Continuous Probability Distribution

In statististics if we have the finite number of set and the probability of the set is called Discrete probability distribution. For an example for a cars on a road, deaths by cancer, tosses until a die shows a first 6. If we measure anything it is called as continuous probability distribution function. For an example we can to say the electric voltage, rainfall, and hardness of steel. In both cases we can determined the distribution by the following distribution function

F(x) = P(X ≤ x)

This is the probability that X will assume any value not exceeding x.

Continuous random variables and probability distribution:

Discrete random variables appear in experiments which is having finite set (defectives in a production, days of sun shines in Chicago, customers standing in a line etc.). Continuous random variables is appear in the experiments which we can't count but we can measure (length of screws, voltage in a power etc.). If we are find the the value probability for  the continuous  variable then it is called continuous probability distribution. By the definition of a random variable of X and its for the distribution are of continuous type will be defined as the following integral.

F(x) = `int_-oo^xf(v)dv`

it is called density of the distribution, is non negative, and is continuous perhaps except for finitely many x-values.

The continuous random variables are simpler than the discrete ones with respect to intervals. Indeed  in the continuous case the continuous probability distribution of the four probabilities corresponding to a < X ≤ b, a < X < b, a ≤ X < b, a ≤ X ≤ b with any fixed a and b (> a) are all same.

Sample problem for probability distribution:

Continuous probability distribution problem 1:

Let X have the density function f(x) = .75(1 - x2) if  -1 ≤ x ≤ 1 and zero otherwise. Find the value of distribution function. Find the continuous probability distribution of P(-1/2 ≤ X ≤ 1/2). Find x such that P(X ≤ x) = 0.95.

Solution:

F(x) = `int_-1^x`(1 - v2) dv= 0.5 + 0.75 x2 -.25x3              if  -1 ≤ x ≤ 1,

And F(x) = 1 if x > 1

P(-1/2 ≤ x ≤ 1/2) = F(1/2)  - F(-1/2) =  `int_(-1/2)^(1/2)`(0.5 + 0.75 x2 -.25x3) dx =  68.75%

Because for continuous probability distribution    P(-1/2 ≤ x ≤ 1/2) =    P(-1/2 < x ≤ 1/2)

Continuous probability distribution P(X = x) = F(x) = 0.5 + 0.75x -1.25x3  = 0.95

If we solve this we will get x = 0.73

Domain and Range of Trigonometric Functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant   Since tan ?=sin?/cos?  and sec?=1/cos?

,cos? being in denominator tan? and sec?  is not defined when cos? iszero

that is ? being odd multiple of ?/2 R- (2n + 1)/2| n`in`  Z

that is these are not defined at odd multiple of pi /2

and range of these functions is set of  all real numbers

Domain of cotangent and cosecant  since cot?= cos?/sin? and cosec?=1/sin?

sin? being in the denominator ,cot ? and cosec? is not defined when sin? is zero

that is  ? being multiple of ?

therefore for cot and cosec to be defined multiple of ? are dicarded

and range of these functions is all real numbers except interval  [-1,1]

Trignometric functions domain and range

Domain and Range of Trigonometric Functions




Explanation for domain and range of some trignometric functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant is R- {`pi` (2n + 1)/2| n `in`  Z } that is these are not defined at odd multiple of `pi`/2

and range of these functions is all values except (-1,1) interval

same can be explained  for cosecant and cotangent.

Learn Divisibility Tests

Divisibility tests means that to discover a number is divisible by a particular number (divisor) without perform the division operation. The rules shown below are used to translate a given number into a normally smaller number while preserve divisibility by the divisor. There are divisibility test for numbers in any radix, and they are all dissimilar, we represent rules simply for decimal numbers below. In this article we shall learn about divisibility tests rules with examples.

Divisibility tests rules

To learn a number is divisible by 2:

The number is end with even number (that is 0,2,4,6) the number are divisible by 2.

To learn a number is divisible by 3:

The sum of the digit is multiple of 3. The given number is divisible by 3.

To learn a number is divisible by 4:

The numbers are formed by the final pairs of digits is divisible by 4.

Example:

4672 => 72 ÷ 4 = 18

So 4672 ÷ 4 = 1168

A number is divisible by 8:

The numbers are formed by the final three digits are equally divisible by 8.

Example:

53,104 => 104 ÷ 8 = 13

So 53,104 ÷ 8 = 6638

To learn a number is divisible by 9:

The sum of the digits is a multiple of 9.

