used to describe data; the mean, median, and mode are measures of central tendency.
- Arrange the data in ascending numerical order.
- The sum of all data, divided by how many numbers there are in the data.
Median
- Arrange the data in ascending numerical order.
- The middle number of the data.
- If there are 2 numbers in the middle, add them together and divide by 2.
Mode
- Arrange the data in ascending numerical order.
- The most common number.
Wednesday, 29 June 2016
STATISTICAL REPRESENTATION
CUMULATIVE FREQUENCY CURVE
Cumulative Frequency Graph , also known as an Ogive, is a curve showing the cumulative frequency for a given set of data. The cumulative frequency is plotted on the y-axis against the data which is on the x-axis for UN-grouped data.
You can construct a frequency polygon by joining the midpoints of the tops of the bars.
Frequency polygons are particularly useful for comparing different sets of data on the same diagram.
STEM AND LEAF PLOTS
Stem-and-leaf plots are
a method for showing the frequency with which certain classes of values
occur. You could make a frequency distribution table or a histogram for
the values, or you can use a stem-and-leaf plot and let the numbers themselves
to show pretty much the same information.
For instance, suppose you have the following list of values: 12, 13, 21, 27, 33, 34, 35, 37, 40, 40, 41. You could make a frequency distribution table showing how many tens, twenties, thirties, and forties you have:
The downside of frequency distribution tables and histograms is that, while the frequency of each class is easy to see, the original data points have been lost. You can tell, for instance, that there must have been three listed values that were in the forties, but there is no way to tell from the table or from the histogram what those values might have been.
On the other hand, you
could make a stem-and-leaf plot for the same data:
Note that the horizontal leaves in the stem-and-leaf plot correspond to the vertical bars in the histogram, and the leaves have lengths that equal the numbers in the frequency table.
That's pretty much all there is to a stem-and-leaf plot. You're just listing out how many entries you have in certain classes of numbers, and what those entries are. Here are some more examples of stem-and-leaf plots, containing a few additional details.
Stem-and-Leaf Plots: Examples
Cumulative Frequency Graph , also known as an Ogive, is a curve showing the cumulative frequency for a given set of data. The cumulative frequency is plotted on the y-axis against the data which is on the x-axis for UN-grouped data.
Cumulative Frequency Graph
If we are
given a table containing continuous data, we can find a running total of the frequency. This is called Cumulative Frequency.
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We can
therefore plot a graph:
· The x-axis (horizontal)
will be Height(m)
· The y-axis will (vertical)
will be Cumulative Frequency
· We plot the numbers in red
· The graph starts at 1.2m,
because this was the lowest value
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We can
now use this graph to estimate a number of key values
In this case
n=40 (the total number of people
Median: On a cumulative frequency
graph we find value
We can read this off
the graph to get 1.64m
Lower Quartile: On a cumulative
frequency graph we find value
We can read this off
the graph to get 1.48m
Upper Quartile: On a cumulative
frequency graph we find value
We can read this off
the graph to get 1.73m
From this we
can deduce the IQR = UQ - LQ =
1.73m-1.48m=0.25m
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Histogram – the A/A* graph!
Suppose you
are given a table of continuous data (see below).
Given a
class eg. the class width is
In the table
below, you can see how the class widths change.
In this
case, we will need to construct a histogram to
represent the data.
In a
histogram, the AREA of the bars equals the FREQUENCY
To achieve this,
we need to calculate a value called the FREQUENCY
DENSITY
We now draw
a graph with:
· Height(m) on the x-axis
· Frequency density on the
y-axis
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The next section considers how to read
graphs to find an average
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Finding the mean and median from a
frequency graph
A frequency graph to show
the frequency of scores in a test
This graph can be turned into a frequency table
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Frequency diagrams and polygons
Bar charts and frequency diagrams. Pie charts are useful for showing proportions, but different types of chart have to be used for representing other kinds of data. A number of these charts are described in this section..This frequency diagram shows the heights of 200 people:Frequency polygons are particularly useful for comparing different sets of data on the same diagram.
Constructing a frequency polygon
STEM AND LEAF PLOTS
For instance, suppose you have the following list of values: 12, 13, 21, 27, 33, 34, 35, 37, 40, 40, 41. You could make a frequency distribution table showing how many tens, twenties, thirties, and forties you have:
Frequency
Class |
Frequency
|
10 - 19
|
2
|
20 - 29
|
2
|
30 - 39
|
4
|
40 - 49
|
3
|
The downside of frequency distribution tables and histograms is that, while the frequency of each class is easy to see, the original data points have been lost. You can tell, for instance, that there must have been three listed values that were in the forties, but there is no way to tell from the table or from the histogram what those values might have been.
Note that the horizontal leaves in the stem-and-leaf plot correspond to the vertical bars in the histogram, and the leaves have lengths that equal the numbers in the frequency table.
That's pretty much all there is to a stem-and-leaf plot. You're just listing out how many entries you have in certain classes of numbers, and what those entries are. Here are some more examples of stem-and-leaf plots, containing a few additional details.
