Stepwise Calculation of Standard Deviation

In Statistics a branch of Mathematics, Standard Deviation is the mean of mean. It is the measure which helps us to know how the data is spread out.  It is denoted by the Greek symbol sigma. Formula for Standard-Deviation is given by square root of variance, variance is defined as the average of the squared differences from the mean given by the formula sigma^2 = 1/(n-1)[summation(i=1 to n)(xi – x bar)^2]. So, the Formula Standard Deviation is given as sigma = sqrt{ [summation(i=1 to n)(xi – x bar)^2]/(n-1)}
Equation for Standard Deviation is given by sigma = sqrt[summation(k=1 to n)(xk – x bar)^2/(n-1)]
 n=total number of values, x bar = mean of the data, xk= each of the data values

How to find Standard Deviation for a given data? The following are the steps involved to find Standard-Deviation of a given data:
1. Mean of the given data is calculated
2. Then the deviations are calculated
3. Find the square of these deviations
4. Find the sum of the squares of the deviations
5. Divide the sum by one less than the total numbers in the data
6. Finally find the square root of the value got from the step 5, this result is the standard-deviation
Let us consider an example to understand how to find standard-deviation
Given data, 91, 23, 47, 62, 76, 38, 82, 29

Step1: First we find the average of the given data, [91+23+47+62+76+38+28+82+29]/8= 59.5
So, the mean =  59.5
Step2: The deviations are calculated by subtracting the mean from each given data value.
(91-59.5), (23-59.5), (47-59.5), (62-59.5), (76-59.5), (38 – 59.5), (82-59.5), (29 – 59.5)
      So, the deviations are, 31.5, -36.5, -12.5, 2.5, 16.5, -21.5, 22.5, -30.5
Step3: Squares of the above deviations are,
992.25, 1332.25, 156.25, 6.25, 272.25, 462.25, 506.25, 930.25
Step4: Sum of the squares of the deviations is 4658
Step5: Divide the sum 4658 with (n-1) = one less than total number of values = (8-1) = 7 which gives,
             4658/7 = 665.43
Step6: Standard-Deviation is the square root of the value got in the step5,
Standard-deviation, sigma = sqrt(665.43) = 25.79

Calculate Standard Deviation of the given data,  7, 9, 12, 6, 4, 13, 21, 14, 22, 16
Total Number of values = n = 10
]
Step1: Mean of the given data, (7+9+12+6+4+13+21+14+22+16)/10 = 124/10 = 12.4
Step2: Deviations are calculated by subtracting the mean from each of the data values which are,
            -5.4, -3.4, -0.4,-6.4, -8.4, 0.6, 8.6, 1.6, 9.6, 3.6
Step3: Squares of the deviations are, 29.16, 11.56, 0.16, 40.96, 70.56, 0.36, 73.96, 2.56, 92.16, 12.96
Step4: Sum of the squares of the deviations is 334.4
Step5: divide the sum with (n-1) = (10-1= 9) which gives 334.4/9 = 37.15
Step6: Standard-deviation, sigma = sqrt(37.15) = 6.095

Functions in math

What is a mathematical function? A function is a technical term used to define relation between variables. Let us more understand what is a Math Function? A variable y is called a function of a variable if for every value of x there is a definite value of y. example  of mathematical functions y = x^2.x is called the independent variable as it takes any arbitrary assigned value whereas y is called the dependent variable as its value depends upon the value of x. The set of all possible value of the independent variable in a function is called the domain of the function and the set of values of the dependent variable is called the range of the function.

