More about Quartiles

Quartiles are the values that divide the given data arranged in ascending order into subdivisions of twenty five percent, fifty percent and seventy five percent. First quartile is the twenty fifth percentile also known as the lower quartile. Second quartile is the 50th percentile also known as the median and the third quartile is the seventy fifth percentile also known as the upper quartile. The lower quartile or the first quartile is the middle value or the median of the first half of the data values arranged in the numerical order. It is denoted as Q1. Q1 =( ¼). (n+1)th value of the data set, here n is the total number of data values. The second quartile denoted by Q2 is the median of the data set arranged in the numerical order. Q2 = median = (1/2)(n+1)th value of the data set. The third quartile or the upper quartile is denoted by Q3. Q3 is the (3/4)(n+1)th value which is the middle value or median of the upper half of the data set.

We know that range of a given data set is the value got by calculating the difference between the highest and lowest values in the data set. So, range = highest score – lowest score.  Now Quartile Range is a bit similar to range, it is the difference between the upper quartile (Q3) and lower quartile (Q1). Quartile Range is given as (Q3 – Q1) for a particular data set which is also called the Inter Quartile Range denotes as IQR. So, IQR = (Q3 – Q1). Let us learn about Quartile Deviation which is the absolute measure of dispersion. It is also called the semi Inter Quartile Range and is half of the Inter Quartile Range. It is written as Q.D in short, Q.D. = (1/2) (Q3 – Q1)

For example let us find the lower quartile, median, upper quartile, inter quartile range and quartile deviation of the data set 15, 18, 14, 20, 26, 16, 18.
First arrange the data values in the numerical order:
14, 15, 16, 18, 18, 20, 26
The number of values, n = 7
Lower Quartile = (1/4)(n+1) = (1/4)(7+1) = 8/4 = 2nd value which is 15(Q1) in the data set
Second Quartile = Median = (1/2)(n+1)= (1/2)(7+1)=8/2= 4th value which is 18(Q2)in the data set
Third Quartile= (3/4)(7+1)=3. 8/4 = 3.2 = 6th value which is 20(Q3) in the data set
Inter Quartile Range = IQR = (Q3 – Q1)= (20 – 15) = 5
Semi Inter Quartile Range = Quartile Deviation= Q.D. = (1/2)(IQR)= (1/2) (Q3 – Q1) = 5/2 = 2.5

Step by step math solution

Line Plot Graph made simple
Line Plot Definition
A data represented on a number line with marks like ‘x’ or any other mark which shows the frequency of a value in the data is defined as the line plot. For example:
The line plot below shows the marks of 20 pupils in a class.

The ‘x’ marks show the frequency of the marks obtained by the pupils
Let us now take a quick look at how to make a line plot
• First we need to gather the information. Once the information is ready we look for the data sets which occur often that is the data which is frequently shown. Something like the favorite flavor of certain people or the number of pets a group of people have.
• The data is to be sorted and then a chart is created so as to organize the list. We then name the chart for convenience.
• This is an important step which involves determining the scale. The scale might not have the labels that are not the data values as per the given information and hence we need  to decide the scale depending upon the frequency of the data items for which a numerical scale is used which begins with the least number and ends in the highest number in the data set.
• Now we draw a horizontal line which is similar to  a number line according to the chosen scale.
• Finally we start marking ‘x’ above the line corresponding to the number on the scale as per the data we have. Once the markings are done, the line plot for the given data is ready for further analysis.

Let us now make line plot graphs using a line plot example
Given are the costs of 15 books sold in a book store, represent a data as a line plot graph.
$20 $15 $9 $15 $9 $20 $9 $20 $20 $35 $25 $20 $9 $30 $15
We need to decide on the scale, the lowest value is 9 and the highest value is 35. So, the scale should start from 5 and end in 35

Solve by the addition method

Addition is a mathematical operation that represents combining collections of objects together into a larger collection. It is signified by the plus sign (+). For example, in the picture on the right, there are 3 + 2 apples—meaning three apples and two other apples—which is the same as five apples. Therefore, 3 + 2 = 5. (Source: Wikipedia)

Example Problems for Solve by the Addition Method:-

Problem 1:-

Solve 453 + 213 by the Addition method.

Solution:-

In the following step by step process of addition method

Step 1:-

               453
            + 213
           -----------
           -----------
The above equation 453 is adding to 213 in between (+) plus operation. In basic addition process start with right side value to left side value

Step 2:-

               453
            + 213
           -----------
                   6
           -----------
Adding the right side values 3 and 3. 3 is equal with 3 in 3+3 =6. Then move to next value.

Step 3:-

               453
            + 213
           -----------
                 66
           -----------
Adding the next two values 5 and 1. 5 adding with value 1 in 5+1 =6. Then move to next value.

