P a B Conditional Probability

Conditional Probability Definition:-
If B is any arbitrary event in a sample space and P ( B ) > 0. The probability of A given B is defined by
P ( A / B ) = P ( A n B ) / P ( B ).
This is called conditional probability of A given B and is written as P ( A / B ).

Proof of Conditional Probability
Let A and B be any two events in a random experiment associated with the sample space "S" then
if n ( S ) = number of elements in S
n ( B ) = number of elements in B
n ( A n B ) = number of elements in A n B.
P ( B ) = n ( B ) / n ( S )
P ( A n B ) = n ( A n B ) / n ( S )

Now for the conditional event A after B the favorable outcomes must be one of the sample points "B"
In other words, the sample space for the event A after B is B.
P ( A / B ) = n ( A ) / n ( B )
P ( A / B ) =n ( A n B ) / n ( B )                [ A n B = A ]
P ( A / B ) =  [ P ( A n B ) / n ( S ) ] / [ P ( B ) / n ( S ) ]
P ( A / B ) = P ( A n B ) / P ( B )
 Hence Proved.

Example of Conditional Probability
If the probability that a communication system will have fidelity is 81 and the probability that it will have fidelity and selectivity is 18. What is the probability that a system will high fidelity will also have high selectivity?
Solution:-
Let the event B will be the communication system to have high fidelity
P ( B ) = 81
Let the event to have high fidelity and selectivity is A n B
P ( A n B ) = 18
By the definition of Conditional probability
P (A / B) = P (A n B) / P (B)
Probability to have selectivity if it will have fidelity
P (A / B) = 18 / 81
P (A / B) = 2 / 9  = 0.22.....

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Statistics Answers Explained

In Statistics, a branch of mathematics there are numerous Statistics Problems for which we can calculate Statistics Answers. The following are some of the Statistics math Answers which use stepwise explanation for the given problems.

Consider the scores obtained by 10 students in mathematics test is given as, 55 36 95 73 60 42 25 78 75 62. Calculate and give the statistics Answer for the mean, median, mode and range of the given scores. First we need to arrange the data in the ascending order (preferably). That gives us, 25 36 42 55 60 62 73 75 78 95. The mean of the scores is given by adding the scores and dividing it by the number of scores, which would be, [25+36+42+55+60+62+73+75+78+95]/10 = 601/10=60.1, the median is the middle value in the ordered data set. Here, the data has even number of scores and hence there are two middle values, the median in such case would be the mean of the two middle values, [60+62]/2= 61. The mode is the number that occurs the most number of times, here there is no such score and hence there is no mode for the given data set. Range is the difference of the highest value and the lowest value of the ordered data set, range = 95-25= 70

Find the mean deviation about the mean for the following data: 12,3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0,  5
Let us first find the mean of the (x bar) of the given data
(xbar) = 1/20*summation(i=1to20)[x(i)]= 200/20 = 10
The respective absolute values of the deviations from the mean , modulus[x(i) – xbar]are
2, 7, 8,7, 6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5
Summation[(i=1to20)[modulus(x(i) – xbar)] = 124
So, the mean deviation about the mean would be, M.D.(x bar) = 124/20 =6.2

Given the frequency table of the monthly salaries of 20 employees of a company:
Monthly salary(in dollars)   frequency of number of employees[f(i)]
     3500 5
     4000 8
     4200 5
     4300 2
Find the statistics answer of the mean of the salaries of the 20 people and also calculate the standard deviation
Let x(i) be the i th salary and f(i) be the corresponding frequency
Mean of the grouped data can be calculated as mu= summation [x(i)f(i)]/summation[f(i)] which gives us
= [3500x5 + 4000x8 +4200x5 +4300x2]/5+8+5+2
= [17500+32000+21000+8600]/20
= $3955
Now, let us find the standard deviation of the 20 people which can be calculated by the formula
Sqrt of {[summation [x(i)-mu]^2 times f(i)]/summation[f(i)]}.
= sqrt {[5*(3500-3955)^2+ 8*(4000-3955)^2+5*(4200-3955)^2+2*(4300-3955)^2]/20} on simplification, we get 282, rounded to the nearest unit.

Solving indefinite integrals

The following theorems and standard forms are used for finding indefinite integrals:
Theorem I: If f and g are differentiable on (a,b) and f’(x) = g’(x) for all x belonging to (a,b), then f(x) = g(x) + c, where c is a constant called the arbitrary constant.

