Examples of Poisson Distribution

Definition: A random variable X is  a Poisson distribution if the probability mass function of X is P(X = x) =e−λ  λx / x!,                                       x = 0,1,2, …for some λ > 0

The mean of  Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.

The Poisson distribution is a restrictive case of Binomial distribution under the following conditions.
(i)   Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii)  The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may be assumed to follow a Poisson distribution. The example problems of poisson distribution is   given below

Examples of Poisson distribution:

Some Examples of poisson distributions are given below

(1) The number of gamma particles emitted by a radio active source in a given time interval.
(2) The number of phone calls received at a telephone exchange in a given time interval.
(3) The number of defective articles in a packet of 250, produced by a good industries limited.
(4) The number of printing errors at each page of a book by a good publication centre.
(5) The number of road accidents reported in a city at a particular time.

Example Problems for Poission distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page

(i) what is the probability of its 400 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].


Solution :

n = 400 , p = 0.005
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= `sum` e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569

Learning Simple and Compound Events

Probability of any event  is the ratio of several favorable outcomes of an event to the total number of outcomes of an event. Otherwise if we toss a single dice then it is called as a simple event. If we toss a two dice then it is called as a compound event. Inclusive means when two events are happen at a same time. Mutually exclusive means when two events cannot happen at a same time.

Learning Simple and Compound Events Formulas :

learning  simple and compound events formulas in the probability of an event can be expressed as

P(a) = number of a favorable outcomes / total number of a possible outcomes

learning  simple and compound events formulas in the probability of two independent events(A and B) are multiplied by the probability of the first event by the probability of the second event. Two events are said to be independent. if P(A and B) = P(A) P(B). P(A)and P(B) are are non zero

P(A and B) = P(A) . P(B)

learning  simple and compound events formulas in the probability of two dependent events (A and B ) are multiplied by the probability of A and the probability of B after A occurs.

P(A and B) = P(A) . P(B following A)

learning  simple and compound events formulas in the probability of one or other of two mutually exclusive events (A or B) are added to the probability of the first event to the probability of the second event.Two events are said to be disjoint. if and only if P(A and B) = 0

P(A or B) = P(A) + P(B)

learning  simple and compound events formulas in the probability of one or the other of two inclusive events(A or B) are added to the probability of the first event to the probability of the second event and subtract the probability of both events happening.

P(A or B) = P(A) + P(B) - P(A and B).

practice problems in learning simple and compound events

Example 1:Abraham is going to the Super market to pick a new sports bats. Today, the shopkeeper has 25 cricket bats and 50 Tennis bats are available for sales. If Abraham randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 25 / (25 + 50 )

= 25 / 75

= 1 / 3

= 0.33

if Abraham randomly picks a bat (which is a Cricket bat) having a probability  0.33.

Example 2:Lenin is going to the Super market to pick a new sports bats. Today, the shopkeeper has 20 cricket bats and 40 Tennis bats are available for sales. If Lenin randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 20 / (20 + 40 )

= 20 / 60

= 1/ 3

if Lenin randomly picks a bat (which is a Cricket bat) having a probability  = 0.33

Exponential Growth Formula

A function is said to be Exponential growth that including exponential decay when the growth rate of that mathematical function is proportional to the function's current value. In a discrete domain of definition with equal intervals of the function is called as geometric growth or geometric decay. The exponential growth model is also called as the Malthusian growth model.


Exponential growth formula:

Exponential formula defines the  X as exponentially on time t.

X(t) = a . b(t/r)

"a" denotes the initial value

a = x,

X(0) = a,

b= a

It denotes the positive growth of the factor, t = time required

Example for exponential growth formula:

If a power doubles every 5 minutes, initially there’s only one doubles, how many powers would be there after 2 hours?

Here, a= 1, b= 2, t = 5 minutes.

X(t) = a . b(t/r) = 1 . 2{(120 minutes)/(5 minutes)}

X(2 hour) = 1 . 2 24 = 16777216

After two hours, there would be 16777216 powers.


Exponential growth Problem:

A microbiologist is researching a newly-discovered species of fungi. At time t = 0 hours, he puts one hundred fungi into what he has determined to be a favorable growth medium. Six hours later, he measures 200 fungi. Assuming exponential growth, what is the growth constant "i" for the fungi? (Round i to two decimal places.)

