Substitute A with B

Introduction to substitute a with b:

Substitution is one of the methods used in math to find the value of unknown variable. When this process takes place some of the arithmetic operations like addition, subtraction, multiplication, division can be done. Finally the result will be a finite. Substituting a with b means the equation to find the value of a should have a variable b. Is this topic Linear Equation Solver hard for you? Watch out for my coming posts.

Example Problem – Substitute a with B.

Example 1 – Substitute a with b.

Find the value of a from the expressions 2a+10b = 18 and 5b = 20.

Solution:

The given expressions are 2a+10b = 18 and 5b = 20

To find the value of a it needs to calculate the value for b

5b = 20

Divide by 5 on both sides

b = 4.

Now plug in the value of b in the equation 2a+10b=18

2a+10(4) = 18

2a+40 = 18

Subtract 40 on both sides

2a+40 -40= 18-40

2a = -22

Divide by 2 on both sides

a = -11

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Two more Example Problem – Substitute a with B.

Example 2 – Substitute a with b:

Find the value of a from the expression 10a+4 = 14a+8b and 3b+4 = 10

Solution:

The given expressions are 10a+4 = 14a+8b and 3b+4 = 10

To find the value of a it needs to find the value of b from the equation 3b+4 = 10

Subtract 4 on both sides

3b+4 -4= 10-4

3b = 6

Divide by 3 on both sides

b = 2

Now plug in b = 2 in the equation 10a+4 = 14a+8b

10a+4 = 14a+8(2)

10a+4 = 14a +16

Subtract 14a on both sides

10a+4-14a = 14a +16-14a

-4a+4=16

Subtract 4 on both sides

-4a =12

Divide by -4 on both sides

-a = 3

a = -3.

Example 3 – Substitute a with b:

Find the value of a from the expressions 5a+10 = 12b and b = 17

Solution:

The given expressions are 5a+10 = 12b and b = 17

Now plug in b= 17 in the equation 5a+10 = 12b to find the value of a

5a+10 = 12(17)

5a+10 = 204

Subtract 10 on both sides

5a = 194

Divide by 5 on both sides

a = 38.8

Box and Whisker Graph Example

Introduction to box and whisker graph example:
A box and whisker graph is used to show a set of data so that you can simply see where most of the numbers are. Box plots are helpful in Quality Analysis for interpreting the division of data since it can easily show whether the data is distorted and if there are abnormal observations (outliers) in the dataset. Box plots are also very useful when huge numbers of observations are involved and when two or more datasets are being compared. The box and whisker plot  is a graphical representation of key values from abstract figures.

Methodology for Box and Whisker Plot:
In Box-and-whisker plot, the interior box represents the values from the lesser to higher quartile. The core line represents the median. The straight line extends from the minimum to the maximum value. Box plots show differences between datas without building any assumptions of the underlying statistical allocation. The spacing’s between the different parts of the box help indicate the degree of dispersion (spread) and skewness in the data, and identify outliers. Box plots can be drawn either parallel or perpendicularly. I have recently faced lot of problem while learning Median Statistics, But thank to online resources of math which helped me to learn myself easily on net.

Steps to Solve Box and Whisker Plot:

Step 1:The first step involves arranging the given data in an ascending order

Step 2:The second step involves find the median (Q1) for the entire data.

Step 3:After finding the median ,find the upper quartile and lower quartiles.

Step 4:Find the median for lower quartile (Q2) and upper quartile(Q3).

Step 5:Fiind the Minimum value and Maximum value.

Step 6:Plot these values for Q1,Q2,Q3 in the graph.

Step 7:Draw a box which joins all the median values.

Step 8:Draw a straight line from Minimum to maximum range.

Example:

Draw a  Box and whisker plot for the given data:

7,8,5,3,12,14,6,8,4,15,10

Step1: The first step involves arranging the given data in an ascending order

3,4,5,6,7,8,8,10,12,14,15

Step 2:The second step is to find the median (Q1) for the entire data.