Basics of Limit

What is limit?
A number suppose (x) is said to be limit of function at a particular point if for any arbitrary chosen for any positive integer, however small but not zero there exist a corresponding number greater then zero for all value of that particular point. In simple way we can understand limit that the absolute value at that particular point without any regard to sign.

Limits are both right handed limit and left handed limit. For better knowing if any function is at x
=a then, (x) approaches (a) from the right that is(x) is larger than a then the limit of function is known as right hand limit.
Similarly if(x) approaches from the left that is (x) is lower than a then the limit of function is known as a left hand limit.

Laws of limits
As we know about limit. We have basic knowledge of limits. Laws of limits are very simple means we easily understand the laws.

Addition law-if limit is for any function such as addition of two or more function then by using this law limits goes in right hand side individually for each function.
Subtraction law-this is similar as addition law but only difference is that here we have to choose subtract function.
Constant law- if (x) is a constant and limit for any function is existing with constant then constant should me out from the function.
Multiplication law-if limit exit for multiplication of two functions then by using this law limits goes to each function separately.

Division law-this law is similar to multiplication law but here the difference is that the second function should not equal to zero.
Power law-if power is only in function then by using this law power should be in function with limit.

Limit laws calculus
Calculus may be derivatives and integrations. Limit laws calculus means by using the laws of limits we have to sole various derivative function. In this process we have to know all laws of limits because when we solve derivative function may be one law or more than one law is used for same derivative function.  In Limit laws calculus which law should simplify calculus more accurately, for this purpose we must have clear understanding of law.

Limit Laws Examples
There are various limit laws examples such as if we take power of (x) is 2 separately with limit in left hand side. Then by using power limit law we get in right hand side power in both limit as well as (x). This is very simple limit laws example.

Derivatives of inverse trigonometric functions : derivative of arc tan function

The tan function:
The trigonometric function tan which is the acronym of tangent function is defined as the ratio of the opposite side to the adjacent side of the angle of which the tan is to be found. Alternatively, if we have a point P on the unit circle such that the co-ordinates of the point are (x,y), then tan theta of angle subtended at the centre by the arc AP would be y/x. See picture below:

The inverse of tan function:
Inverse of the tan function is also called the arctan function. We know that the tangent function is a periodic function. So it is not a one to one function. However if we suitably restrict the domain, then the tangent function is a one to one function and then its inverse function i.e. arctan can also be defined.
The arctan function is: arctan (y) = {(x,y) | tan x = y}.
Thus, if tan (pi/4) = 1, then we say that arctan(1) = pi/4.

What is the derivative of arctan?
The derivative of arc tan can be calculated as follows:
Let y = arctan x,
Then, x = tan y
So, dx/dy = sec^2 (y)
So, dx/dy = 1 + tan^2(y)
So, dx/dy = 1+x^2
So, dy/dx = 1/(1+x^2)
Thus the derivative of arctan is 1/(1+x^2)

The derivatives of other arc tan function such as derivative of arctan x 2, derivative of arctan 2 ^x, arctan(x^2) etc, can all be found using the chain rule along with the formula for derivative of arc tan. The derivative of arctan function represents the rate of change of the function y with respect to change in x. In other words it is the slope of the tangent at any point x. The graph of the derivative of arc tan can be shown as follows:

In the above picture, the red curve is the graph of arc tan (x) and the blue curve is the graph of derivative of arc tan (x). Note that at the point x=0, the graph of arc tan (x) changes from concave up to concave down. Therefore that is the point of inflection. We see that at that point the graph of the derivative of arc tan (x) has a maximum. Also the graph of the derivative is always positive, therefore arc tan is an always increasing function.

How to Find Quartiles?

What are Quartiles? 
Quartiles are used in descriptive statistics.  It is used to describe one-fourth of the given data. Quartiles refer to the 3 points, which divide the whole data taken for statistics into four equal parts.  Here is an overview about the three points or quartiles:

• The first quartile i.e. the first point will separate the first quarter of the data. It is also called as lower Quartile or 25th percentile.
• The second quartile is the median or the middle point, which divides the whole data into two halves. This can be termed as 50th percentile.
• The third quartile is the point that splits the highest quarter of the data from the data set. This is called as the upper quartile or the 75th percentile.

The data values till Q1 is the first quarter, the data values in between the points Q1   and Q2   is the second quarter, the data values in between the points Q2  and Q3  is the third quarter and the data values after Q3  is the fourth quarter. This is the way the three quartiles divide the whole data set into four quarters.

