Study Correlation Coefficient

In this article we will discuss about study correlation coefficient. The correlation coefficient, an idea from statistics calculates of how well trends in the expected values follow trends in past real values. The correlation coefficient is a value between 0 and 1. If there is no association between the predicted values and the definite values the correlation coefficient is 0 or very low.

Formula for study correlation coefficient:

`"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

Where
              N = Total number of values
              X = 1st achieve
              Y = 2nd achieve
            sum XY = Addition of the product of 1st and 2nd achieve
              sum X = Addition of 1st achieve
              sum y = Addition of 2nd achieve
              sum x² = Addition of square 1st achieve
              sum y² = Addition of square 2nd get achieve

Example problems for study correlation coefficient:

Study Correlation coefficient – Example 1:
Calculate the Correlation co-efficient of following table.

X	60	61	62	63	64
Y	3.1	3.6	3.8	4	4.1

Solution 1 for study correlation coefficient:
            Step 1:  Count the number of values.
                         N = 5
            Step 2:  Calculate XY, X², Y²
                         See the below table

X	Y	x*y	x*x = x^2	y*y = y^2
60	3.1	60*3.1=186	60*60=3600	3.1*.31=9.61
61	3.6	61*3.6=219.6	61*61=3721	3.6*3.6=12.96
62	3.8	62*3.8=235.6	62*62=3844	3.8*3.8=14.44
63	4	63*4   =252	63*63=3969	4*4 = 16
64	4.1	64*4.1=262.4	64*64=4096	4.1*4.1=16.81

       Step 3:

       Find sumX, sumy, sumxy, sumx², sumy².

        Sum x = 310
        sum y = 18.6
        sum xy = 1155.6
        sum x² = 19230
        sum y² = 69.82

      Step 4:

       Now, Substitute in the above formula given.

            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

             = `[5(1155.6) - ((310)(18.6)) / ((sqrt([5(19230)-(310)^2][5(69.82)-(18.6)^2]))]]`
              = `5778 - 5766 / sqrt(([96150 - 96100] [349.1-345.96]))`
              = `12 / sqrt(50 x 3.14 )`
              = `12/ sqrt(157)`
            = `12/12.5299`
            = 0.9577
            Answer is 0.9577
Study Correlation coefficient – Example 2:
         Find the Correlation co-efficient of following table

X	Y
50	3
51	3
52	3
53	4
54	4

        


 Solution for study correlation coefficient:
        Step 1: Count the number of values.
            N = 5
        Step 2: Find XY, X², Y²
            See the below table

X value	Y value	x* y	x*x	y*y
70	3	70*3=210	70*70=4900	3*3=9
71	3	71*3=213	71*71=5041	3*3=9
72	3	72*3=216	72*72=5184	3*3=9
73	4	73*4=292	73*73=5329	4*4=16
74	4	74*4=296	74*74=5476	4*4=16

          Step 3: Find sum X, sum y, sum xy, sum x², sum y².
            sum x = 360  
            sum y = 17
            sum xy = 1227
            sum x² = 25930
            sum y² = 59
           Step 4: Now, Substitute in the above formula given.
            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`
              = `[5(1227) - ((360)(17)) / ((sqrt([5(25930)-(360)^2][5(59)-(17)^2]))]]`                
              = `(6135 - 6120)/((sqrt([129650 -129600] [295-289])))`
              = 1`5/sqrt(50 xx6)`
              = `15/sqrt(300) `
              = `15/17.32`
              = 0.8660
              Answer is: 0.8660

Least Common Multiple of 8

In mathematics, the least common multiple of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder. (Source: From Wikipedia).

Least common multiple of two numbers can be found by their multiples. Here we are going to learn how to find the least common multiple of two or more numbers.

Example problems to find the least common multiple of two numbers

Here we will see some example problems to find the least common multiple of two numbers.
Example 1
Find the least common multiple of 6 and 8
Solution
The least common multiple of 6 and 8 can be found by finding the multiples 6 and 8.
The list of multiples of 6 and 8 are given below
The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 72, 78, 84.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.
Here 24 is the lowest common number. So, 24 is the lowest common multiple of 6 and 8.
Example 2
Find the least common multiple of 5 and 8
Solution
The least common multiple of 5 and 8 can be found by finding the multiples 5 and 8.
The list of multiples of 5 and 8 are given below
The multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.