Median Q1=8

Step 3:After finding the median ,find the upper quartile and lower quartiles

Lower quartile =3,4,5,6,7

Upper quartile = 8,10,12,14,15

Step 4:Find the median for lower quartile (Q2) and upper quartile(Q3).

Q1=5

Q­2=12

Step 5: Draw the graph and mark the positions of Q1,Q2 and Q3.

Step 6: Draw the box which joins Q1,Q2, and Q3.

Step 7: And draw the line which joins Minimum and Maximum value.

Probability Distribution Coin Toss

Introduction to probability distribution coin toss:

In probability, a probability distribution used to identify with in a particular interval falling value of a probability. The probability distribution describes the range of the possible value of a random variable. The probability distribution is entirely described by the cumulative distribution function. In this article we shall discuss probability distribution coin toss example problem.

Probability Distribution Coin Toss Example Problem

Example: A coin is tossed for 65times. Find the expectation value of tail with the help of using the binomial distribution.

Solution:

Consider 65 times the coin is tossed, so n = 65

When toss a coin, we get probability of tail is p = 1/2

Probability of getting tail is represent by q

q = 1 – p                        [p + q = 1]

q = 1 – 1/2

q = 1/2

Expectation value:

μ = E[x] = np

= 65(1/2 )

= (65/2)

μ = E[x] = 32.5

Answer:

Expectation value = μ = E[x] = 32.5

Example 2: An unbiased die is rolling for 45 times. By using the binomial distribution find the expectation value of getting four?

Solution:

Let p be probability of getting success

Let q = 1 – p probability of getting failure

Let n be number of times the dice is thrown

An unbiased dice is rolling for 45 times, the value of n = 43

When we throe a unbiased dice, probability of getting four is p = 1/6

Probability of not getting four is represent by q

q = 1 – p

= 1 – 1/6

= 6/6– 1/6

= 6-1/6

q = 5/6

Expectation value:

μ = E(x) = np

= 45 (1/6)

= 45/6

= 7.5

μ = E(x) = 7.5

Answer:

Expectation value = μ = E[x] = 7.5

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Probability Distribution Coin Toss Practice Problem

Problem:

A coin is tossed for 42 times. Find the expectation value of tail with the help of using the binomial distribution.

Answer:

Expectation value = 21

Problem:

An unbiased die is rolling for 35 times. By using the binomial distribution find the expectation value of getting four?

Answer:

Expectation value = 5.8

Mixed Probability Distribution

Introduction to mixed probability distribution:

In this article we shall discuss the mixed probability distributions. Probability plays the main part in the mathematics and in all new fields. Probability is used to discover number of incidence of an event out of total outcomes. The probability of certain two random variables X and Y define on the equivalent probability space, the mixed distribution for X and Y defines the probability of events distinct in expressions of together X and Y.

Basic for Mixed Probability Distribution:

We found with a random test among probability measure P on a primary sample space. Now, we will confer two "mixed" cases for the probability distribution of a random variable in this case where the distribution is partly discrete and partly continuous and the case where the variable has both discrete coordinates and continuous coordinates.

Mixed Probability Distributions Consist of Following Cases:

Cumulative distribution case:

The cumulative distribution function for a couple of random variables is distinct in terms of their mixed probability distribution,

`F(x,y)=P(Xlt=x,Ylt=y)` .

Discrete distribution case:

For discrete random variables, the mixed probability distribution is,

`P(X=x and Y=y)=P(Y=y|X=x).P(X=x)` .

`=P(X=x|Y=y).P(Y=y)` .

Since these are probabilities, we have

`sum_(x)` `sum_(y)`` P(X=x and Y=y)=1` .

Continuous distribution case:

Likewise for continuous random variables, the mixed probability distribution function can be denoted as fX,Y(x, y) and this is,

`f_(X.Y)(x,y)=f_(Y|X)(y|x)f_(X)(x)=f_(X|Y)(x|y)f_(Y)(y)` .

Anywhere fY|X(y|x) and fX|Y(x|y) provide the conditional distribution of Y known X = x and of X known Y = y in that order, and fX(x) and fY(y) give the secondary distribution for X and Y correspondingly. Is this topic math 4th grade word problems hard for you? Watch out for my coming posts.