- Complete a stem-and-leaf plot for the following list of grades on a recent test:
- 73, 42,
67, 78, 99, 84, 91, 82, 86,
94
- I'll use the tens digits
as the stem values and the ones digits as the leaves. For convenience
sake, I'll order the list, but this is not required:
- 42, 67,
73, 78, 82, 84, 86, 91, 94,
99
Stem-and-Leaf Plots: Examples
- Subjects in a psychological study were timed while completing a certain task. Complete a stem-and-leaf plot for the following list of times:
7.6, 8.1, 9.2, 6.8, 5.9, 6.2, 6.1,
5.8, 7.3, 8.1, 8.8, 7.4, 7.7, 8.2
- First, I'll reorder this
list:
- 5.8, 5.9,
6.1, 6.2, 6.8,
7.3, 7.4, 7.6, 7.7, 8.1,
8.1, 8.2, 8.8, 9.2
- Complete a stem-and-leaf plot for the following two lists of class sizes:
- Economics 101: 9,
13, 14, 15, 16, 16, 17, 19,
20, 21, 21, 22, 25, 25, 26
Libertarianism: 14, 16, 17, 18, 18, 20, 20, 24, 29
This example has two lists of values. Since the values are similar, I can plot them all on one stem-and-leaf plot by drawing leaves on either side of the stem. I will use the tens digits as the stem values, and the ones digits as the leaves. Since "9" (in the Econ 101 list) has no tens digit, the stem value will be "0". Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved
- Complete a stem-and-leaf plot for the following list of values:
- 100, 110,
120, 130, 130, 150, 160, 170,
170, 190,
210, 230, 240, 260, 270, 270, 280. 290, 290
- ...but the leaves are
fairly long this way, because the values are so close together.
To spread the values out a bit, I can break each leaf into two. For
instance, the leaf for the two-hundreds class can be split into two
classes, being the numbers between 200 and 240 and the numbers
between 250 and 290. I can also reverse the order, so the smaller values
are at the bottom of the "stem". The new plot looks like
this:
ADVERTISEMENT
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- Complete a stem-and-leaf plot for the following list of values:
- 23.25, 24.13,
24.76, 24.81, 24.98, 25.31, 25.57, 25.89,
26.28, 26.34, 27.09
If I try to use the last digit, the hundredths digit, for these numbers, the stem-and-leaf plot will be enormously long, because these values are so spread out. (With the numbers' first three digits ranging from 232 to 270, I'd have thirty-nine leaves, most of which would be empty.) So instead of working with the given numbers, I'll round each of the numbers to the nearest tenth, and then use those new values for my plot. Rounding gives me the following list:
- 23.3, 24.1,
24.8, 24.8, 25.0, 25.3, 25.6, 25.9,
26.3, 26.3, 27.1
- Then my plot looks like
this:
Saturday, 25 June 2016
linear programming
Linear
Programming: Introduction
In "real life", linear programming is part of a very important area of mathematics called "optimization techniques". This field of study (or at least the applied results of it) are used every day in the organization and allocation of resources. These "real life" systems can have dozens or hundreds of variables, or more. In algebra, though, you'll only work with the simple (and graphable) two-variable linear case.
The general process for solving linear-programming exercises is to graph the inequalities (called the "constraints") to form a walled-off area on the x,y-plane (called the "feasibility region"). Then you figure out the coordinates of the corners of this feasibility region (that is, you find the intersection points of the various pairs of lines), and test these corner points in the formula (called the "optimization equation") for which you're trying to find the highest or lowest value.
- Find the maximal and minimal value of z = 3x + 4y subject to the following constraints:
My first step is to solve each inequality for the more-easily graphed equivalent forms:
y
= –( 1/2 )x
+ 7y
= 3x
|
y
= –( 1/2 )x
+ 7y
= x
– 2
|
y
= 3x
y = x – 2 |
–( 1/2
)x
+ 7 = 3x–x
+ 14 = 6x14
= 7x
2 = x
y
= 3(2) = 6
|
–( 1/2
)x
+ 7 = x
– 2
–x + 14 = 2x – 4 18 = 3x 6 = x
y
= (6) – 2 = 4
|
3x
= x
– 2
2x = –2x = –1
y
= 3(–1) = –3
|
corner point at
(2, 6)
|
corner point at (6,
4)
|
corner pt. at (–1,
–3)
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Somebody really smart proved that, for linear systems like this, the maximum and minimum values of the optimization equation will always be on the corners of the feasibility region. So, to find the solution to this exercise, I only need to plug these three points into "z = 3x + 4y".
- (2, 6): z
= 3(2) + 4(6) = 6 + 24 = 30
(6, 4): z = 3(6) + 4(4) = 18 + 16 = 34
(–1, –3): z = 3(–1) + 4(–3) = –3 – 12 = –15
and the minimum of z = –15 occurs at (–1, –3).
HERE A VIDEO EXPLANATION:
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