Skew Normal Distribution

The skew normal probability distribution refers the normal probability distribution. It is also called as the Gaussian distribution. In normal distribution the mean is μ and the variance is `sigma^2` . Normal distribution is the close approximation of a binomial distribution. The limiting form of Poisson distribution is said to be normal distribution probability. This article has the information about the skew normal probability distribution.

Formula used for skew normal distribution:

The formula used for plot the standard normal distribution is

Z = `(X- mu) /sigma`

Where X is the normal with mean `mu` and the variance is `sigma^2` , `sigma` is the standard deviation.

Examples for the skew normal distribution:

Example 1 for the skew normal distribution:

If X is normally distributed the mean value is 1 and its standard deviation is 6. Determine the value of P (0 ≤ X ≤ 8).

Solution:

The given mean value is 1 and the standard deviation is 6.

Z = `(X- mu)/ sigma`

When X = 0, Z = `(0- 1)/ 6`

= -`1/6`

= -0.17

When X = 8, Z = `(8- 1)/ 6`

= `7/6`

= 1.17

Therefore,

P (0 ≤ X ≤ 4) = P (-0.17 < Z < 1.17)

P (0 ≤ X ≤ 4) = P (0 < Z < 0.17) + P (0 < Z < 1.17) (due to symmetry property)

P (0 ≤ X ≤ 4) = (0.5675- 0.5) + (0.8790 - 0.5)

P (0 ≤ X ≤ 4) = 0.0675 + 0.3790

P (0 ≤ X ≤ 4) = 0.4465

The value for P (0 ≤ X ≤ 4) is 0.4465.

Example 2 for the skew normal distribution:

If X is normally distributed the mean value is 2 and its standard deviation is 4. Determine the value of P (0 ≤ X ≤ 5).

Solution:

The given mean value is 2 and the standard deviation is 4.

Z = `(X- mu)/ sigma`

When X = 0, Z = `(0- 2)/ 4`

= -`2/4`

= -0.5

When X = 5, Z = `(5- 2)/ 4`

= 3/4

= 0.75

Therefore,

P (0 ≤ X ≤ 6) = P (-0. 5 < Z < 0.75)

P (0 ≤ X ≤ 6) = P (0 < Z < -0.5) + P (0 < Z < 0.75) (due to symmetry property)

P (0 ≤ X ≤ 6) = (0.6915 - 0.5) + (0.7734- 0.5)

P (0 ≤ X ≤ 6) = 0.1915+ 0.2734

P (0 ≤ X ≤ 6) = 0.4649

The value for P (0 ≤ X ≤ 6) is 0.4649.

Fundamental Theorem of Calculus Proof

The fundamental theorem of calculus determines the association the two basic operations of calculus called as the differentiation and integration. The first fundamental theorem of integration deals with the indefinite integration and the second theorem of integration deal with the definite integral of the function.

In this article we are going to see about the proof of the fundamental theorem for calculus.

Proof of fundamental theorem for calculus:

Proof of first fundamental theorem:

Let us take a real valued function f which is given by

x
F(x) = int f(t) dt, here f is also a real valued function
a

Then F is said to be continuous on [a,b] and can be differentiated on the open interval (a,b) which is given by

F′(x) = f(x), for all the values of x in the interval (a,b)

Let us consider two numbers x1 and x1+∆x , in the closed interval (a, b), then we have

x1                                                                        x1+∆x
F(x1) = int f(t) dt   -------> (1)     and       F(x1+∆x) = int f(t) dt ----------> (2)
a                                                            a

When we subtract the first equation and the second equation we get

x1+∆x                x1
F(x1+∆x) - F(x1) =  int   f(t) dt  -  int   f(t) dt ------------> (a)
a                    a

a              x1+∆x
=  int f(t) dt + int f(t) dt
x1                 a

a              x1+∆x           x1+∆x
int f(t) dt + int f(t) dt =  int f(t) dt
x1                 a                x1

When we substitute in equation (a) we get,

x1+∆x
F(x1+∆x) - F(x1) =  int   f(t) dt  -------------> (b)
x1

According to the theorem of integration

x1+∆x
int   f(t) dt  = f(c) ∆x  ,  given that there exists c in [x1 and x1+∆x]
x1

When we substitute this value in equation (b) we get,

F(x1+∆x) - F(x1) = f(c) ∆x

When we divide by ∆x on both sides we get,

F(x1+∆x) - F(x1) = f(c)
∆x

Take the limit ∆x  ->0, on both sides we get,

lim      F(x1+∆x) - F(x1) = lim  f(c)
∆x ->0            ∆x                        ∆x ->0

This left side of the equation is the derivative of F at x1

F′(x1)  = lim  f(c) -----> (3)
∆x ->0

We know that

lim    x1 = x1  and lim    x1+∆x  = x1
∆x ->0                 x1+∆x->0

Then according to the squeeze theorem, we get,

lim    c  = x1
∆x ->0

Substituting this in (c), we get

F′(x1) = lim  f(c)
c -> x1

The function f is continuous and real valued and thus we get,

F′(x1) = f(x1)

Hence the proof of the fundamental theorem for calculus.