Step 4:-

               453
            + 213
           -----------
               666
           -----------

Adding the last two values 4 and 2. 4 adding with value 2 in 4+2=6. We get the final answer is 666.

Problem 2:-

Solve 654 + 323 by the Addition method.

Solution:-

In the following step by step process of addition method

Step 1:-

               654
            + 323
           -----------
           -----------

The above equation 654 is adding to 323 in between (+) plus operation. In basic addition process start with right side value to left side value

Step 2:-

               654
            + 323
           -----------
                   7
           -----------

Adding the right side values 4 and 3. 4 adding with value 3 in 4+3 =7. Then move to next value.

Step 3:-

               654
            + 323
           -----------
                  77
           -----------

Adding the next two values 5 and 2. 5 adding with value 2 in 5+2 =7. Then move to next value.

Step 4:-

               654
            + 323
           -----------
               977
           -----------

Adding the last two values 6 and 3. 6 adding with value 3 in 6+3=9. We get the final answer is 977.


Practice Problems for Solve by the Addition Method:-

Problem 1:-

solve 421 + 167 by the addition method.

Answer:- 588

Problem 2:-

solve 217 +171 by the addition method.

Answer:- 388

Problem 3:-

solve 383 + 71 by the addition method.

Answer:- 454

Problem 4:-

solve 152 + 38 by the addition method

Answer:- 190

Problem 5:-

solve 43 + 24 by the addition method.

Answer:- 67

Frequency Distribution in Statistics

In mathematics frequency distribution is used in statistics. Mean of a frequency distribution is that the arrangement in which sets of value occurs and in the values one or more variable takes place. Frequency distribution is in the form of either graphical or tabular. Each value in the table contains frequency or count of values, how many times they occur. The values of frequency in group or interval forms.  After summarizing the entire values frequency distribution table is formed. Mean of frequency distribution is also that it shows the total number of observations within a given interval. The interval is either exclusive or exhaustive. The size of intervals generally depends on the data which we have to analyze and calculate. One thing we have to remind that the intervals must not be overlapped to each other.

Now we discuss that how to construct frequency distribution tables. We use some steps to make a frequency distribution table. In step one; we determine the range of given data. Range of given data means the difference between the higher value and the lower value. In step two, we decide that which data can be grouped means formulation of approximate number of groups. There are no particular rules for step two. It can be 5 groups to 15 groups. But there is one formula for this (K=1+3.322logN), where K is the no of groups, logN is the total number of observations.

In step third, we decide the size of intervals.  The size of interval is denoted by (h). To determine the size we can use a formula (h= range/number of groups). If result is in fraction then we choose next higher value. In step fourth, we decide start point means starting from the lowest value and in the ascending order. In step fifth, we determine the remaining groups. It is determined by adding the interval size corresponding to all values. In step sixth, we distribute all the data into their groups. For this we use tally marks method because it is suitable for tabulating the observations into their respective groups. By using these six steps we can construct a frequency distribution table.

Now we come to standard deviation for frequency distribution. It is a measure of variation or measure of dispersion amongst the data. In place of taking absolute deviation we may square each deviation and obtained the variance. The square root value of variance is known as standard deviation for given values of frequency.

Derivatives of inverse trigonometric functions

Like many functions, the trigonometric functions also have inverse.  Just like how we can find the derivative of trigonometric functions, we can also find the derivatives of inverse trigonometric functions. In this article we shall take a quick look at inverse trig functions derivatives. To be able to find the standard formulae for derivatives of inverse trig functions, we would need the formula for derivative of any general inverse function.

That would be like this:
f’(x) = 1/g’(f(x)), where f and g are the inverse functions of each other.
Let us first try finding the inverse trig function derivatives of the sine function. The sine function inverse is written as arc sin (x). Therefore the function would look like this: y = arcsin(x),
Therefore, x= sin (y) for –pi/2 ≤ y ≤ pi/2. We fix this domain for the sine function to ensure that our inverse exists. If we don’t restrict the domain, then y could have multiple values for same value of x. For example, sin (pi/4) = 1/sqrt(2) and sin (3pi/4) is also 1/sqrt(2). With that in mind, we can write the relationship between sin and arc sin as follows:
Sin(arc sin (x)) = x and arc sin(sin(x))  = x
Thus here our f(x) = arc sin(x) and g(x) = sin (x), then using the derivative of inverse formula that we stated above we have:
f’(x) = 1/f’(g(x)) = 1/cos (arc sin (x))
This formula may to be applicable in practice. So let us make a few changes in there. We had earlier x = sin y so y = arc sin (x). Using that, the denominator of our derivative would become
Cos(arc sin (x)) = cos y
Next we use our primary trigonometric identity which was:
Cos^2 (x) + sin^2 (x) = 1
Thus, Cos^2 (y) + sin^2 (y) = 1 from this we have
Cos (y) =√( 1- sin^2 (y)) = √(1 – (sin y)^2). But we have already established that sin y = x. So replacing that we get,
Cos (y) = √(1 – x^2). So now plugging all that back to our equation of derivative of inverse trig functions for sine f’(x), we have:
f’(x) = 1/cos(arc sin(x)) = 1/cos y = 1/√(1-x^2)
Therefore we see that derivative of arc sin (x) = 1/√(1-x^2).
The derivatives of other trigonometric inverse functions of arc cos, arc tan etc can be derived in a similar way. They are:
d/dx arc cos (x) = -1/√(1-x^2) and d/dx arc tan (x) = 1/(1+x^2)