Example: 3,726

3,726 => 3 + 7 + 2 + 6 = 18

Since 18 = 9 × 2,

Then 3,726 ÷ 9 = 414

To learn a number is divisible by 6:

The numbers satisfy the rule for 2 and 3; that is, first it must be an even number, then a digit sum is a multiple of 3.

To learn a number is divisible by 12:

The numbers satisfy the rules for 3 and 4; that is, the digit sum is a multiple of 3, and its final digit pair is a multiple of 4.

Divisible tests practice problem

Problem1:

Check whether the given number 4652 is divisible by 4

Answer:

4652 => 52 ÷ 4 = 18

So 4672 ÷ 4 = 1163 therefore the given number is divisible by 4.

Problem2:

Check whether the given number 53112 is divisible by 8

Answer:

53,112 => 112 ÷ 8 = 14

So 53,112 ÷ 8 = 6639 therefore the given number is divisible by 8.

Vertical Stretch and Compression

Vertical stretch and compression mean transforming the graph based on the scale factors. Here we will see how we are performing the stretching and compression of graph using the scale factor. Stretching the graph is nothing but we are transforming the graph away from axis. Compression of the graph means squeezing the given graph towards the axis. We will see some example problems foe vertical stretch and compression of the graphs.
Vertical Stretch:

Vertical stretch is nothing but the stretching the graph away from x – axis. If the given function is f(x) then the vertical stretch of the given function is y = a f(x). Where 0 < a < 1

Example for vertical stretch:

Graph the following function and its vertical stretch. Where f(x) = x2 – 1.6x and the vertical stretch scale factor of the function f (x) is 0.4.

Solution:

Given function f(x) = x2 – 1.6x

We can write the given functions like y = x2 – 1.6x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the given function

We have to plug y = 0

So 0 = x2 – 1.6x

x (x – 1.6) = 0

x = 0 and x = 1.6

So the x intercept point is (0, 0) and (1.6, 0)

For y – intercept of the given function

We have to plug x = 0

So y = (x)2 – 1.6(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical stretch function.

The vertical stretch function is

f(x) = a f (x)

So y = 0.4 (x2 – 1.6x)

If we graph both the function we will get the following graph

vertical stretch - stretch graph
Vertical Compression:

Vertical compression is nothing but the squeezing the graph towards the x – axis. If the given function is f(x) then the vertical compression of the given function is y = af(x). Where a > 1

Example for vertical compression:

Graph the following function and its vertical compression. Where f(x) = x2 – 3.5x and the vertical compression scale factor of the function f (x) is 1.25

Solution:

The given function f(x) = x2 – 3.5x

We can write the given functions like y = x2 – 3.5x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the original function

We have to plug y = 0

So 0 = x2 – 3.5x

x (x – 3.5) = 0

x = 0 and x = 3.5

So the x intercept point is (0, 0) and (3.5, 0)

For y – intercept of the original function

We have to plug x = 0

So y = (x)2 – 3.5(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical compression function.

The vertical compression function is

f(x) = a f (x)

So y = 1.25 (x2 – 3.5x)

Solve Vector Spherical Coordinates

This article is to solve vector spherical coordinates. Solve vector spherical coordinates is nothing but solving problems under spherical coordinate with the help of the tutors. Tutor vista tutor have many highly qualified tutors in online. A quantity which has magnitude and direction is a vector. There are three methods  to describe a vector.  Specific lengths, direction, angles, projections or components are the methods. Below we can see about solve vector spherical coordinates.

Solve Vector Spherical Coordinates

From this system any point as the point of intersection of the spherical surface (radius r = constant) conical surface (theta phitheta and phi by dr, rd theta and dphi .The element sides are dr,rd theta and rsin theta dphi . = constant). A differential volume element is obtained in spherical coordinate. This happens by increasing r value,

The differential length dl = sqrt(dr^2+(rd theta)^2+(rsin theta dphi)^2)

The differential areas ds = dr.rd theta = rdrd theta

= dr.rsin theta dphi = r sin theta dphi dr

= rd theta. rsin theta dphi = r2sin2 theta d theta dphi

The differential volume dv = dr.rd theta.rsin theta dphi

dv = r2sin theta dphi dr.

So spherical co ordinates = (r, theta, phi)
Solve Vector Spherical Coordinates

A vector in cartesian co-ordinate system can be converted to spherical coordinate.

