Variance Covariance Formula

Variance: Variance is used in the statistical analysis to find the the extent to which a single variable is varying from its mean value given a set of values.

Variance of a random variable X is often denoted as VAR ( X). It is also denoted by the symbol `sigma` 2X

Covariance: Co-variance is used in statistical analysis to find the extent to which 2 variable are varying together given a set of values for both these variables.

Covariance of two random variable X and Y is denoted as COV ( X,Y). Unlike variance which has the mathematical symbol sigma ( `sigma` ), Covariance doesn't have any kind of symbol to depict it.
Formulae for Variance and Covariance.

Lets consider 2 sets of data namely X and Y such that the values in X are x1 , x2 , x3 , x4 , ......xn and similarly the values in Y are y1 , y2 , y3 , y4 , .......yn

Now lets denote the mean of the X set of variables to be    Xm

Mean Xm = X1 + X2 + X3 +........ + Xn / n

Similarly lets denote the mean of the Y set of variables to be Ym

Mean Ym = Y1 + Y2 + Y3 +.......+ Yn / n

Formulae for Variance of X = ( x1 - xm)2 + (x2 - xm)2 + ( x3 - xm)2 + .......+ (xn - xm)2 / n

VAR ( X ) = (1/n) `sum` (xi - xm)2

Formulae for Variance of Y = ( y1 - ym)2 + (y2 - ym)2 + ( y3 - ym)2 + .......+ (yn - ym)2 / n

VAR ( Y ) = (1/n) `sum` (yi - ym)2

Formulae for Covariance of (X,Y) = ( x1 - xm) ( y1 - ym) + (x2 - xm)(y2 - ym) + ( x3 - xm)( y3 - ym) + .......+ (xn - xm)(yn - ym) / n

COV ( X , Y ) = (1/n) `sum` (xi - xm)(yi - ym)
Example Problem 1 on Variance and co Variance

Lets assume the below set of marks received by Bill and Bob in Maths, Physics, Chemistry, English and Biology for the purpose of solving Variance and Co Variance

table

Based on the Formulae given in the previous paragraph, Lets calculate the Mean Marks for Bill and Bob

Mean Marks for Bill = 16+12+14+18+20 /5 = 16.00

Mean Marks for Bob = 14+18+15+18+20 /5 = 17.00

Now lets apply the formulae of variance and find out the Variance of Bill and Bob

Variance of Bill = (16-16)2 + (12-16)2 + (14-16)2 + (18-16)2 + (20 - 16)2 / 5 = 40 / 5 = 8

Variance of Bob = (14-17)2 + ( 18-17)2 + (15-17)2 + (18-17)2 + (20-17)2 / 5 = 24 / 5 = 4.8

Co variance of ( Bill , Bob ) = (16-16)*(14-17) + (12-16)*(18-17) + (14-16)*(15-17) + (18-16)(18-17) + (20-16)*(20-17) / 5

Co Variance of ( Bill , Bob ) = 21/5 = 4.2

I am planning to write more post on What are Line Segments and Perpendicular and Parallel Lines. Keep checking my blog.

Example Problem 2 on Variance and co Variance

Lets assume a class of 4 student who have been asked to rate their liking toward 2 musical instruments Guitar and Piano on a scale of 10

Based on the above data, lets calculate the Mean of the people liking Guitar and Mean of people liking Piano.

Mean(Guitar) = ( 5+3+7+9)/4 = 24/4 = 6

Mean(Piano) = (5+9+9+5)/4 = 28/4 = 7

Var(Guitar) = ((5-6)2 + (3-6)2 + (7-6)2 + (9-7)2 )/ 4 = (1+9+1+4)/4 = 15/4 = 3.75

Var(Piano) = ((5-7)2 + (9-7)2 + (9-7)2 + (5-7)2 ) / 4 = (4+4+4+4)/4 = 16/4 = 4

Cov(Guitar,Piano) = [(5-6)(5-7) + (3-6)(9-7) + (7-6)(9-7) + (9-7)(5-7) ] / 4 = (2+(-6)+2+(-4))/4 = -6/4 = -1.5

Binomial Probability Function

Binomial probability Problems represented by B(n,p,x). It gives the probability of exactly x successes in ‘n’ Bernoullian trials, p being the probability of success in a trial. The constants n and p are called the parameters of the distribution. A Binomial distribution can be used under the following condition.

(i) any trial, result in a success or a failure

(ii) There are a finite number of trials which are independent.

(iii) The probability of success is the same in each trial.