Again, since these are probability distributions, one has

`int_(x)int_(y)f_(X,Y)(x,y)dydx=1` .

Mixed probability distribution case:

Most of time X is continuous but Y is discrete. Intended for case, in a logistic regression, one might wish to predict the probability of a binary outcome Y conditional on the value of a continuously-distributed X. On the other hand, a "mixed probability distribution" can be distinct in together of two ways.

`f_(X,Y)(x,y)=f_(X|Y)(x|y)P(Y=y)` .

`=P(Y=y|X=x)f_(X)(x)` .

Properly, fX,Y(x, y) is the probability mass function of (X, Y) with value to the product measure on the particular supports of X and Y. each of these two decompositions can then be used to pick up the varied cumulative distribution function.

`F_(X,Y)(x,y)=sum_(tlt=y)int_(s=-oo)^x f_(X,Y)(s,t)ds` .

Grouping of arbitrary statistics of discrete and continuous random variables.

Types of Data Distribution

Introduction to types of data distribution:
The data distribution is also said to be a theoretical distribution. There are three types of data distribution given below. Random variables are distributed regarding to probability law corresponding to probability distribution is called theoretical distribution. Here we are going to discuss about types of data distribution with suitable example problem.

Types of Data Distribution:

There are three types data distribution, such as

Binomial distribution
Normal distribution
Poisson distribution

Example Problems – Types of Data Distribution:

Example problem 1 – types of data distribution:

Obtain the given Poisson distribution where, e = 2.718, x = 10  and `lambda` = 7, x = 10

Solution:

Given: `lambda` = 7; x = 10

Formula to find Poisson distribution = `((e^(-lambda))(lambda^(x)))/(x!)`

First we find the value of `e^(-lambda)` .

e-3 = (2.718)-7

= 0.000911

Then you will find` lambda^(x)`

Given: `lambda` = 7; x = 10

Therefore, `lambda^(x) = ` `7^(10)`

= 282475249

Then you will substitute the values in given formula

`((e^(-lambda))(lambda^(x)))/(x!)` ` = (0.000911(282475249))/(10!)`

= `257334.951839/3628800`

= 0.07091

Answer: Poisson distribution = 0.07091

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Example problem 2 – Types of data distribution:

Obtain the value of p. In a Binomial distribution if n = 4 and P(r = 3) = 2P(r = 2).

Solution:

P(X = r) = nCr pr qn−r

P(X = 3) = 4C3 p3q1 and P(X = 2) = 4C2 p2q2

∴ 4C3 p3q1 = 2 (4C2 ) p2q2

4 p3q1 = 2 * 6 * p2q2

∴ 4 p = 12q

p = 3 q

p = 3 (1 − p) ⇒ 4p = 3; p =`3/4.`

Answer: The value of p = `3/4`

Example problem 3 – types of data distribution

Find the normal distribution of P( X `lt` 200) if the normal random variable has mean(m) = 150, standard deviation(`sigma` )  = 30.

Solutions:

Step 1: Given:

Mean (m) = 150

Standard deviation (`sigma` ) = 30

Step 2: Formula:

X `lt`  mean = 0.5 - Z

X `gt` mean = 0.5 + Z

X = mean = 0.5.

Z = `(X-m)/(sigma)`

Step 3: Solve:

Z = `(X-m)/(sigma)`

= `(200 - 150) / 30`

= `50/30`

= 1.67

Step 3: Find Z = 1.67, Refer Z table,

Z = 1.67 = 0.4525

Step 4: Solving:

X value greater than mean

P(X) = 0.5 + 0.4525

= 0.9525

Result: Normal Distribution P( X `lt` 200) = 0.9525

Solve by Substitution Solver

Introduction to solve by substitution solver:

Generally the solver uses either a elimination method, the matrix method, the substitution method, or the graphing method to solve the system of simultaneous equations. One of the easiest methods is solving simultaneous equations by substitution method. I like to share this How to do the Substitution Method with you all through my article.

