learning statistics probability problems

The word ‘Statistics’ has been derived from either Latin word ‘Status’ or Italian word ‘Statista’ or German word ‘Statistik’ or French word ‘Statistique’ each of which means a political state. The learning statistics defines that the “mathematical method along with numerical style”. Learning statistics is used for solving or finding the problems of the state. Learning probability supplies necessary information for developmental activities in all the departments.Let us see learning statistics probability problems in this article.


Probability definition:


The probability of an event E is a fraction of times we expect E to occur if we show again the same experiment over and over. The expected probability approaches the theoretical probability problems as the number of trials gets bigger and bigger.  If E is a single outcome s, then it is represented as to P (E) and the probability of the outcome is s, and probability of s and E is represented as P(s) for P (E). The groups of the probabilities of all the outcomes are the probability distribution of the problems.


Learning Statistics probability Example problems:


Question 1:

Find x and y so that the ordered data set has a mean of 25.3 and a median of 27.
9, 12, 17, 22, 25, x, 29, 32, 38, y

Answer:

x = 29, y = 40

Question 2:

Given the data set
62 , 60 , 68 , 75 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101,
find
a) the median,
b) the first quartile,
c) the third quartile,

Answer:

1.median = 75

2.first quartile = 69

3.third quartile = 81

Question 3:

The marks scored by the student are,
72, 66, 72, 96, 44, 90, 50
Find
a) the mean
b) standard deviation

Answer:

1.mean = 70

2.standard deviation = 18.6 (rounded to 1 decimal place)

Learn Quadratic Functions

A function of the form f(x)= ax2+bx+c, where a,b,c are real numbers and not equal to zero. It is a polynomial of degree '2'. so it can be called as '2nd degree polynomial'.

examples are          1)3x2-6x+4 and

2)-4x2+6x-7   etc.,

If  we draw the graph of aquadratic function  we get a 'Parabola'.A parabola is "U" shaped symmetrical curve. It may be upward or down ward depending upon the sign of 'a', the coefficient of x2,where it is positive or negative.The point where it changes its shape is called "Vertex".or 'Turning point'.

Various Foms of Quadratic function:

1)The form f(x)=ax2+bx+c,is called "General form"

2)The formf(x)=a(x-h)2+k, where h,k are the coordinates of the vertex, is called "Standard form"

3)The form f(x)=a(x-r1)(x-r2),where r1,r2  the roots of the quadratic function , is called "factored form"

Quadratic Equation: If a Quadratic function is made to equal to zero , then it is called "Quadratic Equation"  i.e.,ax2+bx+c=0 is called Quadratic equation.

examples are  1)2x2+3x+4=0 and -9x2+4x-6=0 etc.,

Since it is  apolynomial of 2nd degree it possess two solutions, which are called 'Root' of the equation.

Quadratic Formula: The formula to find out the roots of given equation is

x=(-b+sqrt(b2-4ac))/2a and x=(-b-sqrt(b2-4ac))/2a

the quantity under root is called "Discriminate" .It isdenoted by "Delta"or 'D'.

i.e., D = b2-4ac .

Discriminate gives the nature of the roots .


Nature of Roots - learn quadratic functions


1) If the discriminate is positive , then both the roots are real and distinct.

For the quadrztic  equation with integer coefficiens. if the discriminate is perfect square then the roots are rational numbers, in all other cases they are irrational.

2) If the discriminate is zero , roots are equal and real and they are equal to x=-b/2a

3) if the discrimionate is negative,roots are complex numbers and they are conjugate to each other.

To understand the concept observe the following problems.