Conversion of  cartesian to spherical system

The Cartesian co-ordinates is (x,y,z). This cartesian coordinate is converted into spherical co-ordinates ( r, theta, phi )

Given                                  Transform

x                                     r = sqrt (x2+y2+z2)

y                                     theta = cos-1(z/(sqrt(x2+y2+z2))) = cos-1(z/r)

z                                    phi = tan-1(y/x)

Problem 1: Convert the cartesian coordinates x = 2, y = 1, z = 3 into spherical co-ordinates.

Given                        Transform

x = 2                           r = sqrt (x2+y2+z2)

= sqrt (4+1+9)

= sqrt (14) = 3.74

y = 1                            cos-1(z/r) =  cos-1(3/sqrt(14)) =  36.7°

z = 3                           phi = tan-1(y/x) tan-1(1/2)26.56°

Spherical co- ordinates are (3.74, 36.7°, 26.56°)

Information for Line Segment

Information for line segment, the division of a line with two end points is called a line segment. Line segment FG which we denoted by the symbol `bar(FG)` .

Note: We shell denote a line segment `bar(FG)` by FG only.

From the above figure, we call it a line segment FG. The points F and G are called end-points of the line segment FG.

We can also name it as line segment FG.

A line segments:

(a) A line segment has a definite length.

(b) A line segment has two end-points
Information for Drawing a Line Segment by Using Ruler:

Example:

Describe the information about to draw a line segment FG of length 11.5 cm?

Solution:

Given:

length of  line segment FG = 11.5 cm.

Steps to draw a line segment:

1. Place the ruler on a drawing paper.

2. Mark a point at the zero of the ruler. Here, we mark point F at the zero of the ruler.

3. Count the divisions in the ruler till we reach the required length. Here, we count the divisions in the ruler till we reach 11.5 cm.

4. Mark the required end point. Here, we mark the end point G.

5. Join points F and G.

information for line segment

Therefore, FG is the required line segment of length 11.5 cm. The required line segment is represented by `bar(FG)`
Information for Line Segment on a Polygon:
Describe the information to find the line segments of the given polygon. The polygon shown below figure,

information for line segment

Solution:

Given:

Polygon Triangle  EFG.

To find the line segments on a polygon:

We know that the line segments are consisting of two end points. Here, the triangle has three end points, such as E, F, and F. In the given polygon, the three end points to form the line segments on a triangle by connecting these end points consecutively, such line segments are EF, FG, and GE. These line segments are represented by `bar(EF)` , `bar(FG)` , and `bar(GE)` . Therefore, the given polygon triangle has three-line segments.

Simple Data Format

Any collection of information that makes a form to giving the required information is called data. Simple data format are used to compare the collection of data. Graphs are helping to analysis the various types of data. For example data analysis is statistics, bar graphs, histogram graphs, pie charts, and line graphs are used to simply analysis the data. In this article, we are going to discuss about simple data format with suitable example problems.
Example Problem for Simple Data Format:

1. Analyze the statistical data using the given data, find the mean, mode, median, and range of the given data.

14, 7, 13, 12, 10, 10, and 11

Solution:

Rearrange the data for ascending order.

7, 10, 10, 11, 12, 13, 14

Mean:

Mean is the sum of data divided by the number of data in the given data.

Mean = `("sum of data")/ ("number of data")`

Mean = `77/7`

= 11

Mode:

Mode is the most common value in given data.

Most common value is 10.

Therefore, mode is 10.

Median:

Median is a middle number else we have two middle value means; we find the average of their two values is median. We need to find it, we should make to arranged in ascending order from the given data.

Median:

In this problem we have a middle number.

Therefore, Median is the middle number 11.

Range:

Range is difference between maximum and least values in the given data.

Range = maximum value – minimum value

= 14 - 7.

Range = 7.
More Example Problem for Simple Data Format:

2. Analyze the data and create histogram the given simple data.

Solution:

histogram

The above histogram analyzed the given data that which one is high range and low range. This is the needed histogram.

Explanation:

Step 1:

First represented, the class intervals in the x-axis with the value 0 to 10, 10 to 20, 20 to 30, 30 to 40, and 40 to 50

Step 2:

Then represented the frequencies in the y-axis, which ranged as 250, 100, 50, 175, and 75

Given frequencies are represents to each area of rectangle in the histogram.

Step 3:

From the table values according to the class intervals and the frequencies are drawn. This is the required histogram for the simple data.