Interval Estimates

Interval Estimate:

• Interval estimation is the process of calculate the interval for possible value of unknown parameter in the population.
• It is calculate in the use of sample data and contrast to the point estimation. It is different from the point estimation. It is the outcome of a statistical analysis.

The most common forms of interval estimations as follows:

• A frequents Method or Confidence interval
• A Bayesian method or credible intervals

The other common methods for interval estimations are

• Tolerance interval
• Prediction interval

And another one is known as the fiducial inference.

Construction of interval estimates parameter:

The normal form of interval estimate of the population parameter is,

• Point estimate of parameter and
• Plus or minus margin of error

Margin of error:

• The amount which is subtracted or added from  the point estimate  of the statistic and produce the parameter interval  estimate is known as the margin of error.
• The margin of error size depends on the following factors:
• Sampling distribution type of sample statistics.
• Area under sampling distribution percentage   that includes the researchers      decision.Usually we consider the confident level as 90%, 95%, 99%.
• The interval of each interval estimates are constructed in the region of the point estimate with its confident level.

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Construction of Interval estimate for Population mean

• Take the point estimate of μ  that is  the sample mean`vecx`
• Define  the mean distribution for the sample.When the  value of n is large we  have to use the central limit  theorem. And   is the normal distribution with the,

                      standard deviation `sigma``vecx` = `sigma/sqrt(n)`  

                      and mean μ.

• Choose the most common confident  level as 95%
• Find the margin of  error  which is related with the confidence level.
• The area  under the curve of  the sample means the normal distribution contains the 95%  of the interval from.

                               z= -1.96 to z= 1.96 

• The interval estimate for 95 % is,   

                            `vecx`- 1.96 (`sigma/sqrt(n)` ) to `vecx``sigma/sqrt(n)`

Common Factors

The common factors of two or more whole digits is the biggest whole digit that equally divides all the whole digits. There are two methods to find common factors in math.

The initial method is to list all the factors of each digit. Then decide the biggest factor.

For Example:

Find the common factors of 12 and 18.

The factors of 12 are 1, 2, 3, 4, 6 and 8.

The factors of 18 are 1, 2, 3, 6, 9 and 18.

The common factors of 12 and 18 are 1, 2 and 3.

Methods for Finding Common Factors

There are two methods to find the common factors of numbers. Listed below are the steps to be followed in finding common factors.
Method I:

• List all the factors of the numbers.
• Collect the common factors among each digit.

Method II:

• Find the prime factors of each digit.
• Merge the common terms of each digit.

Examples

Listed Below are some of the examples in finding the common factors.
Example 1:
What are the Common factors of 34 and 36?
Solution:
The factors of 34 are 1, 2, 17, and 34.
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Now, using method 1,
The common factors to the two numbers are 1 and 2.
The Common factors of 34 and 36 are 1 and 2.
Example 2:
What are the common factors of 40, 45 and 50?
Solution:
Prime factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
Prime factors of 45 are 1, 3, 5, 9, 15 and 45
Prime factors of 50 are 1, 2, 5, 10, 25, and 50.
Now, using method 2,
The common factors of 40, 45 and 50 are 1, 2 and 5.

Practice Problems

Listed below are some of the practice problems in finding the common factors.

Problem 1:

What are the Common factors of 8, 14, 18 and 22?

Answer:    

8   `->` 1, 2, 4, and 8

14 `->` 1, 2, 7 and 14

18 `->` 1, 2, 3, 6, 9 and 18

22 `->` 1, 2, 11 and 22

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Problem 2:

What are the Common factors of 15, 30, 45 and 60?

Answer:    

15 `->` 1, 3, 5 and 15

30 `->` 1, 2, 3, 5, 6, 10, 15 and 30

45 `->` 1, 3, 5, 9, 15 and 45

60 `->` 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60

Review of Word Problems

Introduction on review of word problems are first used to identify the variables present in the given problem. Also the word problems are used for the unit numbers and unit variables. Word problems are used for finding the phrases. The consideration includes in this is, we have to analyse the problem present in the statements.

I like to share this Fun Math Problems with you all through my article.

Types of word problems:

There are many types review word problems present. They are,
1.Word problem for numbers
2.Word problem for mixtures
3.Word problem for Age
4.Word problem for time
5.Word problem for linear
Word problem for numbers:
          In word problem for numbers, we  first review the relationships are identified. The variables present in the word problems are reduced.
Word problem for mixtures:
       Different types of concentrations are included in this word problem mixtures.
Word problem for age:
       The relationship of ages are calculated on this word problem.
Word problem for time:
     In this word problem, we review  the various types of train problems are included. Time taken for a train to go and coming back are included in this word problem.
Word problem for linear:
      Linear word problem mostly review the cost problems. For example, the cost for an grapes and the apple is 14.etc..





