Measures of Central Tendency Simplified

The Mean, Median and Mode are the measures of central tendency. Mean is the average of the values in the given data. Median is the middle term or value in the given data. Mode is the value that occurs most often or most number of times in the given data.

Definition of Mean Median and Mode
Arithmetic Mean can be defined as the sum of the values given in the data divided by the number of values.
Median can be defined as the middle value in a list of data arranged from the smallest to greatest.
Mode can be defined as the value which occurs most often in a given list.
Let us consider the data 6,7,9,6,8,4,3. The Mean of the given data would be (6+7+9+6+8+4+3)/7= 43/7 = 6.14. To find the median we need to arrange the given data in the ascending order and the middle value would be the median. 3,4,6,6,7,8,9. The middle value in the arranged list is 6 and hence the median is 6. Mode is the value that occurs the most number of times, 6 is such value and hence the mode.

To understand the Mean, Median and Mode of the given data let us consider a simple example problem. The responses of nine people as to how many times they visit a grocery store a month were, 8, 10, 9, 5, 8, 4,11, 8, 9.  Here let us first arrange the given data in the ascending order. 4, 5,8,8,8,9,9,10,11. Mean is the average of all the data values given, that would be[sum of the data values/total number of data values]. The sum of the data values is (4+5+8+8+8+9+9+10+11=72) the total number of data values is 9. So, the Mean = 72/9 = 8.  The middle value of an arranged data list is the Median. Here we have odd number of values and hence the middle value in the list, 8 is the median.
Mode is the value that occurs the most number of times in a list, 8 is the value which is occurring most number of times and hence the mode

What Does Mean Median and Mode Mean
Mean is sum of the data values divided by the number of data values. It is most useful when the data set has no outliers. Median is the middle value in a sorted list (If even numbered, 2 middle values-average them). It is most useful when the data set has outliers and there are no big gaps in the middle of the data. Mode is the data value(s) that occurs most often in a set of data. It is most useful when the data set has many identical values.

Percent Deviation

Percent deviation problems bring the fact that the average of a set of measurements, such as weight, does not necessarily reflect the fact none of the data in the sample will be at the calculated average.
The percent deviation gives a number that indicates where the majority of measurements are. Percent deviation is most useful  in statistics and chemistry problems. It finds how far a measurement, on average, will deviate from the  mean value. This problem gives a set of measured  numerical data.
Deviation or Error Percent.

Percent deviation :
In order to find percent deviation first of all you need to find the Average deviation and Mean. After finding the two things you divide the average deviation into the mean then multiply by 100% . To Find the average deviation you need to subtract the mean from a measured value.  Mean value can be measured as sum the data values and then divide that number by the number of data values.
percent deviation = (Average Deviation/Mean) x 100

The following steps are required to find the percent deviation:
ex:  4, 5,9
Step 1: find the mean value of the measurements. Sum the data values and then divide that number by the number of data values
Mean =  (4+5+9)/3  = 18/3  = 6;
Step 2: Find the Average Deviation. Find the deviation of each  Data value, sum them and then divide by the number of measurements. The Deviation is calculated like as absolute values of subtraction of the each value from the mean.
Deviation = ( |4-6|+|5-6|+|9-6|);
= ( 2+1+3)
= 6
Average Deviation = 6/3 = 2  
Average Deviation = 2
Step 3: By using the Formula of percent deviation substitute the Average deviation and Mean then we get Percent deviation
Percent Deviation =  (Average Deviation/Mean) x 100
=  (2/6) x 100
= 33.33%

Exercise on Parent Deviation
1) A student  measured the length of his  desk as  81cm.Later he found  that it was actually 75cm long. What was his percent deviation ?
2) A student the volume of  a piece of rock to be 25 grams and the accepted value is 19 grams. What is the deviation in percentage?

