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.