Corollary of fundamental theorem proof:

The fundamental theorem is used to calculate the definite integral of a given function f if f is a real-valued continuous function on the interval [a, b], then

b
int f(x) dx = F(b) – F(a)

Learn Analytic Geometry Online

Another name of Analytical  geometry is co-ordinate geometry and it describes as a graph of quadratic equations in the co-ordinate plane. Analytical geometry grew out of need for establishing uniform techniques for solving geometrical problems, the aim being to apply them to study of curves, which are of particular importance in practical problems. In this article we shall discuss the learning analytic geometry ans some example problems

Example Problem 1 to learn analytic geometry online:

Find the equation of the parabola if the curve is open upward, vertex is (− 1, − 2) and the length of the latus rectum is 4.

Solution:

Since it is open upward, the equation is of the form

(x − h)2 = 4a(y − k)

Length of the latus rectum = 4a = 4 and this gives a = 1

The vertex V (h, k) is (− 1, − 2)

The equation of parabola is [x-(-1)]2=4*1 [y-(-2)]

[x+1]2=4[y+2]

Example Problem 2 - Learn analytic geometry online  :

Find the equation of the parabola if

(i) the vertex is (0, 0) and the focus is (− a, 0), a > 0

Solution: (i) From the given data of the parabola is open leftward

The equation of the parabola is of the form

(y − k)2 = − 4a(x − h)

Here, the vertex (h, k) is (0, 0) and VF = a

The required equation is

(y − 0)2 = − 4a (x − 0)

y2 = − 4ax.

Example Problem 3 - learn analytic geometry online :

Find the equation of the parabola if the curve is open rightward, vertex is (2, 1) and passing through point (6,5).

Solution: Since it is open rightward, the equation of the parabola is of the form

(y − k)2 = 4a(x − h)

The vertex V(h, k) is (2, 1)

∴ (y − 1)2 = 4a (x − 2)

But it passes through (6, 5)

(5-1)2 = 4a (6 − 2) -- > 16= 4a * 16

4a= 1 ---- > a = 1/4

The required equation is (y − 1)2 = 1/4 (x − 2)

Example Problem 4 - learn analytic geometry online :

Find the equation of the parabola if the curve is open leftward, vertex is (2, 0) and the distance between the latus rectum and directrix is 2.

Solution: Since it is open leftward, the equation is of the form

(y − k)2 = − 4a(x − h)

The vertex V(h, k) is (2, 0)

The distance between latus rectum and directrix = 2a = 2 giving a = 1 and the equation of the parabola is

(y − 0)2 = − 4(1) (x − 2)

or y2 = − 4(x − 2)

Practice problems- learn analytic geometry online:

1.Find the equation of the parabola whose vertex are (1, 2) and the equation of the directrix x = 3.

The required equation is

(y − 2)2 = 4(2) (x − 1)

(y − 2)2 = 8(x − 1)

2.The separate equations of the asymptotes of the hyperbola 4x2-25y2=100

Answer   :x/5-y/2=0 and x/5+y/2=0

Definition to Population Variance

Population variance is defined as the square of the mean deviation value by the total number of data. Population variance is calculated to find the variance in the probability of data.

Formula for finding the

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

Where as

sigma^2 - symbol for population variance

mu - It is the mean of the given data.

N  - Total number of values given in data set.

For finding the population variance we have to find the sample mean. For the take the average for the given values.

 mu = (sum_(K=1) ^n (x_k))/N

This mean value is used in the population variance to find its value.

Population Variance Values - Example Problems:

Population Variance Values - Problem 1:

calculate the population variance for the given data set. 2, 4, 3, 3, 5, 6, 5

Solution:

Mean:  Calculate the mean for the given data

 mu = (sum_(K=1) ^n (x_k))/N

using the above formula find the average for the given values.

 mu = (2+4+3+3+5+6+5)/7

mu = 28/ 7

 mu = 4

Population Variance: Calculate the population variance value from the mean.