Introduction to solve by elimination method

Introduction to solve by elimination method

This is one of the methods used to calculate the unknowns involved in different simultaneous equations. Here we need to have equations equal to the number of unknowns. Now, let us discuss the elimination method by solving few simultaneous equations of two unknowns.
Example to Solve by Elimination Method

Ex 1: Solve the following simultaneous equations using elimination method

          3x + 2y = 18

          4x – y = 2

Sol:  Here let us call the equations as 1 and 2.

         3x + 2y = 18 ---------`|->` (1)

         4x – y = 2------------`|->` (2)

(2) × 2  `rArr` 8x – 2y = 4

(+) (1)  `rArr` 3x + 2y = 18

         `rArr` 11x      = 22

          `rArr`   x = `(22)/(11)` = 2

                   x = 2

Now plug in x = 2 in the equation number (1) as follows:

3x + 2y = 10   `rArr` 3(2) + 2y = 18

                      `rArr` 2y = 18 - 6 = 12

                      `rArr` y = `(12)/(2)` = 6.

                      `rArr` y = 6

Therefore, the solution of the given equations are x = 2, y = 6.
Example to Solve by Elimination Method

Ex 2: Solve the following simultaneous equations using elimination method

          3x + 2y = 23

           x – y = 1

Sol:  Here let us call the equations as 1 and 2.

         3x + 2y = 23 ---------à (1)

          x – y = 1------------à (2)

(2) × 2  `rArr` 2x – 2y = 2

(+) (1)  `rArr` 3x + 2y = 23

         `rArr` 5x   = 25

         `rArr` x = `(25)/(5)` = 5

                x = 5

Now plug in x = 5 in the equation number (1) as follows:

3x + 2y = 23 `rArr` 3(5) + 2y = 23

                   `rArr` 2y = 23 -15 = 8

                   `rArr`  y = `(8)/(2)` = 4

                   `rArr` y = 4.

Therefore, the solution of the given equations are x = 5, y = 4.
Practice Problems to Solve by Elimination Method

Solve the following simultaneous equations using elimination method:

(i) 4x + 3y = 10

        x + y = 3

Answer: x = 1, y = 2.

(i) 5x + 2y = 16

    3x + 4y = 18

Answer: x = 2, y = 3.

Introduction of passport to algebra and geometry

Introduction of passport to algebra and geometry:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Geometry is one of the oldest sciences. Initially a body of practical knowledge concerning lengths, areas, and volumes, in the 3rd century BC geometry was put into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a standard for many centuries to follow.

Concepts of Passport to Algebra and Geometry:

Algebra: The following are some of the concepts of passport to algebra.

Number theory : The number theory has a natural numbers and whole numbers which can show below. The counting of numbers 1, 2, 3, 4, 5…is known as natural numbers and the whole numbers have o along with the natural numbers.

Rational numbers : The rational number can be defined as the integer of values is not equal to 0 which is in the form m / n, where m, n is the integers and n ≠ 0. The rational number is said to be in a proper form.

Linear equation: It is an algebraic equation which has the term of constant and product of a constant in single variable. The values are not fixed it can be represented as p, q and r.

Geometry: The following are some of the concepts of passport to geometry.

Radius: A radius is the distance from center of a circle to any point on that circle's circumference.

Perimeter :  The perimeter is a path which can be surrounds an area and also used for path or length.

Circumference: A circumference is the distance around a circle.

Example of Passport to Algebra and Geometry Answers:

Ex 1: We can be obtained the natural number by subtracting of 1 from in the given numbers as shown.

Predecessor of 13 = 13 – 1 = 12

Predecessor of 43 = 43 – 1 = 42

Sol:  We can be obtained the natural number by adding of 1 from in the given numbers as shown.

Successor of 97 = 97 + 1 = 99

Successor of 18 = 18 + 1 = 19

EX 2:   Determine rational number represented as 65.0.