Algebra is widely used in day to day activities watch out for my forthcoming posts on find the prime factorization of 125 and cbse neet 2013. I am sure they will be helpful.

Example for review word problem:

Question: 1. The difference of twice a number 4 is 8. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 4 equals 8.
Step 2: Next step is to write the above said equation in the equation form.
     2V-4=8.
This is the solution for a word problem.
Question 2. The difference of twice a number 2 is 4. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 2 equals 4.
Step 2: Next step is to write the above said equation in the equation form.
     2V-2=4.
This is the solution for a word problem.
Practice to review word problem:
Practice 1.The difference of twice a number 12 is 14. What is that number?
Practice 2.The difference of twice a number 20 is 40. What is that number?

Factoring With Fractions

Factoring is the process of finding the factor of the given function. It is otherwise called as divide the given function as two separate part. We can factor the function more than two times. Fractions are also used in factoring process. Fraction means it has some numerator and denominator values. Both the numerator and denominator values are different. Now in this article, we see about ac factoring with fractions and their example problems.

Example problems for Ac factoring with fractions

Ac factoring with fractions example problem 1:

Factorize the given polynomial fraction equation `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

Solution:

Given equation is `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 6 on both the sides, we get

x2 - 7x - 120 = 0

Factorize the above equation, we get

x2 - 15x + 8x - 120 = 0

Grouping the first two terms and next two terms, we get

(x2 - 15x) + (8x - 120) = 0

x (x - 15) + 8 (x - 15) = 0

(x - 15) (x + 8) = 0

The factors are (x - 15) and (x + 8)

Answer:

The final answer is (x - 15) and (x + 8)

Ac factoring with fractions example problem 2:

Factorize the given polynomial fraction equation `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

Solution:

Given equation is `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 9 on both the sides, we get

x2 - 12x + 11 = 0

Factorize the above equation, we get

x2 - 11x - x + 11 = 0

Grouping the first two terms and next two terms, we get

(x2 - 11x) + (- x + 11) = 0

x (x - 11) - 1(x - 11) = 0

(x - 11) (x - 1) = 0

The factors are (x - 11) and (x - 1)

Answer:

The final answer is (x - 11) and (x - 1)

Ac factoring with fractions example problem 3:

Factorize the given polynomial fraction equation `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Solution:

Given equation is `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Multiply the given equation by 2 on both the sides, we get

x2 - 11x + 28 = 0

Factorize the above equation, we get

x2 - 7x - 4x + 28 = 0

Grouping the first two terms and next two terms, we get

(x2 - 7x) + (- 4x + 28) = 0

x (x - 7) - 4 (x - 7) = 0

(x - 4) (x - 7) = 0

The factors are (x - 4) and (x - 7)

My forthcoming post is on Rules for Dividing Integers and class 11 cbse books will give you more understanding about Algebra.

Answer:

The final answer is (x - 4) and (x - 7)

Adding Positive Integers

Let us see about adding positive integers. In mathematics the numbers or integers are very essential. The integers are either may be positive or it may be negative. The positive integers are represented as its normal numbers but the negative integers are represented as (-). Adding positive integers is the one of the operation in arithmetic operation. The integers are may be two digits or four digits or it may be more than that.

Definition:

Addition is the process of adding to integers. The positive integers are easy to add. It is also defined as, the group of objects mutually into a larger collection. It is signified by plus sign (+).Process of the addition is one of the easiest numerical tasks.

Examples:

Let us see about examples of positive integers.

Problem 1:

Adding the integers 24 and 35.

Solution:

The given integers are 24 and 35, both integers are positive integers. So, the addition is easy.

24

35  (+)

-------

59

----------.

This is the answer of the given problem.

Problem 2:

Adding then positive integers 12 and 55.

Solution:

The given integers are 12 and 55, both integers are positive integers. So, the addition is easy.

12

55  (+)

-------

67

----------.

This is the answer of the given problem.

Problem 3:

Adding the positive integers 61 and 76.

Solution:

The given integers are 61 and 76, both integers are positive integers. So, the addition is easy.

61

76  (+)

-------

137

----------.

This is the answer of the given problem.

Problem 4:

Adding the positive integers 90 and 40.

Solution:

The given integers are 90 and 40, both integers are positive integers. So, the addition is easy.

90

40  (+)

-------

130

----------.

This is the answer of the given problem.

Problem 5:

Adding the positive integers 35 and 49.

Solution:

The given integers are 35 and 49, both integers are positive integers. So, the addition is easy.

35

49  (+)

-------

84

----------.


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This is the answer of the given problem.

Problem 6:

Adding the positive integers 78 and 42.

Solution:

The given integers are 78 and 42, both integers are positive integers. So, the addition is easy.

78

42  (+)

-------

120

----------.

This is the answer of the given problem.

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c