All about the hypotenuse of a right triangle

Like every triangle, a right angled triangle would also have three sides. However, in a right triangle one of the angles is a right angle. That means one of the angle measures 90 degrees (or 𝛑/2 radians). Since the sum of angles in any triangle has to be 180 degrees, in a right triangle as one angle is already 90 degrees, the sum of the other two angles have to be 90 degrees. That means that the other two angles are compliments of each other. It also means that the other two angles have to be acute angles. A typical right triangle would look as follows:

The longest side is called the hypotenuse. The side that is adjacent to the know angle is called the adjacent side and the side opposite to the known angle is called the opposite side. By default the hypotenuse will always be the side opposite the right angle.

The adjacent and opposite sides together are also called the legs of the right triangle. The length of the hypotenuse of a right angled triangle can be found using different methods, depending on what part of the triangle is given to us.

To find the hypotenuse of a right triangle given the length of the legs:
If the hypotenuse = c and the legs are ‘a’ and b. If a’ and b’ are known, then we can calculate the length of the hypotenuse using the Pythagorean rule as follows:
C^2 = a^2 + b^2
Finding hypotenuse of a right triangle given one of the angles and the adjacent side:
In the picture below,

Suppose the angle marked in red is x and the adjacent side = a, then the length of the hypotenuse H can be given by the formula:
H = a/Cos (x)
Formula for the hypotenuse of a right triangle given one of the angles and its opposite side:
Again from the picture above, if we are given the opposite = b instead of the adjacent side. Then the formula for the hypotenuse can be written as follows:
H = b/sin (x), where x is again the angle marked in red.
Thus as we saw above there are more than one ways to find the length of the hypotenuse of a right triangle.

Trapezoidal Numerical Integration

Introduction to numerical integration:  As we have seen, the ideal way to evaluate a definite integral a to b f(x) dx is to find a formula F(x) for one of the antiderivatives of f(x) and calculate the number F(b) – F(a). But some anti derivatives are hard to find, and still others, like the antiderivatives of (sin x)/x and sqrt(1 + x^4), have no elementary formulas. We do not mean merely that no one has yet succeeded in finding elementary formulas for the anti-derivatives of (sin x)/x and sqrt(1 + x^4). We mean it has been proved that no such formulas exist. Whatever, the reason, when we cannot evaluate a definite integral with an anti-derivative, we turn to numerical methods such as the trapezoidal rule and Simpson’s rule.
Numerical Integration Methods are Trapezoidal rule and Simpson’s rule. Let us now describe Trapezoidal numerical integration.

Trapezoidal Numerical Integration: When we cannot find a workable anti-derivative for a function f that we have to integrate, we partition the interval of integration, replace f by a closely fitting polynomial on each sub interval  integrate the polynomials and add the result to approximate the integral of f. The higher the degrees of the polynomials for a given partition, the better the results are.  For a given degree, the finer the partition, the better the results, until we reach the limits imposed by round-off and truncation errors.

The polynomials do not need to be of high degree to be effective. Even line segments (graphs of polynomials of degree 1) give good approximations if we use enough of them  .To see why, suppose we partition the domain [a, b] of f into n subintervals of length delta x = h = (b – a)/n and join the corresponding points on the curve with line segments.

The vertical lines from the ends of the segments to the partition points create a collection of trapezoids that approximate the region between the curve and the x-axis. We add the areas of the trapezoids, counting area above the x-axis as positive and area below the axis as negative.
T = ½ (y0 + y1) h + ½ (y1 + y2)h + ….. + ½ (yn-2 + yn-1) h + ½ (yn-1 + yn)h= h (1/2y0 + y1 + y2 + ….. + yn-1 + ½ yn) = h/2 (y0 + 2y1 + 2y2 + …. + 2yn-1 + yn).Where, Y0 = f (a), y1 = f(x1), Yn-1 = f(x n-1), yn = f(b).

The trapezoidal rule says: use T to estimate the integral of f from a to b. Numerical Integration in R As we have seen , the definition of the Riemann integral is not very efficient way to prove that a function is Riemann integral . However once it is known that a function f is Riemann integral on some interval [a, b] a modification of the definition makes it possible to evaluate the integral of simple limit.