Statistics homework answers

Statistics is the branch of applied mathematics which deals with scientific analysis of data. The subject statistics had been started in early days as arithmetic to aid a ruler who needed to know the wealth of his subjects to levy new taxes. Now a days statistics  plays an important role in all organizations in their decision-making and planning. Statistics homework deals with mean,median and mode.

Classification of Statistics Homework Answers:

In statistics homework answers has following topics are

• Arithmetic mean
• Median
• Mode

The arithmetic mean;

In Statistics the arithmetic mean (A.M) or simply the mean or average of n observations x1, x2, …, xn is defined to be number x such that the sum of the deviations of the observations from x is 0.

         x1 +x2 +x3.......xn
x =  --------------------
                     n

the symbol Σ, called sigma notation is used to represent summation.

x=Σ xi
   -------
      n

Homework Example:

Calculate  mean of the data 9, 11, 13, 15, 17, 19.

Sol :

X =   Σxi/n = `[9+11+13+15+17+19]/[6] ` =` [14]/[6]`

Median :

Median is defined as central most or middle value for given series of data it should be arranged in ascending or descending order.

Homework Example:

Find the median of 23, 25, 29, 30, 39.

Sol:

The given values are already in  ascending order. No. of observations N = 5.

So the median = ` (N+1)/(2)`  =`(5+1)/(2)`
= 3 rd term =29
∴ Median = 29.

Mode:

In Statistics answers,Mode is also a measure of central tendency.

In a set of each observations, mode is defined as value which occurs most often.

If the data are arranged in the form of a frequency table, the class corresponding to  greatest frequency is called  modal class.

Homework Example:

Find the mode of 7, 4, 5, 1, 7, 3, 4, 6,7.

Sol:

Arranging the data in  ascending order, we get 1, 3, 4, 4, 5, 6, 7, 7, 7.

In above data 7 occurs several times. Hence mode = 7.

Homework Problems and Answers for Statistics:

Calculate mean :    
Problem 1:
Calculate the mean of the data 7, 12, 18, 14, 19, 20.
Problem 2:
Calculate the mean of the data   16,18,14,15,21,30,26,44
Problem 3: 
Calculate the mean of the data   22,77,55,11,22,26,38,72
Answers:
1) 15
2) 23
3) 40.35

Calculate median:
Problem 1: 
Find the median of 3, 4, 10, 12, 27, 32, 41, 49, 50, 55, 60, 63, 71, 75, 80.
Problem 2:
Find the median of 29, 23, 25, 29, 30, 25, 28.
Problem 3: 
Find the median of 26, 25, 29, 23, 25, 29, 30, 25, 28, 30.
Answers:
1) 49
2) 28
3) 27

Calculate mode:
Problem 1:
Find the mode of 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, 15
Problem 2:
Find the mode of 3,5,8,3,9,3
Problem 3: 
Find the mode of 3,5,8,5,6,7
Answers:
1) 12 and 15
2) 3
3) 5
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An Introduction to Calculating the Area of a Right Triangle

The Area of a Right Triangle
A Right Triangle or a Right Angled Triangle is a triangle which has right angle or 90 degrees as one of its interior angles. We have two types of Right Triangles namely, scalene right triangle in which one angle is right angle and the other two are unequal angles and the corresponding sides are also unequal. The other type is the Isosceles Right Triangle in which one angle is right angle and the other two angles are equal and also their corresponding sides are of equal length.  A right triangle is very useful in trigonometry, a branch of mathematics. Area of a triangle is given by half times the base times the height of the triangle. We can deduce the formula for Area of a Right Triangle using the following method. Let us consider a rectangle of length l cm and width w cm. Let us now cut the rectangle into two equal halves diagonally as shown in the figure below.

 
The triangles ABC and ADC are congruent triangles which means when placed one over the other they overlap exactly. Also they are right triangles with 90 degrees as one of their angles. They are of the same size and hence we can say that they have the equal area. From this it is clear that the area of a right triangle will be half the area of the rectangle.
The Area of the right triangle = ½ [Area of the Rectangle]
               = ½ [length x width]
             = ½ [l w]

From the figure, we can see that the length of the rectangle is one of the sides of the triangle which is called the base of the triangle. So, we have base of the triangle equals length of the rectangle.  Also the distance from the vertex A of the triangle ABC to the vertex B is the height of the triangle ABC which is the width of the rectangle. So, we have height of the triangle equals width of the rectangle. Substituting these in the above equation, Formula for Area of a Right Triangle = ½ [base x height]. So, to calculate area of a right triangle we use the formula ½ times [base x height]
Calculate the Area of a Right Triangle with base and height 12 cm and 17 cm respectively
Solution:  Base of the right triangle   = 12 cm
         Height of the right triangle = 17 cm
       Area of the right triangle   = ½ x base x height
                      = ½ x 12 x 17 = 6 x 17 = 102 cm^2

Characteristics of Right Angle Triangle

Triangle is a basic shape in geometry. A right triangle means 90 - degree angle that is one angle is right angle. It also called as right angled. It shows the relations between the sides and angles of a right triangle and it is the basic concept of trigonometry. The opposite side of the right angle is known as hypotenuse. The adjacent side to the right angle is known as legs or sometime known as catheters. This phenomenon is the basis of Pythagorean triangle theorem. This theorem states that the length of all three sides of a right triangle are integer and the side length are known as Pythagorean triple.