Model Problems of learning quadratic functions


Find The  roots and  their nature of the following quadratic equations:

1) x2+2x-3=0

sol: Comparing the given equation with the general form ax2+bx+c=0

a=1;b=2;c=-3

now the disciminant D=2*2-4*1*(-3) = 4+12 = 16, positive so the roots arereal and disticnt

they are x1 = -b+sqrt(D) / 2a = -2 + sqrt(16)/2*1 = 1   and

x2 = -b-sqrt(D) / 2a = -2 - sqrt(16) / 2 * 1 = -3

so the roots are real ,distinct and roots are 1,-3

2) -x2+2x-1 = 0

sol:  comparing it with the standard form  ax2+bx+c = 0, we get

a = -1;b = 2;c = -1

discriminant D = 2*2-4*(-1)*(-1)=4-4=0 so the roots real and equal ,given by

x = -b/2a = -2/2*(-1) =1

hence the roots are real equal and given by 1,1

3) -2x2+2x-2 = 0

Sol: Comparing it with the standard form ax2+bx+c=0,we get

a=-2;b=2;c=-2

D = 2*2-4*(-2)*(-2 ) = 4 -16 = -12 negative  so the roots complex conjugates

x1=-b+sqrt(D)/2a = -2 + sqrt(D) / 2(-2) = (1-2isqrt(3)) / 2

x2=-b-sqrt(D)/2a=-2-sqrt(D)/2(-2)=(1+2isqrt(3))/2 .

In this way we find the roots of the Quadratic equations .

The Unit Circle Learning

A unit circle is a circle with a radius of one. Especially in trigonometry, the unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane.The point x, y the unit circle in the first quadrant, they are the lengths of the legs of a right triangle whose hypotenuse has length 1.

x2 + y2 = 1.

(Source: Wikipedia)



Definition for unit circle learning:


A reference value of a quantity used to express other values of the same quantity. Circle with radius unity. Its area is П and circumference 2 П.

It is a plane curve formed by the set of all points of a given fixed distance from a fixed point. The fixed point is called the centre and the fixed distance the radius of the circle. A circle is a locus of a point whose distance from a fixed point is constant.

The diameter of a circle is double of its radius. The circumference is 2 П r and its area is Пr2. The equation of a circle with radius r and centre at origin in Cartesian coordinate is x2 + y2.= r2




Explanation and application for unit circle learning:


Explanation for unit circle learning:

1 Length of circumference:

The circumference length is related to the radius (r) by,

C = 2 `pi`r

Diameter d = 2r

`pi` = Circumference of a circle / diameter of the circle

`pi` = 3.14

C =  `pi` d

2. Area enclosed the circle:

Area of the circle =  `pi` r2

Area of the circle interms of diameter

A =  `pi` (d / 2)2          (d – diameter, r – radius)

Area =  `pi` d2 / 4         ( `pi`– 3.14)


Application of the Unit Circle learning:

The most important function of unit circle is the implementation of trigonometric ratios over the unit circle.
And it shows the relative of unit circle and the trigonometric ratios.
It is used to understand the relationship between the unit circle and the trigonometric (sine and cosine) functions.
Because of this explanation it is use to solve a real-world application.

Division Solver

I think, you are involved in the math division of numbers, so you need to do undertaking has of finding the quotient and the remainder in the division of two positive numbers. Do most of “quotients, remainders” and you bring to aptitude an image like this:

                                       6        

                          200 | 1220

                                   1200   

                                       20

 

In this example, we are solver dividing 1220 by 200 and we are find the quotient to be 6 and the remainder to be 20 using dividing 1220 by 200. The division problem has the dividend and the divisor as its given value; in the example, these are, in that order, 1220 and 200. Results have of the quotient and the remainder, which, in the example, are 6 and 20, in that order.

Think about the solver division of number are 27 % 5= 5 remainder 2. It may also be written as

                                   `27/5`=2+(`25/5`)

In solver division using algorithm, 27 is called the dividend, 5 the divisor, 5 the quotient, and 2 the remainder. The relation between these four quantities can also be representation as 27=5x5+2.

In fact, given any integers g and h (h not equal to 0) there exist only one of its kind integers q and r, where 0≤ r<h, such that                            g=hq+r

This solver division algorithm is referred to as the math division decimals or the division independence.

If you require to do divide g by h (b not equal to 0), we get `g/h`=q=`r/h`, which makes the solver  division algorithm almost clear.

When g is divided by h we get a quotient q and a remainder r, 0 ≤r<h, and this cannot be ended in more than one way (q and r are only one of its kind).

Division solver need to follow the step:

     Given: solver division positive numbers g and h.        
  
     Result: non-negative numbers q and r such that g=hq+r, and 0≤r<h.