Let us more understand about functions math. We can classify functions in math as follow:
1. Into functions
A function f : X -> Y is called an into function if there is atleast one element in Y which has no pre-image in X. A function is an into function if its range is a proper subset of codomain.
2. Onto functions(Surjection)
A function f : X -> Y is called an onto function if every element in Y has atleast one pre-image in X i.e. every element of Y is image of some element of X under f. i.e. for every y belongs to Y, there exists an element x in X such that
f(x) = y.
3. One-one function (Injection)
A function f : X -> Y is called a one-one function if no element of X has more than one image in Y. In other words, if the images of distinct elements in X under f are distinct i.e. for every x1, x2 belongs to X.
4. Many-one function
A function f : X -> Y is called a many-one function if two or more elements of X have the same image in Y.
5. One-one into function
When a function is both 1-1 and into function, then it is called one-one into function. It satisfies the following properties:
(i) No two elements of the domain have the same image.
(ii) There is atleast one element in codomain which is not the image of any element of the domain.
6. One-one onto function
It is a function which is both 1-1 and onto. It satisfies the following properties:
(i) No two elements of the domain have the same image.
(ii) It is one-one i.e. f(x) = f(y)
(iii) It is onto i.e. for all y belongs to Y, there exists x belongs to X such that f(x) = y.
7. Many one into function
It is a function which is both many-one and into. It has the following properties:
(i) There are atleast two elements of the domain which correspond to the same element of the codomain.
(ii) There is atleast one element of the codomain which is not the image of any element of the domain.
8. Many-one onto function
It is a function which is both many-one and onto. It satisfies the following properties:
(i) There are atleast two elements of the domain which correspond to the same element of the codomain.
(ii) Every element of the codomain corresponds to some element of the domain.
9. Identity function
A function f : X -> X such that f(x) = x for all x belongs to X is called the identity function. In an identity function each element of the domain corresponds to itself.
10. Constant function
A function f in which all elements of X are associated with the same element of Y is called a constant function.
11. Equal function
Two functions f and g are said to be equal if their domains are same and f(x) = g(x) for all x. If f and g are equal we write them as f = g.

Know more about Continuity definition and function in calculus. If you have problem in solving calculus problems leave a comment with the problem. I will try to solve.

First order linear differential equation

First order differential equations: A first order differential equation is a relation dy/dx = f(x, y).....(1)in which f(x, y) is a function of two variables defined on a region in the xy-plane. A solution of the equation (1) is a differential function y = y(x) defined on an interval of x-values such that d/dx y(x) = f(x, y(x)) on that interval.
The initial condition that y(x0) = y0 amounts to requiring the solution curve y = y(x) to pass through the point (x0, y0).The general form of a first order and first degree differential equation is f(x, y, dy/dx) =0....(i).we know that the tangent of the direction of a curve in Cartesian rectangular coordinates at any point is given by dy/dx .

so the equation in (i)  can be known as the equation which establish the relationship between the coordinates of a point and the slope of the tangent ie dy/dx to the integral curve at a point .Solving the differential equation  given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field.All the curves represented by the general solution when takes together will give the locus of the differential equation .Since there is one arbitrary constant in the general  solution of the equation of the first order. The locus of the equation can be said to be made up of single infinity of curve.

First Order Linear Differential Equation A first order differential equation that can be written in the form  dy/dx + P(x) y = Q(x),where P and Q are functions of x, is a linear first order equation. The above equation is the standard form.  Let us understand First Order Differential Equation Solver using some example. Suppose we have the equation dy/dx = 1 – y/x is a first order differential equation in which f(x, y) = 1 – (y/x).
now let us take one more example show that the function y = 1/x + x/2 is a solution of the initial value problem dy/dx = 1 – y/x,y(2) = 3/2 . The given function satisfies the initial condition because y(2) = (1/x + x/2)x = 2= ½ + 2/2 = 3/2To show that it satisfies the differential equation, we show that the two sides of the equation agree when we substitute (1/x) + (x/2) for y.On the left: dy/dx = d/dx(1/x + x/2) = -1/x^2 + ½.On the right: 1 – y/x = 1 – 1/x(1/x + x/2) = 1 – 1/x^2 – ½ = -1/x^2 + ½.The function y = (1/x) + (x/2) satisfies both the differential equation and the initial condition, which is what we needed to show.

What is a Mode in Math?