Sol :   Let x = 65.0. Then x = 0.656565…

 100 x = 65.656565…

 100 x − x = (65.656565…) − (0.656565…)

                = 65.0000…

 99x = 65

 x = 65 / 99

    = 13 / 33

Ex 3:  Solve the linear equation for a variable:      x + 7 = 19

Sol :  Take away 7 from the both sides using the property 1

x + 7 - 7 = 19 - 7

x = 12

Ex 4: Solve: y - 3 = 16

Sol :  Add 3 to both sides using property 2

y - 3 + 3 = 16 + 3

y = 19

Q 5 : Find the area and perimeter of rectangle with length 10cm, width 2 cm.

Sol :  Area of rectangle = Length x width

                            = 10 * 2

                            = 20cm^2

Perimeter of rectangle = 2 (Length + width)

                                   = 2 (10 + 2)

                      Answer = 24cm

Q 6:  Find the area and circumference of the circle when the radius is 5cm.
 

Sol : Circle Area= (r= 5) (Pi=3.14 constant)

                                  = 3.14 * 5 * 5

                                  = 78.5cm^2

       Circumference = 2* pi * r

                                 = 2 * 3.14 * 5

                  Answer = 31.4 cm

Permutations and combinations probability

Math Permutations and Combinations We have learnt some techniques of counting the objects so that we need not count them one by one. Such techniques have been of interest since thousands of years. We have learnt to solve several counting problems through the fundamental principles of counting and the concepts of permutations and combinations.

Permutation is used when we are interested in different arrangements of the given objects while combination is used when we are interested only in selection of the given objects and it matters little which object is selected first. Note that ab and ba are two different permutations but the same combination. Let us now define permutations and combinations.

Permutation: The word permutation stands for arrangement. An arrangement that can be made with a given number of distinct objects by taking some or all of them is called a permutation.
The notation nPr or P(n, r) stands for permutation of n objects taking r at a time, i.e., arrangements of r objects out of n objects. nPr makes sense only when n > 0, r = 0 and n = r.

Combination: Sometimes we are not interested in arrangement butt only in selection. For example, out of the three persons a, b, c two can be presented in 6 ways; ab, ba, ac, ca, bc, cb. However there will be only 3 combinations: ab, ac, bc. Each of the different groups or selections which can be made by taking some or all of a number of things (irrespective of order) is called a combination.

Calculating permutations and combinations: We will use nPr = n!/(n-r)! for calculating permutations and we use nCr = nPr / r! for calculating combinations . Let us take an example for calculating permutations.
Example: Evaluate 7P3.7P3 = 7!/(7 – 3)! = 7!/4! = (7 × 6 × 5 × 4!)/4! = 210.and if we will Evaluate 5C2  to solve  5C2 = 5! / (2! (5 – 2)!) = 5! / (2! 3!) = (5 × 4 × 3!) / (2 × 3!) = 5 × 2 = 10.

Let us take some examples to understand permutation and combination probability.
Example: One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn.
Find the probability that the card drawn is an ace. Here we will use combination as out of 52 cards, one card can be drawn in 52 C 1 ways.

So, total number of elementary events = 52 C 1 = 52.
There are four aces in a pack of 52 cards, out of which one ace can be drawn in 4 C 1 ways.
So, favourable number of elementary events = 4 C 1 = 4.
So, required probability = 4/52 = 1/13.

Statistics Problems with solutions

Following are some of the few Statistics Problems from the wide ranging problems in statistics
The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does it signify?

Number of wickets 20-60 60-100  100-150    150-250   250-350     350-450
Number of bowlers    7              5             16                 12             2                3

Solution: the class size varies and the x(i)’s are large. We shall apply the step deviation method with a=200 and h=20. Let us tabulate the values

No. of wickets Number of x(i) d(i)=x(i)-200 u(i)=d(i)/20 u(i)f(i)
   Taken bowlers
   20-60 7 40         -160     -8    -56
   60-100 5 80       -120     -6    -30
   100-150       16         125        -75             -3.75             -60
   150-250       12         200 0       0        0
   250-350        2 300 100               5      10
   350-450 3 400 200      10      30
   ------------------------------------------------------------------------------
   Total       45    -106

So, u(bar) = -106/45 and hence, x(bar) = 200 + 20[-106/45]= 200 – 47.11 = 152.89
This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89

