Solving Sampling Distributions Calculator

Introduction to solving sampling distributions calculator:

In statistics, sampling distribution is based on a random sample size of n. The sampling distribution is also called as finite sample distribution. The sampling distributions depend upon the distribution of population and the sample size used. With the help of the sampling distribution calculator we can easily solving the sampling distributions problems.

Solving Sampling Distributions Calculator Example Problem:

Properties of sampling distribution:

The mean value of the sampling distribution = mean of the sample population
standard deviation of the sampling distribution = standard deviation of the sampled population
square root of the sample size
The distribution of the populations normal, then the sampling distributions is `barX`
Example1 – solving sampling distributions calculator:

A lamp manufacture claim that the lifespan of the lamp mean value of the lamp is 56 months and the standard deviation is 6 month. The consumer group 50 of them test, what probability that it find mean lifetime of less than 54 month.

Solution:

We seeking the value of P(`barX` <54 br="br">
`barX` is approximately normally distributed by the Central Limit Theorem and

The mean value is µ = 56 months and

The standard deviation of  `sigma``barX` = 6/`sqrt50` = 0.85 months to find the required probability need to convert the z- scores.

Z = `barX` - µ`barx` /`sigma` `barX`

Z = `(54- 56)/(0.85)` = -2.35

We need to use the table to find the value of P(Z ≤-2.35).

0.5 - P(0≤ Z ≤ 2.35) = 0.5 - 0.4906 = 0.0094.

The probability is 0.0094, or 0.94%

Please express your views of this topic Empirical Probability by commenting on blog.

Example 2– Solving Sampling Distributions Calculator:

A lamp manufacture claim that the lifespan of the lamp mean value of the lamp is 58 months and the standard deviation is 6 month. The consumer group 50 of them test, what probability that it find mean lifetime of less than 56 month.

Solution:

We seeking the value of P( `barX` ≤ 56),

`barX ` is approximately normally distributed by the Central Limit Theorem and

The mean value is µ = 58 months and

The standard deviation of  = 6/ `sqrt50` = 0.85 months to find the required probability need to convert the z- scores.

Z=`barX` µ `barx` / `sigma` `barx`

Z = `(56 - 58) /( 0.85) ` = -2.35

We need to use the table to find the value of P(Z ≤-2.35).

0.5 - P(0≤ Z ≤ 2.35) = 0.5 - 0.4906 = 0.0094.

The probability is 0.0094, or 0.94%

Lagrange's Mean Value Theorem

Introduction to Lagrange’s Mean Value Theorem:

In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (differentiable) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.

(Source – Wikipedia)

Lagrange’s Mean Value Theorem – Proof:

The followings statements are the proof for the Lagrange’s Mean Value Theorem.

If a function f(x) is derivable in the interval [a, b], then, there exists at least one value ‘c’ of x lying within [a, b] such that `(f(b)-f(a))/(b-a)` = f’(c).

Proof:

To prove the theorem, we defined a new function `phi` (x) involving f(x), which is designed so as to satisfy the condition of Rolle’s Theorem.

Let `phi` (x) = f(x) + A x where A is a constant to be determined such that `phi` (a) = `phi` (b).

Thus, f(a) + Aa = f(b) + Ab

Therefore, A = `(-f(b)-f(a))/(b-a)`

Now, f(x) is derivable in [a, b]. Also x is derivable and A is a constant. Therefore, `phi` (x) is derivable in [a, b] and its derivative is f’(x) + A.

Thus, `phi` (x) satisfies all the conditions of Rolle’s Theorem.

There is, therefore, at least one value ‘c’ of x, lying within (a, b) such that `phi` ’(c) = 0

Therefore, 0 = `phi` ’(c) = f’(c) + A, that is –A = f’(c)

Or

` (-f(b)-f(a))/(b-a)` = -f’(c)

Or

`(f(b)-f(b))/(b-a)` = f’(c) →(i)

In certain books Lagrange’s Mean Value Theorem is stated as follows.

If a function f(x) defined in the closed interval [a, b] is such that

(i) f(x) is closed interval in continuous as [a, b], and

(ii) f(x) is derivable in the open interval (a, b), then there exists at least one value ‘c’ of x within the interval [a, b] such that f’(c) = `(f(b)-f(a))/(b-a)` .

This seems to be unnecessary as every finitely derivable is continuous and therefore the definition given by us is quite sufficient. I have recently faced lot of problem while learning define rational number, But thank to online resources of math which helped me to learn myself easily on net.

Hence we proved the Lagrange's Mean Value Theorem.

Lagrange’s Mean Value Theorem – Example Problem:

The followings examples are the proof for the Lagrange’s Mean Value Theorem.

Verify Lagrange’s Mean Value Theorem for the f(x) = 2x – x^2 in the interval [0, 1].

Solution:

The function 2x – x^2, being a polynomial is derivable in [0, 1].

f(x) = 2x – x^2 also f’(x) = 2 – 2x.

By Lagrange’s Mean Value Theorem, there exists c in (0, 1) such that

` (f(1)-f(0))/(1-0)` = f’(c)

Or

f(1) – f(0 = f’(c)

Now f(1) =` 2 * 1` – 12 = 1, f(0) = ` 2 * 0` – 0, f’(c) = 2 – 2c

1 – 0 = 2 – 2c `rArr` c = ` 1/2` which lies in(0, 1).

Hence we can verify Lagrange’s Mean Value Theorem.

Linear

Introduction to linear:

In linear function, the linear models are Maximized or minimized which is used to set a linear constraints.Objective function variable,Set of constraints for linear,set of decision variables for linear are the important components of linear models.Most of the real world problem are leads to linear models components. Most of the real world problem in linear algebra can be approximated by using linear models.