Here 40 is the lowest common number. So, 40 is the lowest common multiple of 5 and 8.
Example 3
Find the least common multiple of 7 and 8
Solution
The least common multiple of 7 and 8 can be found by finding the multiples 7 and 8.
The list of multiples of 7 and 8 are given below
The multiples of 7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.
Here 56 is the lowest common number. So, 56 is the lowest common multiple of 7 and 8.
Example 4
Find the least common multiple of 11 and 8
Solution
The least common multiple of 11 and 8 can be found by finding the multiples 11 and 8.
The list of multiples of 11 and 8 are given below
The multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88 , 99, 110
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88.
Here 88 is the lowest common number. So, 88 is the lowest common multiple of 11 and 8.

Practice problems to find the least common multiple of two numbers

Practice problem 1

Find the least common multiple of 10 and 8
Answer: The least common multiple of 10 and 8 is 40
Practice problem 2
Find the least common multiple of 12 and 8
Answer: The least common multiple of 12 and 8 is 24
Practice problem 3
Find the least common multiple of 15 and 8
Answer: The least common multiple of 15 and 8 is 120

Prime Number

• In mathematics, a prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself. The first twenty-five prime numbers are: Prime Numbers:- 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

• Condition: - If X is the prime number next the factors of the number x is 1 and X. (Source:Wikipedia)                                                                                                                                           

Prime number of 7?

• The prime number 7 has only two divisors 1 and 7.
• Here we are going to discuss about some prime number problems.

Example problems of prime number:

Example: - 1

          Find the number 17 is the prime number or not?

Solution:-

          Given number is 17 we have to find out it is prime number or not.

          The factors of the number 17 is the 1, 17 only.

          In the number 17 have only two factors so it is a prime number.

Example: - 2

          Find the number 331 is prime number or not?

Solution:-

         Given number is 331 we have to find out it is prime number or not.

          The factors of the number 331 is the 1, 331 only.

          In the number 331 have only two factors so it is a prime number.

Example:-3

        Find the number 1259 is prime number or not?

Solution:-

          Given number is 1259 we have to find out it is prime number or not.

          The factors of the number 1259 is the 1, 1259 only.

          In the number 1259 have only two factors so it is a prime number.

Example: - 4

        Find out the given number 8 is prime number or not?

Solution:-

        Given number is 8 we have to find out it is prime number or not.

        The factors of the number 8 are 1, 2, 4, and 8 so it is more than two.

        In the number 8 has two or more factor so it not a prime number.

Example: - 5

        Find the number 300 is prime number or not?

Solution:-

        Given number is 300 we have to find out it is prime number or not.

        The factors of the number 300 are 1, 3, 5, 15, 30 and 300 is more than two.

        In the number 300 have more than two factor hence it is not a prime number.

Ellipse Parabola Hyperbola

An ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve.

A parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point and a corresponding line on the plane, the locus of points in that plane that are equidistant from them is a parabola.

A hyperbola is a curve, specifically a smooth curve that lies in a plane, which can be defined either by its geometric properties or by the kinds of equations for which it is the solution set.(Source: Wikipedia)



Example ellipse problem :

Find the equation of ellipse with its center at the origin and having foci at (`+- 2sqrt(6)`, 0 )and an eccentricity

equal to the  ,`(2sqrt(6))/(7)`

Solution:

The focal points on the X axis is the ellipse is oriented, view A, and the standard form of the equation is ` (x^2)/(a) + (y^2)/(b)`

The center of the origin, the numerators of the fractions on the left is x2 and y2. The problem is find the values of a and b.

The distance from the center to be either of the foci is equal to c. so in this problem

C = `2sqrt(6)`

from the given coordinates of the foci.

The values of a, c, and e (eccentricity) are related by

c=a*e

or

` a = (c)/(e)`

Substitute the values from c and e,

a = `(2sqrt(6))/((2sqrt(6))/(7))`

and

a = 7

a2= 49

Then, using the formula

b = `sqrt(a^2-c^2)`

or

b2 = a2 - c2

and substituting for a2 and c2,

b2 = 49 - (`2sqrt(6)` )2

b2= 49 - (4)(6)

b2= 49 - 24

gives the final required value of

b2 = 25

Then the equation of the ellipse is

`(x^2)/(49) + (y^2)/(25) ` = 1

Example parabola problem :

Reduce the equation y2-6y = 8x – 1 to standard form.