Let us solve Statistics problems involving confidence interval; A sample of 16 students is taken.  The average age in the sample was 22 years with the standard deviation of 6 years. Construct a 95% confidence interval for the average age of the population

Solution: The formula to construct a 95% confidence interval for the mean population can be given as,
  C.I. = [{Xn(bar)+z(alpha/2) sigma/sqrt(n)} , { Xn(bar)-z(alpha/2) sigma/sqrt(n)}]
Where Xn(bar)=22 is the sample mean, sigma=6 is the standard deviation given, n=16 is the sample size, z(alpha/2) is the cutoff point to the standard normal deviation. Since we want a 95% confidence interval , we consider alpha =0.05 and with the help of the normal distribution table we can arrive to z(alpha/2) = 1.96. Plugging in all the data we have in the formula, we get
        C.I = {[22+1.96(6/sqrt(16)], [22-1.96(6/sqrt(16)]= (24.94, 19.06) is the required confidence interval

Statistic problem to find the variance and variation for the data given, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
In this problem, the mean is calculated by step-deviation method taking 14 as the assumed mean. The number of observations n=10. Here the x(i) values are 6,8,10,12,14,16,18,20,22,24; d(i) values we get are, -4, -3,-2,-1, 0, 1,2,3,4,5 and summation of d(i) is 5. The deviation of mean [x(i)- x(bar)] values are -9, - 7,-5,-3, -1, 1, 3,5, 7, 9 and finally the [x(i)- x(bar)] are 81, 49, 25, 9, 1, 1, 9, 25, 49,81, summation of which is 330. Mean is calculated using the formula,
Mean x(bar) = assumed mean + [summation(i=1ton) d(i) *h]/n
       = 14 + (5/10)x 2 = 15
                      Variance (sigma^2) = 1/n summation (i=1to10)[x(i)-x(bar)]^2 = (1/10) x 330 = 33
Standard deviation = sqrt(sigma) = sqrt(33) = 5.74

Introduction toVector Calculus

The vector calculuswas developed by Willard Gibbs and Oliver Heaviside from quaternion analysis. In 1901, most of the notation and terminology used in multivariable calculus was established by Gibbs and Edwin Bidwell Wilson in book called vector analysis.

Vector Calculus
Vector or multivariable calculus is primarily expressed in three dimensional Euclidean space R^3, and is the branch of mathematics that is concerned with differentiation and integration of vector field. The word multivariate calculus is used as a synonym for the broader subject of multivariable calculus. This multivariate calculus includes multivariable calculus as well as partial differentiation and multiple integration.
In differential geometry and in the study of partial differentiation equation, multivariable calculus plays a very important role. It is also used in the description of the electromagnetic field, fluid flow and gravitation field, extensively used in physics and engineering.
In vector or multivariate calculus, the basic objects are scalar field and vector field.

Algebraic operation
In multivariable calculus, the basic algebraic operations are referred to as vector algebraic being defined for a vector space and the globules applied to a vector field, and consist of scalar multiplication, vector addition, dot product, cross product, scalar triple product and vector triple product.

What is Vector Calculus?
Vector calculus is used to measure all the three dimensions of the space. For example to measure the variation of temperature, fluid velocity, fluid force, magnetic flux, etc., multivariable calculus gives the necessary mathematical notation and technique.

Vector fields
A vector field is a vector valued function. Vector field ?(x, y, z) is defined by a function that takes a vector and returns a vector. Example for vector field is the special variation of fluid is the spatial variation of fluid velocity ?(x, y, z) in a steady flow.

Vector field is an assignment of a vector to each point in a subset of Euclidean space. Vector field in a plane can be visualized as a collation of arrows of varying length or thickness with a given magnitude, and direction and each arrow attached to a point in the plane. Often, vector fields are used to model, for example, the direction and speed of a moving fluid through out space, or the direction and strength of some fields, like magnetic or gravitational force, as it change from point to point. Vector fields are generates the flow and vice versa.

Vector Calculus Solutions