Math teacher, Ms Grace assigned the following number of problems for homework on 8 different days; 10, 12, 8, 10, 7, 10, 8, 10. Let us arrange the data in the increasing order, 7,8,8,10,10,10,10,12. What do you observe in the data? 10 is the number which is seen most number of times. This number 10, which occurs most often, is called the Mode in Math or Mode Math. So, Mode Definition Math would be, the Mode of a set of data is the value in the set that occurs most often (most number of times)

By definition, Statistics Mode or statistical mode is the value that occurs most frequently in the data set. For example, what would be the mode of the data set, 12, 11, 13, 9, 11, 8, 6, 11. Let us first arrange the data in the ascending order. That gives us, 6, 8, 9, 11, 11, 11, 12, 13. As you can see the value or the number 11, occurs most number of times than any other number in the data and hence 11 is the mode of the given data set.

To learn the steps involved in finding mode of a given set, let us consider an example. Let us assume that there is a basket ball match going on and the scores of the game are as listed below. Let us determine the mode of the scores.

Scores: 19, 6, 3, 22, 19, 9

Step1: list in the scores in the ascending order
3, 6, 9, 19, 19, 22
Step2: identify the number which is occurs most often, 19
  Hence,  19 is the mode of the score

So, the steps involved in finding mode are, first we need to order the list in ascending order and then identify the number or value which occurs most often that will give us mode.

Note: If there is no number which occurs most often, then we can say that the given data has no mode
On a cold winter day in December, the temperature of 7 cities in North America is recorded in Fahrenheit. How do we find the mode of these temperatures?  Temperatures recorded:  -9,-12, 0, -6,      - 10, -3, 5
To find the mean, the first step would be to order the given temperatures in ascending order and then identify the temperature which occurs most often
-12, -10, -9, -6,-3, 0, 5
From the above ordered list, we can see that there is no value or number which occurs most often. So, we can conclude that there is no mode for the given temperatures.

This article gives basic information about Online Statistics Tutoring. Next article will cover more Statistics concept and its problems and many more. Please share your comments.

Definition of Absolute Value

Definition of Absolute Value
The absolute value is defined as the distance of ‘a’ from zero and is denoted by the symbol |a|. The absolute value only tells how far from zero and not in which direction is the location of a.

From the above figure, the absolute value of |3| = 3 since it is 3 units from the right side of the zero, and |-3|= 3 since it is 3 unit from the left side of the Zero. So the absolute unit should not have the negative sign, and is always positive or zero.

Absolute value equation

To solve absolute value equations, first split the equation into two equations. Then solve the equation to get the solution of absolute value equation.
Example 1:
|x| = 7
X = 7 x = -7
The solutions are {7, -7}
Example2:
|3x-4| = 5
3x-4  = 5 3x-4 = -5
3x = 9 3x = -1
X = 3 x = -1/3
The solutions are {1/3, 3}


Absolute Value Inequality
To solve absolute value inequalities first isolate the one side on the inequality symbol. Then write the first equation with out absolute sign and solve the inequalities and write second equation without absolute sign, reverse the inequalities and then solve the problem. The absolute value should be greater than any negative number and it should not be less than a negative number, sincethe absolute value should be always positive.
Example1:  (greater than)
|x-20| > 5
x-20> 5 X-20 < -5
x> 25 x < 15
The solution is 15 > x > 25
Example2: (less then or equal to)
|X-3| = 4
X-3 = 4 X-3 = -4
X = 7 X = -1
The solution is -1 = x = 4

Absolute value of a complex number
The definition for absolute value ofa real number is not directly generalized for definition for complex numbers, since the complex numbers are not ordered. But the geometric interpretation of the absolute value of a real number is as its distance from ‘0’ to be generalized.

The definition for absolute value ofa complex number is, it is a distance in the complex plane from the origin ‘0’ by using of Pythagorean Theorem. The absolute value of the difference of the complex number is equal to the difference between them.
The definition for absolute value ofa complex number is,
Z = x + iy
Where,
Z - Absolute value or modulus
x, y -  Real numbers
|z| = v(x^2+y^2 )
In above equation, the absolute value of a real number is x, when the complex part becomes 0.
In polar form the complex number z is expressed like z = re i?
If the absolute value is (r = 0, ?-real)
|z| = r
So, the absolute value of a complex number in the complex analogue equation is,

The properties of absolute value of a complex number are as same as absolute values of a real numbers.