Solution:

Rearrange the equation of the 2nd -degree term and any 1st -degree terms of the same unknown are on the left side. Then group the unknown term appearing only in the 1st degree and all constants on the right:

y2 - 6y = 8x -1


Then complete the square in y:

y2 - 6y + 9 = 8x – 1 + 9

(y - 3)2 = 8x + 8

To get the equation in the form

(y - k)2 = 4a(x-h)

Factor an 8 out of the right side. Thus,

(y - 3)2= 8(x + 1)

is the equation of the parabola with its vertex at (-1,3).

Example hyperbola problem :

Find the equation of the hyperbola with an eccentricity of 3/2, directories x = ± 4/3, and foci at ( ± 3,0).

Solution:

The foci of the X axis at the points are (3,0) and ( - 3,0), so the equation is of the form ` (x^2)/(a^2)+(y^2)/(b^2)` = 1

First, the foci are given as (± 3,0); and since the foci are also the points (± c,0), then c=3

The eccentricity is given and the value of a2 can be determined from the formula

c = a * e

a = c/(e)

a = `(3)/((3)/(2))`

a =` (6)/(3)`

a = 2

a2 = 4

The relationship of hyperbola a, b and c are b2 = c2 – a2

And

b2 = (3)2 – (2)2

b2 = 9 – 4

b2 = 5

When these values are substituted in the equation `(x^2)/(a^2)+(y^2)/(b^2)` = 1 the equation `(x^2)/(4)+(y^2)/(5)` = 1

results and is the equation of the hyperbola.

The directories are given as x = `(4)/(3)` and, since

d = `(a)/(e)` or

a=de

Substituting the values given for d and e results in

a = `(4)/(3)((3)/(2))`

therefore,

a=2

and

a2 = 4

The value of c can be determined by the given equation,

c = ae

and a has been found to equal 2 and e is given as(3)/(2); therefore,

c =` 2((3)/(2))`

With the values of a and c computed, the value of b is found as before and the equation can be written.

Solving Translation

“Transformation” is the term which is used in advanced and intermediate geometry. Transformation of image is defined as the transforming the shape from one position to another position without any change in the angle or dimension of the image. Study of transformations is applicable in high school. Transformations learning are very interactive and interesting. In online, animated study of transformations is available for the following topics.  There are four types of transformations they are,


1) Translation Transformation

2) Reflection Transformation

3) Rotation Transformation and

4) Dilation Transformation.

Study of Translation:

Translation:

Translation is one of the types of transformation. Translation means shifting the image up or down and shifting the image left or right. In translation, the image is translated or moved in a straight line with respect to the original image. The translated image doesn’t vary from original image. The dimension, shape and angle of the translated image are same as the original image. The translated image and the original image will be facing the same direction. Solving the problems involving the translation is easy when the concept of the translation is known very well.

Problems on solving translation:

Solving translation Example 1:

Move the object 2 units to the right and 4 units up.

Solution:

The given problem is translation.

In this problem, the object is moved up 4 units and then 2 units to the right side i.e. towards the y axis.



Solving translation Example 2:

Move the object 4 units to the right and 2 units up.

Solution:

The given problem is translation.

In this problem, the object is moved up 2 units and then 4 units to the right side i.e. towards the y axis.

Solving translation Example 3:

Move the object 3 units to the left and 5 units down.

Solution:

The given problem is translation.

In this problem, the object is moved up 2 units and then 4 units to the right side i.e. towards the y axis.

Probability Distribution Curve

The Normal distribution or Gaussian distribution is the continuous probability distribution that often gives a good description of data that cluster around the mean. The graph of the associated probability density function is a bell-shaped with a peak at the mean and is known as the Gaussian function or bell curve. The shape of normal distribution resembles that of a bell. So it is referred as the "bell curve".

Definition of probability distribution curve:

A continuous random variable X is said to follow a probability distribution curve with parameter μ and σ (or μ and σ2) if the probability function is
Y = [1/σ *   ] * e-(x - μ)2/2σ2
X -μ(μ, σ) denotes that the random variable X follows bell curve distribution  Even we can write the probability distribution curve as X-μ(μ, σ2) symbolically. In this case the parameters are mean and standard deviation with mean μ and standard deviation σ. Where X = normal random variable, μ= mean, σ = standard deviation, π = 3.14159, e = 2.71828.
A probability distribution is evaluated by the probability of random variable X in open interval (-∞, x), which is given by,
F(x) = P [X ≤ x]
“Probability distribution curve is often called as normal distribution or bell curve or Gaussian curve”
Types of probability distribution:

Discrete probability distribution,
Continuous probability distribution.