In a Binomial distribution function mean is always greater than the variance. The binomial probability function example problems and practice problems are given below.
Example Problems - Binomial Probability Function:

Ex 1:  By using the binomial distribution. If the sum of mean and variance is 4.8 for  5 trials find the distribution

Solution:  np + npq = 4.8 , np(1 + q) = 4.8

5 p [1 + (1 − p) = 4.8

p2 − 2p + 0.96 = 0 , p = 1.2 , 0.8

p = 0.8 ; q = 0.2 [p cannot be greater than 1]

The Binomial distribution is P[X = x] = 5Cx (0.8)x (0.2)5−x,  x = 0 to 5

Practice Problem - Binomial Probability Function:

Pro 1: Find the value for p by using the Binomial distribution if n = 5and P(X = 3) = 2P(X = 2).

Ans: p = `(2)/(3)`

Pro 2: Find the probability values by using the binomial distribution.  A pair of dice is thrown 10 times. If getting a doublet is considered a success (i) 4 success (ii) No success.

Ans: (a) (35/216)(5/6)6

(b) (5/6)10

Proportion Equation

The data is obtaining by the comparison of two ratios is called proportion data. Proportion data is represented as a:b = c:d. This proportion data can be written in the form of fraction as `a/b` = `c/d` . Where the pairs of data (a,b) and (c,d) are in proportion. When the proportions are equal, the cross product of the proportion will be also equal. That is, `a/b` = `c/d` can be written as ad=bc.

Examples for Proportion Equation:

Example 1 for proportion equation:

Martin read 63 pages of the book in 33 minutes. How many pages will he be able to read in 43 minutes?

Solution:

Martin takes 33 minutes to read 63 pages.

Martin will take 43 minutes to read x pages.

This can be written as,

`63/33` = `x/43`

Now we have to do the cross multiplication.

63 `xx ` 43 = x `xx` 33

2709 = 33x

This can be written as,

33x = 2709

Now we have to divide both sides by 33.

`(33x)/33` = `2709/33`

x = 82.09 now we have to round it to the unit place.

x = 82

Therefore, Martin will read 90 pages in 45 minutes.

Example 2 for proportion equation:

Paul bought 12 apples for dollar 48.  How many apples will he be able to buy in $ 93?

Solution:

Paul spends $48 for 12 apples.

Paul will spend $93 for x apples.

This can be written as,

`12/48` = `x/93`

Now we have to do the cross multiplication.

12 `xx` 93 = x `xx` 48

1116 = 48x

This can be written as,

48x = 1116

Now we have to divide both sides by 48.

`(48x)/48` = `1116/48`

x = 23

Therefore, Paul can buy 23 apples for $ 93.

Practice Problems for Proportion Equation:

Problem 1 for proportion equation:

Martin read 40 pages of the book in 28 minutes. How many pages will he be able to read in 52 minutes?

Solution: Martin will read 74 pages in 52 minutes.

Problem 2 for proportion equation:

Paul bought 8 apples for dollar 22. How many apples will he be able to buy in $ 66?

Solution: Paul can buy 24 apples for $ 66

Polynomials Calculator

An Algebraic expression is of the form axn is called a monomial. The variable a is called the coefficient of xn and n, the degree of monomial. For example, 7x3 is monomial in x of degree 3 and 7 is the coefficient of x3. The combination of two monomials is called a binomial and the combination of three monomials is called a trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is trinomial. The sum of n number of monomials, where n is finite and x is called a polynomial in x.
Illustration to Polynomials:

Polynomial Calculator Example 1:

The polynomial calculator of the equation x2 + ax + b gives the remainder 18, when divided by x – 2 and leaves the polynomial calculator of remainder –2 when that is been divided by (x + 3).

Find the values of a and b.

Solution to the polynomial calculator:

P(x) = x2 + ax + b.

In tyh is polynomial calculator,

When x – 2 divides P(x) then the remainder is P (2).

∴P (2) = 4 + 2a + b.

But remainder = 18 ⇒ 4 + 2a + b = 18;

2a + b = 14 (1)

When (x + 3) divides P(x)

, the remainder is P (–3).     ∴ P (–3) = (–3)2 + a (–3) + b

= 9 – 3a + b.