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

substitute the mean values to find the deviation values from the given values.

 sigma^2 = ((2-4)^2+(4-4)^2+(3-4)^2+(3-4)^2+(5-4)^2+(6-4)^2+(5-4)^2)/7

sigma^2 = 12/7

sigma^2 = 1.7142857142857

Hence the population variance value is calculated using the mean value.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Differentiation Math and What are Composite Numbers. I am sure they will be helpful.

Population Variance Values - Problem 2:

Find the value for the population variance of the given data set. 367, 378, 365, 366.

Solution:

Mean: Calculate the mean for the given data set using the formula,

 mu = (sum_(K=1) ^n (x_k))/N

Find the mean value that is average of the giave data set

mu = (367+378+365+366)/4

mu = 1476 / 4

mu = 369

Population variance: Calculate the population variance value from the mean.

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

substitute the mean values to find the deviation values from the given values.

sigma^2 =((367-369)^2 + (378-369)^2 + (365 - 369)^2+(366-369)^2) / 4

sigma^2 = 110/4

sigma^2 = 27.5

Hence the population variance value is calculated using the mean value.

Population Variance Values - Problem 3:

Find the value for the population variance of the given data set. 36, 37, 36, 39.

Solution:

Mean: Calculate the mean for the given data set using the formula,

 mu = (sum_(K=1) ^n (x_k))/N

Find the mean value that is average of the giave data set

mu = (36+37+36+39) / 4

mu = 148 / 4

mu = 37

My previous blog post was on Natural Numbers please express your views on the post by commenting.

Population variance: Calculate the population variance value from the mean.

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

substitute the mean values to find the deviation values from the given values.

"sigma^2 = ((36-37)^2 +(37-37)^2+(36-37)^2+(39-37)^2) / 4

sigma^2 = 6/4

sigma^2 = 1.5

Hence the population variance value is calculated using the mean value.

Population Variance Values - Practice Problems:

Find the value for the population variance of the given data set. 45, 47, 49, 43.

Answer: Population Variance = 5

Find the value for the population variance of the given data set. 87, 89, 78, 82, 89.

Answer: Population Variance = 18.8

Polynomial Matrices

A polynomial matrices or matrix polynomial is a  matrix whose elements are univariate or multivariate polynomials.A univariate polynomial matrix P of degree p is define like:sum_(n=0)^pA(n)x^(n)=A(0)+A(1)x+A(2)x^(2)+....+A(p)x^(p) where A(p) is non-zero and A(i) indicate a matrix of constant coefficients. Hence a polynomial matrix is the matrix-equivalent of a polynomial, by means of every one element of the matrix satisfying the classification of a polynomial of degree p.

Properties of polynomial matrices

A polynomial matrix in excess of a field with determinant equivalent to a non-zero constant is called unimodular, and have an inverse, which is also a polynomial matrix.
Note, that the simply scalar unimodular polynomials are polynomials of degree 0 - nonzero constants, for the reason that an inverse of an arbitrary polynomial of high degree is a rational function.
The roots of a polynomial matrix in excess of the complex numbers are the points in the complex plane wherever the matrix loses rank.

Characteristic polynomial of a product of two matrices

If A and B are two square n×n matrices then,attribute polynomials of AB and BA match:

PAB(t)=PBA(t).

If A is m×n-matrix and B is n×m matrices such that m
PAB(t)=tn-mPAB(t)

Polynomial in t and in the entry of A and B is a general polynomial identity. It consequently suffice to verify it on an open set ofparameter value in the complex numbers.

The tuples (A,B,t) wherever A is an invertible complex n by n matrix,

B is any complex n by n matrix,

and t is any complex number since an open set in complex space of dimension 2n2 + 1. When A is non-singular our result follow from the fact that AB and BA are similar:

BA=A-1(AB)A.

Example 1:

An example the 3x3 polynomial matrices

P=[[1,x^(2),x],[0,2x,2],[8x+2,x^2-1,0]]

=[[1,0,0],[0,0,2],[2,-1,0]]+[[0,0,1],[0,2,0],[3,0,0]]x+[[0,1,0],[0,0,0],[0,1,0]]xx^(2)

Example 2:

Find the eign value of given polynomial matrices

P=[[3,3],[0,6]]

The polynomial has the characteristic equation

0=det(P-λI)

=det[[3-lambda,3],[0,6-lambda]]

18-6lambda -3lambda + λ2

18-9lambda +18

λ2-3λ-6λ+18

λ(λ-3)-6(λ-3)

(λ-3)(λ-6)

λ=3,andλ=6

The eigenvalues of these matrices are 3,6

Example 3: Find the product of the given matrices M1=[[1,2],[3,4]] and M2=[[8,3],[2,7]]

The given polynomial is

M1=[[1,2],[3,4]]


M2=[[8,3],[2,7]]

The product of the given matrices M1 and M2  =M1xM2

M1xM2    =[[1,2],[3,4]]xx [[8,3],[2,7]]

The product of the given  matrices is=[[12,17],[32,37]]

Least Common Denominator

If the denominators are unlike, then we can find LCD (Least common Denominator) of the given denominators.