Steps you need to do for division solver:

        Step 1: Begin by set Q=0 and R=g.

       Step 2: if R<h, write down, “The quotient is Q and the remainder is R” and stop; or else, go to do step 3.

       Step 3: If R≥h, take off b from R, boost Q by 1, and go back to step 2.

Problems on Division Solver:

Problem 1:- Division problem solver 23 by 68

Solution:-

In following steps for how to do divide 23 by 68 

Step 1:-
        -------
    23| 68

In the above equation 23 is divisor and 68 is dividend. In the divisor has two decimal numbers put the value dividend of 23.

Step 2:-
         2
       -------
   23| 68
        46
      ---------
        22
In 23 x 2 = 46 the divisor number 23 is multiplied with 2 to get an answer 46. In 46 is less than from 68.So use the value then subtract the value and get 22. 

Step 3:-
         2.9
        -------
    23| 68
         46
       ---------
         220
         207
      ---------
          13
The value 22 has no more value in the right side. So put 0 to get 220 and put decimal point on the quotient. Then normal dividing 23x 9 = 207 the divisor number 23 is multiplied with 9 to get an answer 207. In 207 is less than from 220.So use the value then subtract the value we get 13 

Step 4:-
         2.95
        -------
    23| 68
         46
       ---------
        220
        207
      ---------
         130
         115
      ---------
           15 (continued)

The value 13 has no more value in the right side. So put 0 to get 130 and put decimal point on the quotient. Then normal dividing 23x 5 = 115 the divisor number 23 is multiplied with 5 to get an answer 115. In 115 is less than from 130.So use the value then subtract the value we get 2.95


Problem 2:- Division problem solver 96 by 4 Solution:-

In following steps for how to divide 96 by 4

Step 1:-
       -------
     4| 96
In the above equation 4 is divisor and 96 is dividend. In the divisor has two decimal numbers put the value dividend of 96.

Step 2:-
       2
     -------
   4| 96
      8
     ---------
      16

In  4 x 2 = 8 the divisor number 4 is multiplied with 2 to get an answer 16. In 8 is less than from 9.So use the value then subtract the value and get 1.

Step 3:-
       24
     -------
   4| 96
      8
    ---------
      16
      16
   ----------
       0
    ----------

In 4 x 4 = 16 the divisor number 4 is multiplied with 4 to get an answer 16. In 16 is equal to 16.So use the value then subtract the value and get remainder is 0.

solving shape area formulas

Area is nothing but a region bounded by a closed curve. Area is a quantity which expresses the 2D size of a defined part of a surface. In terms of differential geometry of surfaces, area is an important variant. In this article solving shape area formulas, we are going to discuss about solving the basic shape area formulas.


Solving shape area formulas types:


Area of a Rectangle.

The Formula for Area of a Rectangle is given below:

     Rectangle shape Area = Length x Breadth   ( A = LB )

Area of a triangle.

The formula for the area of a triangle is given below :

         Triangle shape Area = ½ x Base x Height

                                    A = ½ BH

Area of a Trapezium.

A Trapezium is closed shape which has  two sides that are parallel and two sides that are not parallel.

Now we are going to find a formula for the area of the trapezium.

    Trapezium shape Area = A1 + ( A2 + A3 )

                              Area =  b x h + ½ x (a - b) x h

                               Area = bh + ½ h(a - b)

                               Area = bh + ½ ah – ½ bh

                               Area = ½ ah + ½ bh

                               Area = ½ h ( a + b )

The Area of a Circle.

The area of a circle with the radius r is given by the formula                

                              Area = `pi` r2.


Examples for solving shape area formulas:


Example 1:

Find the area of a triangle with the base length of 20 mm and a height of 6 mm.

Solution:

               Area of a triangle =  ½ b h

                                        =  ½ (20) (6)

                                        =  60 mm2  

Example 2:

Find the area of rectangle given the length is 10 cm and width is 7 cm.