How to find Quartiles?
Now that you know what a quartile is, the next step is to know how to find quartiles.  For finding quartiles, we have to know where the three points or the quartiles lie on the given data set. In order to find quartile, sort the values in the given data set in the ascending order.

The quartiles or the points can be found out by using the formula: L is equal to y multiplied by (x divided by 100). Here, x refers to the percentile for the first, second and third quartiles respectively and y refers to the number of data values present in the given data set in which the data values are arranged in ascending order. L refers to the location of the quartile.

Examples of Quartiles
Let us consider a class of 20 students whose English marks have to be compared to find the number of students who are below average, average and excellent in English subject.
The marks obtained by the students are 80, 73, 64, 99, 67, 96, 87, 53, 43, 97, 78, 90, 69, 81, 40, 35, 100, 96, 93, and 98.

The above data is arranged in ascending order as: 35, 40, 43, 53, 67, 64, 69, 73, 78, 80, 81, 87, 90, 93, 96, 96, 97, 98, 99, and 100.

As per formula the 25th percentile will be 20 multiplied by 25 divided by 100 which will be 5.  As the value of L25   is 5, the number in the 5th location is the first quartile which is 67.  By similar calculations we get the 50th percentile or L50 i.e. 80 as the second quartile which is the 10th position and the third quartile is 96 which is in the 15th position or L75 in the data set. Based on the three quartiles, the four quarters are identified as highlighted below in red:

35, 40, 43, 53, 67, 64, 69, 73, 78, 80, 81, 87, 90, 93, 96, 96, 97, 98, 99, 100

Students within 67 marks are below average. Students who scored from 67 to 96 are average scorers and students scoring above 96 to 100 are excellent performers.

Know more about the statistics tutor. This article gives basic information about finding quartiles . Next article will try to cover more statistics help topics and its problems and many more. Please share your comments.

Absolute value of an integer

The absolute value of an integer is defined as the numerical value with out its sign. The absolute value of an integer is the distance between the number and zero on a number lineand is not considered which direction from zero the number lies.

Example:
The absolute value of |5| = 5
        |-5| = 5
How to Solve Absolute Value Equations
Solve the absolute value equation can be using the following steps.
1. First absolute value is isolated on any one side of the equation.
2. Then write two equations with out absolute symbol or bars.
3. First equation will fix the expression within the absolute value symbol equal to the given expression on the other side of the equal sign.
4. Second equation will fix the provided expression within the absolute value symbol equal to given expression on the other side of the opposite sign.
5. Finally solve the equation and get the solution.
Examples 1:
|2x-1| + 3 =6
First isolate the above equation
|2x-1| = 6-3
|2x-1| = 3
Then write in to two equations with out absolute symbol.
2x-1 = 3 2x-1 = -3
2x = 3+1 2x = -3+1
X = 2 x = -1
The solutions are {-1, 2}

Give some example to solve the Absolute value problems
Example 1:
|3(x+4)| = 24
3(x+4) =24 3(x+4) =-24
3x = 24-12 3x = -24-12
3x = 12 3x = -36
X = 4 x = -12
The solutions are{-12, 4}
Example 2:
6 / |x+3| = 3
First isolate the equation
3|x+3| = 6
3x+9 = 6 3x+9 = -6
3x = -3 3x = -15
X = -1 x = -5
The solutions are {-5, -1}
How to solve the inequalities with absolute value
To solve the inequalities with absolute value first isolate the equation on one side. Then the inequalities can be spilt in to two equations as positive and negative as per properties of the absolute value. Finally solve the equations to get the solution.
Example 1:
|x-1| = 2
X -1 = 2 x-1 = -2
X = 3 x = -1
The solution is -1 = x = 3
Example 2:
|3x+1| = 2x+3
3x+1 = 2x+3 3x+1 = -(2x+3)
3x-2x  = 3-1 3x + 2x = -3-1
X = 2 x = -4/5
The solution is -4/5 = x = 2
The absolute value inequalities

Here some of the absolute value inequalities can be explained with the absolute value equation and picture of real number line.

For example to graph the solution to |x|<5, the solutions of the absolute value of all the point should be less then 5 unit away from the zero. The solution of the absolute value inequalities are -5 < x < 5.

If |x| > 2, the solutions of the absolute value of all the point should be greater then 2 unit away from the zero.  The solution of the absolute value inequalities are -2 > x >