But remainder = –2;      ∴ 9 – 3a + b = –2;

⇒ –3a + b = –11 (2)

(1) ⇒   2a + b = 14

(2) ⇒ –3a + b = –11 (subtracting)

5a        = 25

(Or) a     = 5

Substituting a = 5 in equation (1) we get

10 + b = 14; b = 4, ∴ a = 5, b = 4
Subtraction of Polynomials Calculator:

Example for Polynomials calculator:

Subtract        2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Solution:

Using associative and distributive properties, we have

( x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the signs of the polynomial in Line (2), we get

Line (3): –2x3 + 3x2 + 1.

Adding the polynomials in Line (1) and Line (3), we get

–x3 + 8x2 – 4x – 5.

Integral Part Definition

In geometrical and other applications of integral calculus it becomes necessary to find the difference in the valves of an integral of a function f(x) for two assigned values of the independent variable x, a and b. this difference is called the definite integral of f(x) over the range (a,b) and is denoted by $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where F(x) is an integral of f(x), F(b) is the variable of F(x) at x=b, F(a) is the value of F(x) at x=a.

It is often written thus  : $\int_{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)$.

Note 1. The integral $\int_{a}^{b}f(x)dx $ is read as the integral of f(x) from a to b.The number a is called the lower limit and the number b, the upper limit of integration.

Note 2. It should be seen that the value of a definite integral is perfectly unique and is independent of the particular form of integral which may employ to calculate it. Considering F(x)+c instead of F(x), we get

$\int_{a}^{b}f(x)dx=[F(x)+c]_{a}^{b}=[F(b)+c]-[F(a)+c]=F(b)-F(a)$

so that the arbitrary constant disappears in the process and we get the same values as on considering F(x). This is why the name is given as definite integral.

Note 3. It is assumed that a and b are finite.
Definition of Integration by Parts

When the given function cannot be integrated directly by using standard formulae, we try other methods. The process of integration is largely of tentative nature and no systematic procedure can be given as in differentiation. However, the following are two important methods opf integration.

1. Integration by Substitution           2. Integration by Parts

Here, we will discuss about Integration by parts

An Integration is the inverse process of differentiation.By differentiaon we find the derivative of the given function, whereas by integration we find the function whose derivative is known.

If the derivative of F(x) is f(x) then we say that the antiderivative or integral of f(x) is F(x),such that

int f( x ) dx= F( x )

Thusd/dx F(x) = f(x) => intf(x) dx = F(x)

Integration by Parts:

Theorem: if u and v are two differentiable function of x then

int ( u v ) dx = [ u * int v dx ] - int { du/dx * int v dx } dx .

We can express this result as given below:

Integral of product of two function

= ( 1st function ) * ( integral of 2nd function ) - int { ( derivative of 1st function ) * ( integral of 2nd function ) } dx .

We should choose u and v in such a way that the second function v on the right hand side is easy to integrate. Sometimes, this rule has to be used repeatedly. This rule is also useful in integrating logarithmic and inverse t - functions of the type log x, log(ax2 + bs + c), sin-1 x, tan-1 x etc.The following guidelines will help you to see which of the two functions in the product should be taken as the first function.

Notes:

If the integrand is of the form f(x) * xn, we consider xn as the first function and f(x)as the second function.
If the integrand contains a logarithmic or an inverse trigonometric function, we take it as the first function. If the second function is not given in any case we take that as one.

Example Problems on Integration by Parts:

Pro 1: Evaluate, int xsin 2x dx

Sol : Given,

int   xsin 2x dx

=     int sin 2x dx - int { d/dx ( x ) * int sin 2x dx } dx

=     x * ( -(cos 2x)/2 ) - int 1 * ( -(cos 2x)/2 ) dx

=  (-x cos 2x)/2 + 1/2 int cos 2x dx

"-(x cos  + 1/2 * "(sin

= -(x cos 2x)/2 + 1/4 sin 2x + C , which is the Answer.

Pro 2: int logx/x^2 dx

Sol :   Integrating by parts, taking logx as the fi rst function and 1/x^2  as the second function, we get

int "log   dx

=  int (log x) * 1/x^2 dx

=   ( log x ) * int 1/x^2 dx - int { d/dx ( log x ) * int 1/x^2 dx } dx

=    ( log x )(-1/x )   - int 1/x  * ( -1/x ) dx

= - log x/x + int 1/x^2 dx

= - log x/x - 1/x + C, which  is the required Answer.