Give example how to solve the absolute value equations and inequalities
Example 1: absolute value equation
|x - 7| = |2x-2|
Write in to two equations with out absolute symbol.
X-7 = 2x-2 x-7 = -(2x-2)
X-2x = -2+7 x+2x = 7 + 2
-x = 5 3x = 9
X = -5 x = 9
The solutions are { 9, -5}
Example 2: Absolute value inequalities
|2x-3| > 5
Write in to two inequalities with out absolute symbol.

2x-3 > 5 2x-3 < - 5
2x > 8 2x< -2
X < 4 x < -1
The solution is -1 > x < 4

Equation of a line which is passing through two points

Question :-

Find the equation of the line which is passing through two points (-3,7)(5,-1)

Answer:-

We have to use the point formula to find the equation of the line,this is much similar to midpoint formula

y-y1     x-x1
------ = ------
y2-y1     x2-x1

We have 2 points

( -3 , 7 )  and ( 5 , -1 )
x1  y1          x2  y2

So the equation is

y-7      x-(-3)
------ = ------
-1-7      5-(-3)

y-7      x+3
------ = ------
-8        8

We can further simplify it by cross multiplication.which is a part of indices maths

similarly we can find all points having an x-coordinate of 2 whose distance from the point 2 1 is 5

Statistics: Mean

Mean Math Or Mean Statistics
In Mathematics in the branch of Statistics, the expression for the mathematical mean of a statistical distribution is the mathematical average of all the terms in the data. To calculate this, we add up the values of the terms given and divide the sum by the number of terms in the data. This expression is also called the Arithmetic Mean.
Example: Find the Arithmetic Mean of the following data 5, 5,10,10,15,15,20,30
Solution: Arithmetic Mean = Regular average = sum of the values of the terms/number of terms
Sum of the values of the terms =5+5+10+10+15+15+20+30=110
Number of terms = 8
Mean = 110/8 = 13.75
Sample Mean
The sample mean in statistics branch of Mathematics is the sum of all observed outcomes from the sample divided by the total number of events. It is denoted by the symbol x with a (bar) above it. The formula used to compute the sample mean is as follows:
X (bar)= (1/n) (x1+x2+x3………xn)
 If we consider the sampling of some non-indigenous weed in a land of five acres in Springfield and came up with the counts of this weeds in that area as 38, 56, 84,105,116
We calculate sample mean for the above sampling by adding the weed counts and divide by the number of samples, 5
(38+56+84+105+116)/5= 79.8
So, we get the sample mean of non-indigenous weeds = 79.8
Weighted Mean
While computing Arithmetic Mean all the terms or values have equal importance. But at times we come across some situations where all the terms or values do not have equal importance. For instance, when we compute the average number of marks per subject, as per the different subjects like English, Science, Mathematics, Social Sciences; each subject has different levels of importance.
So, weighted mean is the arithmetic mean calculated by considering the relative importance (weight) of each term.
Each item is assigned a weight in proportion to its relative importance
Formula to compute Weighted Mean: xw (bar) = sigma(wx)/sigma(w)
(here x =value of the items  w= weight of the item)
Example:  A student scored 50, 70, 80 in English, Science and Math respectively and assume the weights of each subject to be 4,5,6  respectively. Find the weighted arithmetic mean of each subject.
Solution: let us tabulate the given data
Subjects              Marks obtained          weight              wx
English                50                          4                  150
Science                       60                          5                  300
Math    80                          6                  480
         Sigma(w)=15  sigma(wx)=930
Arithmetic Weighted mean = 930/15= 62 marks/subject
Short-cut Method
A short-cut method of calculating the arithmetic mean is based on the property of arithmetic average. In this method the deviations (D) of the items from an assumed mean are first calculated and then multiplied with their respective frequencies (f). Then, the total of these products [sigma(fD)] is divided by the total frequencies [sigma(f)]and added to the assumed mean(A). The figure we get is the actual arithmetic average or the Arithmetic Mean.
Formula used in the short-cut method of calculating the arithmetic mean:
X(bar) = A + sigma(fD)/sigma(f)