The basic principle properties of a right triangle are area, altitude, Pythagorean Theorem, inradius and circumradius. First we discuss about area of a right triangle. Area of a right triangle is equal to the one half of the base multiplied by the corresponding height. It is true for all right angled triangles. In a right triangle from the two legs if one leg is taken as a base then other leg is taken as height. So in other way we can say that area of a right triangle is one half of the product of the two legs means one half products of base and height. Mathematically suppose in a right triangle two legs are (a) and (b) and hypotenuse is (c) then formula for the area of a right triangle is expressed as [A=(1/2 ab)].

For finding area of a right triangle we have to calculate perimeter also. The formula we can see above. Suppose in a right triangle two legs (a=5) and (b=12) where a is base and b is height. Hypotenuse (c=13) is given. We know the formula area of a right triangle so we substitute all the values in formula. Finally we get area= (1/2 *5*12) = (1/2*60) =30 and unit of right angle triangle is square meter or square centimeters.

Similarly we can find surface area of a right triangle.  Formula for surface area of right angle triangle is [Area (A) =bc/2] where (a=√b^2+c^2), (b=√a^2-c^2) and (c=√a^2-b^2). Suppose for above problem we have to find surface area. We substitute all the values in formula that is (A=12*13/2).
From the right angle triangles we also find the Pythagorean Theorem. It is very useful in all parts of mathematics. It states that in any right angle square of hypotenuse is equal to sum of square of base and square of height that is (c^2=a^2+b^2).

Trigonometric equations

Trigonometric equations
In this article we shall study some basics of how to solve trigonometric equations. A trigonometric equation is an equation that involves trigonometric ratios in addition to algebraic terms. As trigonometric  functions are many to one type of functions, it is quite possible that such trigonometric- equations may have many roots (solutions).

This can be made clear by solving trigonometric equations examples:
For instance, sin ( pi/6) = ½ but the equation sin x = ½ has not only the solution x =  pi/6, but also x = 5 pi/6, x = 2 pi +  pi/6, x = 3 pi -  pi/6 etc. Thus, we can say that x =  pi/6 is a solution of sin x = ½ but it is not the general solution of the equation.

A general solution gives all the roots of an equation. When attempting to solve trigonometric equations online it is very imperative that the equations fed in are valid. For example, and equation like sin x = 2 would not have any solution. That is because we know that the range of the sine function is [-1,1]. So for any value of angle x, the value of sin x can never be = 2. Therefore the solution of such an equation would be : no solution.

The objective should be to develop methods to find general solutions of trigonometric-equations.
We know that sine, cosine and tangent functions are all periodic. Sine and cosine functions have a period of 2 pi and the tangent function has a period of  pi. Therefore the general solutions of the equations, sin t = 0, Cos t = 0 and tan t = 0 can be found as follows:
Sin t = 0 ? t = k pi, k belongs to the set of integers ------------- (1)
Cos t = 0 ? t = (2k+1)/2, k belongs to set of integers ----------- (2)
Tan t = 0 ? t = k pi, k belongs to the set of integers -------------- (3)

These results can be used to solve the general trigonometric-equations which are as follows:
Cos t = a, |a| = 1
Sin t = a, |a| = 1
Tan t = a, a’ belongs to set of all real numbers.
By solving a trigonometric equation we intend to find a set of solutions for the given set of trigonometric-equations such that each member of the solution set satisfies the set of equations.

Learn more about how to solve Trigonometry Problems.

Limit of a function

For understanding the limit of a function we must know the fundamental concepts of calculus and analysis which belonging to that particular function near a define input. In another way we can simplify by taking a simple example. Suppose function is f(x) and limit is x tends to c where c is a constant, then it means that function f(x) is get closer to limit as x get closer to c. more accurately when function is applied to each input , the result in an output value that is close to limit.

Limits of a sequence
As we know about the limits. Limits of a sequence means a value in term of sequences. If limit value exist then such sequence occurs. Limits of a sequence of any function can be understood by applying the function on a real line. This is one way to know the limits of function in term of sequence.

Limits of sequences
For knowing the limits of sequences we choose such type of example where limits value exist then the sequences will convergence. Limits are first apply for the real numbers and then for others such as metric spaces and topological spaces.

Limits of functions
In limits of functions we take any calculus function like additive or subtractive functions, because for simplify purpose these functions are easy. Functions are with limits where in limit x is approaches to any real number. When we solve such function in last we must put the limits. So in limits of functions output result are moves according to limits.