Practice problems on Integration by Parts:

Pro 1: Evaluate, int (  x cos x) dx    ( Answer:   x sinx +cosx+c )

Pro 2: Evaluate, int e2x sin x dx            ( Answer: 1/5 e2x ( 2 sin x - cos x ) + C )

Natural Logarithmic Calculator

In mathematics, the natural logarithmic function is defined as the function contains three ports, namely the number, the base and logarithm itself,  now the natural logarithmic calculator is the function of  the logarithm function calculator which means we are using to the base of ‘e’ where e is constant, base of  ‘e’ value is given by

Understanding Definition of Logarithm is always challenging for me but thanks to all math help websites to help me out.

e = 2.718281828. and the numbers, and the logarithmic itself.

The logarithmic function calculator is represented as y = log_b^(x).

We take the value of base of ‘e’ is approximately e = 2.71.
Natural Logarithmic Calculator:

Here, we introduce the natural logarithmic calculator. The function of natural logarithmic calculator is represented as  y = ln(x)

Explanation about natural logarithmic calculator:

Input:

calcutor             given value

The given value is conversions by logarithmic calculator

Conversions:

 x =  conversion value.

Solution:

 y = calculated value will be displayed

Here we can the change the equation by x and y.

y = ln(x) for y i n natural logarithm

y = ln(x) for x  in natural logarithm

Note: we can calculate or perform the positive numbers only not negative numbers by natural logarithmic calculator.
Some Problems about Natural Logarithmic Calculator:

Problem1:

To solve the natural logarithmic function of ln(2)

Solution:

Input:

x = 2  given value

Conversion:

x = 2  the given value conversion by logarithmic calculator.

Solution:

y = 0.69 approximately

Problem 2:

To solve natural logarithmic function of  span style="font-size: small; " mce_style="font-size: small; ln(4)

Solution:

Input:

 x = 4 given value

Conversion:

 x = 4  the given value conversion by logarithmic calculator.

Solution:

 y = 1.38 approximately

Problem 3:

To solve natural logarithmic function of ln(6)

Solution

Input:

 x = 6 given value

Conversion:

 x = 6  the given value conversion by logarithmic calculator.

Solution:

 y = 1.79 approximately

Problem 4:

To solve the natural logarithmic function of in ln(10)

Solution:

Input:

x = 10 given value

Conversion:

 x = 10 the given value conversion by logarithmic calculator.

Solution:

 y = 2.30 approximately

Problem 5:

To solve the natural logarithmic function of ln(15)

Solution:

Input:

x = 15 given value

Conversion:

 x = 15 the given value conversion by logarithmic calculator.

Solution:

 y = 2.70  approximately


Practices problems for natural logarithmic function:

Problem1;

To solve the natural logarithmic of ln (20)

Answer is y = 2.99

Problem2

To solve natural logarithmic function of ln (50)

Answer is y = 3.91

Problem3

To solve natural logarithmic function of ln(100)

Answer is y = 4.60

Problem 4

To solve the natural logarithmic function of in ln(99)

Answer is y = 4.59

Hence, here we obtain the natural logarithmic calculator.

Scientific Notation Rules

 Often we come across abnormally large numbers or abnormally small numbers like 400,000,000,000 or very, very small number like0.00000000002. Especially in the field of science such abnormal size of numbers is quite common.

In calculations involving such numbers, the work is cumbersome and possible errors may occur.

A special type of notation is used to simplify the dealings with these type of numbers and it is named as scientific notation as such numbers are very predominant in the field of science.

However a scientific notation must be used following a set of rules, universally accepted.

Let us take a closer look.
Scientific Notation Rules – Description

Since very large or very small numbers can be rounded to the appropriate place value, the power of 10 is used as base in scientific notation. The established general form is,

                                       a x 10n, where,  1= I a I < 10

If  ‘n’ is positive, it represents the number of zeroes from ‘a’ to the right of unit place and hence  it denotes large numbers.

For very, very small decimal numbers ‘n’ is negative and represents the number of zeroes before ‘a’ up to the decimal point.
Scientific Notation Rules – Examples

The velocity of light is about 310,000,000 kilometers per second.

We find there are seven 0s after the digits 1. That is the number is same as 31 times 10 to the power 7. But as per scientific notation rules, 31x 107 is not allowed and hence the given is modified as 3.1 x 108

Thus in scientific notation, 310,000,000 = 3.1 x 108

Similarly it is equally incorrect to mention (as a scientific notation) 310,000,000 as 0..31 x 109

Let us consider another example with a very  small number.

The wave lengths of a certain color is experimentally found to be 0.0000015 millimeter.

There are 6 decimal places to cross the digit 1from right. That the given number is (1/1000000)th of 1.5.

Hence in scientific notation it is written as 1.5 x 10-6