Least Common Denominator is the smallest positive(least) integer which is common in multiples of the denominators.

For example, given fractions are 1/3 and 1/6. Find LCD.

List the multiples of 3:   3, 6, 9, 12, 15, 18, 21,...

Multiples of 6:   6, 12, 18, 24,...

Here, 6 is the lowest common term for both the multiples of 3 and multiples of 6.

The answer is 6, and that is the Least Common Denominator.

There are five examples for least common denominator. From these examples for least common denominator, we can get clear view about least common denominator. Let us see the examples for least common denominator in the following section.

Examples on examples for least common denominator:

We are going to explain examples for least common denominator.

Example 1:

Find the least common denominator of the fractions; 1/5,1/3

Solution:

Here, the denominators are 5 and 3.

The common denominator of 5 and 3 is 15.

Multiples of 5: 5,10,15,20,…..

Multiples of 3: 3,6,9,12,15,18,….

Here, 15 is the lowest common term for both the multiples of 5 and multiples of 3.

The answer is 15, and that is the Least Common Denominator.

Example 2:

If the given fractions are 3/4,1/3, find the least common denominator.

Solution:

Here, the denominators are 4 and 3.

The common denominator of 4 and 3 is 12.

Multiples of 4: 4,8,12,16,20….

Multiples of 3: 3,6,9,12,15….

Here, 12 is the lowest common term for both the multiples of 4 and multiples of 3.

So, the Least Common Denominator is 12.

Example 3:

Find the least common denominator of ; 5/6, 2/15

Solution:

Here, the denominators are 6 and 15.

The common denominator of 6 and 15 is 30.

Multiples of 6:  6,12,18,24,30…

Multiples of 15: 15,30,45,60….

Here, 30 is the lowest common term for both the multiples of 6 and multiples of 15.

So, the Least common denominator is 30.

Example 4 on examples for least common denominator:

What is the least common denominator of the fractions;  5/12, 11/18

Solution:

Here, the denominators are 12 and 18.

The common denominator of 12 and 18 is 36.

Multiples of 12:  12,24,36,48,….

Multiples of 18: 18,36,54,….

Here, 36 is the lowest common term for both the multiples of 12 and multiples of 18.

So, the Least common denominator is 36.

Example 5:

1/5 + 1/6 + 1/15  What is the LCD?

Solution:

First we list the multiples of each denominator.

Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,...

Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,...

Multiples of 15 are 30, 45, 60, 75, 90,....

Here, 30 is the lowest common term for both the multiples of 5 and multiples of 6 and multiples of 15 .

So, the Least common denominator is 30.

Therefore, Examples for least common denominator are explained.

Learning Parallel Lines

Learning parallel lines is very easy. As the name it self indicates that there will be some lines that are parallel to each other. Think logically if two lines are parallel will they ever meet at some point. The answer is no. Because the two parallel lines maintain equal distance between them so there is no chance that they meet at some point. So we can conclude that two lines are parallel if and only if they are coplanar and never meet each other that is they maintain same distance apart.

Transversals are very important while learning parallel lines.

Tranversal can be defined as a line that intersects two or more  lines, at different points. Simply a line that crosses two or more lines  is called transversal. With the help of transversals we can say whether the two given lines are parallel or not. Figure below shows how a transversal looks.



How can we know the lines are parallel or not?

Answer is with the help of some angles that are formed when transversal passes through pair of coplanar lines. These angles are very important while learning parallel lines.The angles are

Corresponding Angles
Alternate Interior Angles
Alternate Exterior Angles
Co-Interior Angles


Corresponding angles:

In our figure the corresponding angles are " a,e " , " d, h" , "b , f" , "c,g".

Alternate Interior angles:

The name it self indicates that alternate means on the other side and interior means inner angles. The Alternate Interior angles are "d,f" and "c,e".

Alternate exterior angles:


The name it self indicates that alternate means on the other side and exterior means outer angles. Alternate exterior angles in the given figure are "a,g" and "b,h".

Co-interior angles:

In the given figure Co-interior interior angles are "d,e" and "c,f".

Now for two lines to be parallel corresponding angles, alternate Interior angles, alternate exterior angles must be equal and sum of Co-interior  angles must be equal to 180. Even if one condition is satisfied it is enough as automatically all the other will satisfy.

This is all about learning parallel lines. Hope you enjoyed it......