Solving steps used by the substitution solver:

Step 1: Let us take any one of the equation. From this equation, write one variable in terms of the other variable.

Step 2: Next, substitute this in another one equation to get the single variable equation. Simplify and then find a one variable value.

Step 4: Substitute this variable value in any one of the equation, so we get the other another one variable value.

Example Problem to Solve by Substitution Solver:

Example problem 1:

Solve the following two simultaneous equations by substitution method:

x + y = 2---------Equation (1)

x – y = 4---------Equation (2)

Solution:

Step 1: Let us consider the equation (1)

x + y = 2

Subtract x on both sides of the equation

x + y - x = 2 – x

y=-x + 2------------------------Equation (3)

Step 2: Substitute this value of y in Equation (2). We get

x – y = 4

x – (-x + 2) = 4

x + x – 2 =4

2x – 2 = 4

Add 2 on both sides of the equation

2x – 2 + 2 = 4 + 2

2x = 6

Divide by 2 on both sides of the equation

2x/2 = 6/2

x = 3

Step 3: Plugging this value of x in Equation (3), we get

y=-x + 2

y=-3 + 2

y = -1

So, the solution of the simultaneous equations is (3, -1). Is this topic Derivatives Calculus hard for you? Watch out for my coming posts.

Additional Problem to Solve by Substitution Solver:

Solve the following two simultaneous equations by substitution method:

y = 2x ------------Equation (1)

x + 2y = 5------------Equation (2)

Solution:

Step 1: Let us consider the equation (1)

y = 2x

Step 2: Substitute the value of y in Equation (2). We get

x + 2y = 5

x + 2(2x) = 5

x + 4x = 5

5x = 5

Divide by 5 on both sides of the equation

5x/5 = 5/5

x = 1

Step 3: Plugging this value of x in Equation (1), we get

y= 2x

y = 2(1)

y = 2

So, the solution of the simultaneous equations is (1, 2).

Linear Equation in One Variable Definition

Introduction to linear equation in one variable definition:

The linear equations are the equations which contain one or more variables which contains the order 1. The linear equation is to find the roots for the given linear equations. The linear equation has only the simple terms for their equation. The linear equations of the two variables are calculated by the elimination method, substitution method and the graphing method. In this, first find any one of the values for the linear equations. Using that value find the value for other linear equation. Solve the linear equation is to find the values for the given linear equations. This article has the information about the linear equation in one variable definition.

Examples for Linear Equation in One Variable Definition:

Example 1 to linear equation in one variable definition:

Find the value for the linear equation 12x+ 14 = 566.

Solution:

The given linear equation is 12x+ 14 = 566.

Step 1: 12x+ 14 = 566

Step 2: 12x+14 - 14 = 566 - 14

Step 3: 12x = 552

Step 4: `(12x)/12` = `552/12`

Step 5: x = 46

The value for the linear equation 12x+ 14 = 566 is x = 46.

Example 2 to linear equation in one variable definition:

Find the value for the linear equation 10x+ 24 = 746.

Solution:

The given linear equation is 10x+ 24 = 746.

Step 1: 10x+ 24 = 746

Step 2: 10x+24 - 24 = 746 - 24

Step 3: 10x = 722

Step 4: `(10x)/10` = `722/10`

Step 5: x = 72.2

The value for the linear equation 10x+ 24 = 746 is x = 72.2.

Example 3 to linear equation in one variable definition:

Find the value for the linear equation 18x+ 17 = 852.

Solution:

The given linear equation is 18x+ 17 = 852.

Step 1: 18x+ 17 = 852

Step 2: 18x+17 - 17 = 852 - 17

Step 3: 18x = 835

Step 4: `(18x)/18` =` 835/18`

Step 5: x = 46.39

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Practice Problems for Linear Equation in One Variable Definition:

Find the value for the linear equation 2x+ 7 = 157.
Answer: x = 75

Find the value for the linear equation 5x+ 18 = 248.
Answer: x = 46