Limit of sequence
In limit of sequence, first take a sequence such as (Xn) with limit where x is tens to a and a is a constant. When we apply limit to this function if and only if for all sequences(Xn) means with (Xn) with all value of n but not equal to constant a.

Limits of a function
Limits of function means limits should apply to a function. Function may be different types. In limits of function we explain that what it means for any function which tends to real number, infinity or minus infinity. Limits of function can be both right handed and left handed.
Suppose in any case if both right hand and left hand limits of  function as x approaches to constant exist and are equal in value, their common value evidently will be the limits of a function. If however either or both of these limits do not exist then the limits of a function does not exist.

Triple integral solver

We know that when solving double integrals, we divide the two dimensional region into very small rectangles. Then we multiply the area of the rectangles with the value of the function at that point. Then we sum up these areas and then apply the following limit: lim (size of rectangle -> 0). Doing all that gives us the double integral of the said function.

Let us now try extending this concept to three dimensions. Just like in double integrals we had some region in a plane (say the xy plane), in triple integral we will consider a solid in space (i.e. xyz space). Just like in double integrals we had split the region into rectangles, in solving triple integrals we break down the solid into numerous rectangular solids (or cuboids). Extending further on same lines, just like how in double integrals we multiplied the value of the function by the area of the respective rectangle, in a triple integral example, we would multiply the value of the function at each of the point by the volume of the rectangular solid at that point. Instead of the limit of size of rectangle tending to zero, in case of triple integrals we have the limit as the volume of rectangle tending to zero.

With that we come to the definition of a triple integral. Which is like this: Triple integral is defined as the limit of the sum of product of volume of rectangular solids with the value of the function.
We call the double integral as an equivalent to double iterated integral. In the same way we can understand the triple integral as a triple iterated integral.

Symbolically the definition of a triple integral can be stated as follows:
Consider a function f(x,y,z). It is of three variables. It is continuous over any solid S. Then the triple integral of this function over the solid S can be symbolically stated as:

where the sum is taken over all the rectangular solids that are contained in the solid S. The limit is for the side length of the rectangles.
The above definition of triple integrals is useful only when we are given a set of data in the form of a table of values of volume and value of function. When a function is defined symbolically, then we use the Fubini’s theorem to solve triple integral examples.

First and Second Fundamental theorem of calculus

First Fundamental Theorem of Calculus: If f is continuous on [a, b], then F(x) = integral a to x f(t) dt has a derivative at every point of [a, b] and dF/dx = d/dx integral a to x f(t) dt = f(x), a is less than and equal to x is less than equal to b. Let us understand Proof of Fundamental Theorem of Calculus: We prove the fundamental theorem of calculus by applying the definition of derivative directly to the function F(x).

This means writing out the difference quotient F(x + h) – F(x) /h and showing that its limits as h approaches to 0 is the number f(x). When we replace F(x + h) and F(x) by their definite integrals, the numerator in above equation becomes F(x + h) – F(x) = integral a to x+h f(t) dt – integral a to x f(t) dt. The additive rule for integrals simplifies the right hand side to Integral x to x+h f(t) dt So that the above equation becomes  F(x + h) – F(x) /h = 1/h [F(x + h) – F(x)] = 1/h integral x to x+h f(t) dt. According to the mean value theorem for definite integrals, the value of the last expression in the above equation is one of the values taken on by f in the interval joining x and x + h. That is for some number c in this interval, 1/h integral x to x + h f(t) dt = f(c).

We can therefore find out what happens to (1/h) times the integral as h approaches to 0 by watching what happens to f(c) as h approaches to 0. As h approaches to 0, the endpoint x + h approaches x, pushing c ahead of it like a bead on a wire. So c approaches x, and since f is continuous at x, f(c) approaches f(x) .Lim h approaches 0 f(c) = f(x).Going back to the beginning, then we have dF/dx = lim h approaches to 0 [F(x+h) – F(x)] /h= lim h approaches 0 1/h integral x to x+h f(t) dt = lim h approaches 0 f(c)= f(x). This concludes the proof.

Let us more understand this through Fundamental Theorem of Calculus problems: let us take few fundamental theorem of calculus examples .suppose we have Find dy/dx if y = integral 1 to x^2 cos t dt.

Now to understand this solution suppose let us notice that the upper limit of integration is not x but x^2. To find dy/dx we must therefore treat y as the composite of y = integral 1 to u cos t dt and u = x^2 and apply the chain rule: dy/dx = (dy/du).(du/dx) = d/du integral 1 to u cos t dt . du/dx = cos u . du/dx= cos x^2 . 2x = 2x cos x^2.

Second Fundamental Theorem of Calculus: If f is continuous at every point of [a, b] and F is any anti-derivative of f on [a, b], then Integral a to b f(X) dx = F(b) – F(a). Let us understand this by second Fundamental Theorem of Calculus Examples suppose we have Evaluate integral 0 to pi cos x dx. Now let us solve this integral 0 to pi cos x dx = sin x 0 to pi = sin pi – sin 0 = 0 – 0 = 0.