Definition to Linear

Line segment present in the linear algebra are having only one dimension.  This line segment is characterized by using the composition of the dimension present in the vector and linear numbers. The linear segment present in the linear algebra should be narrow.The linear segment also be elongated by using the parallel margins.These paralel margins are present in a linear leaf.

Types of Linear

Intersect
Parallel
Coincide


Intersect

For an intersecting line, have to set a cross line or we can overlap the two lines. For a model draw a two intersecting lines which are placed from a shape named  "X."
Learn more about the intersect lines in the class Geometry section.


Parallel

The electrical system one of the conductor are having positive poles, electrodes present in the conductor,terminals used in the conductor and the negative poles, are joined in an another conductor present in an electrical system. This is also called as multiple effect and this is also opposed to one of the series.
Parallel planes have same direction in all parts.This is also same as the parallel lines.
The line which having all dimensions are of equal length from another line.This is also similar to the parallel line and the parallel plane.
For setting a parallel line, the direction should be confirmed first. The same procedure is followed for the parallel plane also.Parallel lines are also to be confirmed for setting a line.I like to share this differentiation rules with you all through my article.


Coincide

For  occupying  the same place in the space, the line should be placed to one another. For example we have to place the two triangles one another.
All the lines present in the linear are equal and clear means then they are said to be a coincide line.

Solving Quadratic

Introduction for solving quadratic:

The quadratic equation is also called polynomial equation whose degree is  second. The general format for solving quadratic equation is   ax2 + bx + c =0. where x is variable  and a,b and c are constants.

Generally, quadratic equation will give two solutions. These solutions may be either real or complex.The real solutions may be rational or irrational.

Quadratic Formula for Solving:

The roots can be found by completing the square,

x2 + ( b / a ) * x = - c / a

( x + ( b / 2 a )) 2=- c / a + b2 / 4 a2 = ( b2 – 4 a c ) / ( 4 a2)

( x + ( b / 2 a )=  ± v( b2 – 4 a c ) / 2 a

Solving for x then gives

X = ( -b ± v( b2 – 4 a c )) / 2 a

This equation is known as the quadratic formula.

Discriminant for Solving Quadratic Equation

From the above quadratic formula, the expression below the square root is discriminant .

Discriminant=b2 – 4ac

When the discriminant is positive, the two roots will be real and distinct.
When the discriminant is negative, the two roots will be non real and complex.
-(b/2a) + i v(4ac – b2) / 2a and  -(b/2a) - i v(4ac – b2) / 2a

Where i represents imaginary part

When the discriminant is zero, the equation has exactly one real solution.
X= -(b/2a)

I have recently faced lot of problem while learning what is an equivalent fraction, But thank to online resources of math which helped me to learn myself easily on net.

Example for Solving Quadratic Equation:

Example 1:

Solve x2 + x – 4 = 0.

The quadratic formula is

X = ( -b ± v( b2 – 4 a c )) / 2 a

Here, a=1, b=1 and c= -4

substitute the values of  a,b, and c in quadratic equation for solving

X=-1 ± v( 12- 4 * 1 * -4)/ 2 * 1

X=-1± v(1+16)/2

X=-1± v(17)/2

The solution is X=-1± v(17)/2

Example 2:

Solve  2x2 + 3x + 5 = 0.

The quadratic formula is

X = ( -b ± v( b2 – 4 a c )) / 2 a

Here, a=2, b=3 and c= 5

Substitute the values of a,b, and c in quadratic equation for solving

X = ( -3 ± v( 32 – 4 * 2 *  5 )) / 2 * 2

X = ( -3 ± v( 9 – 40)) / 4

X = ( -3 ± v( – 31)) / 4

X = ( -3 ± 5.6) / 4

X = ( -3 - 5.6) / 4 and ( -3 + 5.6) / 4

X = ( - 8.6) / 4 and ( 2.6) / 4

X= - 2.15 and 0.65

Smart Math Calculator

Introduction to smart calculator:

The word Smart Calculator itself signify the quickness the results are displayed.  A calculator as such is a small digital inexpensive device used for performing basic operations of arithmetic.  The idea of numbers necessitated invention of calculators. Primitive calculators were knots in strings, beads threaded onto strings, balls on wires and even triangles with rows of numbers on them. The breakthrough came in the seventeenth century with the invention of the first digital mechanical calculator and in coarse of time improvements were made to the mechanical design.  The creation of first digital electronic calculator was in 1960's.  The power source for the calculator which was manually operated earlier changed to a battery/solar-powered calculators.

Smart and Modern Calculator:

Modern calculators are electrically powered (usually by battery/solar cell and are small in sizes.  By 1980's calculator prices so reduced that calculator was affordable to most.  Children in school were able to learn much more in a shorter amount of time.  Moreover students were able to complete the homework assignment within less time.  Calculator is a universal tool.  It has become necessary for everyone like a cell phone.  The type and size will vary for each person.  Simple 10-key calculator will do for household purpose and elementary schools.   A  scientific calculator will help the middle school student, and a graphing calculator is necessary for high school and colleges.

Uses of Smart Calculator:

Smart calculator is designed to deal with scientific problems, such as those in Physics, Chemistry, and those in advanced mathematics such as trigonometry, exponential, logarithm and hyperbolic functions.  It is a great tool for anyone needing mathematical help.  Can choose between options like colours, fonts, language.  In-dept help is available which explains how to use the functions, buttons and much more. Thus smart calculator is really wonderful . It is the easiest and most convenient calculator in the world.  Simply type down a math expression and observe how the sub-total is calculated immediately after each stroke.  If there is a problem in the expression it is highlighted in red.  A wide range of advanced  features are available in smart calculator.