Discrete probability distribution:

If the cumulative distribution function of a probability distribution increases in jumps then the probability distribution is said to be discrete.
The discrete probability distribution with the probability mass function p is given by,
P [X = x] = p(x).

Continuous probability distribution:
If the cumulative distribution function F(x) = µ (-∞, x) is continuous then the probability distribution function is called as continuous probability distribution function. These are characterized by the probability density function f given by,

x
F(x) = µ (-∞, x) = `int` f (t) dt
-∞

Example problem based on probability distribution curve:

The mean score of 1000 students for an examination is 34 and S.D is 16.  How many candidates can be expected to obtain marks between 30 and 60 assuming the normality of the distribution?

Solution: The given data are μ = 34, σ = 16, N = 1000

We have to find the probability of marks between P (30 < X < 60)

We can use the formula, Z = `(X - mu)/sigma`

X = 30,

Z1 = `(30 - mu)/sigma`

=>`(30 - 34)/16`

Z1 = `(-4)/(16)`

Z1 = – 0.25

Z2 = `(60-34)/16`

= `(26)/(16)`

= 1.625


So the value of z2 = 1.63 (approximate)

P(−0.25 < Z < 1.63) =P(0 < Z< 0.25) + P(0 < Z < 1.63) (due to symmetry)

= 0.0987 + 0.4484 = 0.547

No of students scoring between 30 and 60 is, 0.5471 × 1000 => 547.

Function Intervals

The function intervals are defined as the funcions that are appliable with intervals of the numbers which are applied in the variables that changes the functions.
Open Interval

This interval is common to denote open intervals in mathematics using parenthesis (  ). Thus the open interval discussed above will often be shown as (a, b),
Closed Interval

This type of interval are in the form of  [ ] closed both sides also.  For example consider that , -2< or equal to x < or equal 2.  This is represented by closed interval [2,2].  So open interval represents that the boundary value are not taken into the considerations, whereas closed interval means boundary value are taken into the considerations.

Intervals on the function:

Example 1:

Verify Lagrange’s law of the mean for the function f(x) = x3 on [−2,2]

Solution : f is a polynomial, hence continuous and differentiable on [− 2, 2].
f(2) = 23 = 8 ; f (−2) = (−2)3 = −8
f ′(x) = 3x2 ⇒ f ′(c) = 3c2
By law of the mean there exists an element c ∈ (− 2, 2) such that
f ′(c) = (f(b)−f(a)) / (b−a)
⇒ 3c2 = [ 8 − (−8) ] / 4 = 4
i.e., c2 = 4 / 3 ⇒ c = ± 2 / 3
The required ‘c’ in the law of mean are 2 / 3  and − 2/3  as both lie in [−2,2].

Example 2:

Suppose that the functions f(0) = − 3 and f ′(x) ≤ 5 for all values of x, how large can f(2) possibly be?
Solution :

Since by hypothesis f is differentiable, f is continuous everywhere. We can apply Lagrange’s Law of the mean on the interval [0,2]. There exist atleast one ‘c’∈(0, 2) such that
f(2) − f(0) = f ′(c) ( 2 − 0)
f(2) = f(0) + 2 f ′(c)
= −3 + 2 f ′(c)
Given that f ′(x) ≤ 5 for all x. In particular we know that f ′(c) ≤ 5.
Multiplying both sides of the inequality by 2, we have
2f ′(c) ≤ 10
f(2) = −3 + 2 f ′(c) ≤ −3 + 10 = 7
i.e., the largest possible value of f(2) is 7.

Example 3:

Determine the interval in the which curve f(x) =sinx ,  X  ∈ ( 0,2∏) only it be the concave downward

Solution :

f(x)  = Sinx

f'(x) = cosx

f''(x) = -sinx

In the interval (0,∏)  ,  -sinx <0

so f''(x) < 0 in the interval (0,∏)

∴f(x) = Sinx is concave downward in the interval (0,∏)

Practical applications on the function intervals

Example 1:

A cylindrical hole 4 mm in diameter and 12 mm deep in a metal block is rebored to increase the diameter to 4.12 mm. Estimate the amount of metal removed.
Solution :


The volume of cylindrical hole of radius x mm and depth 12 mm is given by

V = f(x) = 12 πx2
⇒ f ′(c) = 24πc.


To estimate f(2.06) − f(2) :
By law of mean,
f(2.06) − f(2) = 0.06 f ′(c)
= 0.06 (24 πc), 2 < c < 2.06

Take c = 2.01
f(2.06) − f(2) = 0.06 × 24 π × 2.01
= 2.89 π cubic mm.