Solution:  

         Area of a Rectangle  =  l * w

                                       = 10 * 7

                                       = 70 cm2

Example 3:

Find the area of a circle given the radius is 44 inches

Solution:

                 Area of  Circle = `pi` r2

                                      =  3.14 (44)2

                                      =  3.14 (1936)

                                      =  6079.04 in2

Coefficient Matrix

Introduction for Matrix:

A matrix is a rectangular array or arrangement of entries or elements displayed in rows and columns put within a square bracket or parenthesis. The entries or elements may be any kind of numbers (real or complex), polynomials or other expressions. Matrices are denoted by the capital letters like A, B, C…

A = [ai, j] m * n

i, j represent the numbers of m, n.

i, j, m, n all represented in the suffix of the considered terms.

A = `[[a11,b12],[c21,d22]]`

order of above matrix 2 * 2

Definition for Coefficient matrix:


The coefficient matrix is formed from the linear equations. There can be any number of linear equations. In the linear equation we are taking the coefficient of the variables in the linear equation to forms the coefficient matrix. Consider  linear equation from it we are going to form a coefficient matrix. The general linear equation as follows:

a11x1 + a12x2 + . . . +a1nxn = b1
a21 x1 + a22x2 + . . . +a2nxn = b2
a31x1 + a32x2 + . . . +a3nxn = b3
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
am1x1 + am2x2 + . . . +amnxn = bn

The coefficient matrix for the linear equation will be represented as:

A = `[[a11,a12,. . .,a1n],[a21,a22,. . .,a2n],[a31,a32,. . .,a3n],[.,.,. . .,.],[am1,am2,. . .,amn]]`

m * n

The coefficient matrix is formed from the linear equations.

Order or size of a coefficient matrix:

The order or size of number of Row Matrix and the number of columns that are present in a matrix.


Example for coefficient matrix:


The coefficient matrix is a resultant from the linear equations.

Linear equation as follows as:

x + 2y +3z = 4
2x + 3y + 4z = 5
3x + 4y +5z = 6

The coefficient equation of the above linear equation as follows:

A = `[[1,2,3],[2,3,4],[3,4,5]]`
3 * 3

The coefficient matrix is formed from the set of linear equations.

Study Online Factors

In online study on factors, it is essential to know the definition of factors.The set of numbers which produce the remainder zero by dividing a particular number. These sets of numbers are called as factors. We can express a number by the multiples of another two numbers called as factors.

For Example,

                   10= 2 x 5

                   32=4 x 8

     Here 2, 5 are the factors of 10 and 4, 8 are the factors of 32.

     Types of Factors and Sample problems:


For the online study on factors, knowing the types of factors is an essential one. The types of factors for online study are as follows:

There are two types of factors,

Prime Factors
Composite factors

Online study on Prime Factors:
     The numbers that can be expressed as the multiple of one and that number itself are called as prime numbers. The factors of prime numbers are called as prime factors.

Consider the following Examples,

3= 1 x 3 (1, 3 are the prime factors of 3)

5= 1 x 5 (1, 5 are the prime factors of 5)

7= 1 x 7 (1, 7 are the prime factors of 7)

Online study on Composite Factors:

     The factors that are not a prime numbers are called as composite factors.

Consider the following example,

24=4 x 6

Here the numbers 4, 6 are not a prime number. Therefore, 4, 6 are called as composite factors of 24.

Example:

Find the factors of 130.
Solution:

130= 1 x  130

130= 2 x  65

130= 5 x  26

130= 10 x 13

So the factors of 130 =1, 2,5,10,13,26,65,130.

Important note: We can express all the numbers except “1” as a product of prime numbers are called as prime factorization.

Consider the following Example,

24 = 4 x 6

     = (2 x 2) x (2 x 3)

24 = 23 x 3

Here 2, 3 are the prime numbers. Hence, we conclude that all the numbers can be expressed as a product of prime numbers.

Prime factorization tree:

     By using the prime factorization tree, we can get all the prime factors of the given number.

Example:

Find the prime Factors of 124
Solution:

                                       124

                                         /  \

                                     2     62

                                             /  \ 

                                          2   31

                            124 = 2 x 2 x 31


Practice Problems on factors:


Following are the practice problems given for online study on factors.

1. Find the factors of 524.

2. Find the Prime factors of 100

3. Find the composite factors of 1056.