For the given problem, the units on time t will be hours, because the growth is being measured in terms of hours. The starting amount P is the amount at time t = 0, so, for this problem, P = 100. The ending amount is A = 200 at t = 6. The only variable we don't have a value for is the growth constant i, which also happens to be what I'm looking for. So I'll plug in all the values we know, and then solve for the growth constant:
A = Peit
200 = 100e6i
2 = e6i
ln(2) = 6i
`ln(2)/6` = i = 0.11552453

Derivative Step by Step

erivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity.-  Source : Wikipedia

Calculus is used to solve differentiation of any subjective equation and the equivalent output result. we need to find f(x) where,     d/dx f(x) = g(x).The derivative of constant is zero in calculus. The linear equations are also compared using calculus. Integration is considered as one of most important study of calculus in mathematics. British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. The methods that are applied in continous graphs, curves or fucntions  is calculus.

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Steps to solve the Derivative function:

Derivative  problem 1:

Find the first derivative and second derivative of

f(x) = x4 + 7x3 - 3x2 - 9x +3

Solution:

Step 1:  First differentiate f(x) = x4 + 7x3 - 3x2 - 9x +3

Step 2:  when we differentiate the first term we get  (1 × 4) x (4-1)

Step 3:  The above equation can above simplified as  4 x 3

Step 4:  Like wise we can differentiate  and simplify all the terms in the equation

Step 5:  Finally after the first derivative we will get  4 x 3 + 21 x2 - 6x - 9

Step 6:  Now we will go for Second derivative
End of the second derivative we will get   12x2 + 42 x - 6


First derivative:

f' = (df )/ (dx)

= (1 × 4) x (4-1) + (7 × 3)x(3-1) - (3 × 2)x(2-1) - (9 × 1)x (1-1)

= 4 x 3 + 21 x2 – 6x - 9

Second derivative:

f '' = (df ' )/ (dx)

= 4 x 3 + 21 x2 – 6x - 9

= 12x2 + 42 x - 6

Answer:

(d^2)/(dx^2)  (x4 + 7x3 - 3x2 - 9x +3) = 12x2 + 42 x - 6

Derivative problem 2:

Find the differential for y = sqrt(2x^3 - 9x)

Solution:

Step 1:  First turn sqrt(2x^3 - 9x)to (2x3 - 9x)1/2
Step 2:  Differentiate in the usual method
Step 3:  Then finally we get,

d/(dx) ((2x3 - 9x)1/2 ) = ½ ( 2x3 - 9x ) ½ -1( 6x2 - 9 )

= ½ ( 2x3 - 9x ) -1/2( 6x2 - 9 )

= ( 6x^2 - 9 ) / (2 sqrt (2x^3 - 9x ))

Answer of given calculus test exam problem is

d/(dx) ((2x3 - 9x)1/2 ) = ( 6x^2 - 9 ) / (2 sqrt (2x^3 - 9x ))

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Derivative practice problem:

Derivative practice problem 1:

Find the derivative of f(x) = ( x - 5 ) ( 2x + 1 )

Answer:   d/(dx) f(x) = 4x - 9

Derivative practice problem 2:

Differentiate the given equation with respect to t. y =  8t3 + 12t

Answer:  24t2   + 12

Division with Remainders Word Problems

Division is an arithmetic operation which is the opposite of multiplication. The symbol used for division is (÷).

Notation:

Division is often represented by inserting fraction bar which is nothing but the horizontal line between divisor and dividend. Dividend is placed in numerator and divisor are placed in denominator.  For example, a divided by b is written as,

`a/b`

Key terms - division with remainders word problems

Dividend:

The number which is being divided in a division is known as dividend.

Divisor:

The number which divides the dividend is known as divisor.

Quotient:

Quotient is the solution / answer for the problem

Remainder:

The remainder is the amount "left over" after the division

Example problems on division with remainders problems

Division problem 1:

What is division value and remainder of the given fraction 13 / 4

Solution:

When we divide 13 by 4 it gives 3 ( 3 * 4 = 12) as a quotient and 1 is the (13 - 12) remainder value

Quotient = 3

Remainder = 1

Division problem 2:

What is division value and remainder of the given fraction 6 / 4

Solution:

6 / 4 =   There is one 4 in 6

Hence Quotient = 1

Remainder = 2

Solving Division with remainders Word problems:

Division with remainders word problem 1:

A 40 in stick is cut into pieces of 7 in length. How many pieces we can get out of this stick? What is the length of the scrap? Solve the word problem.

Solution:

The problem describes how to divide a 40 in stick into pieces of 7 in each. The remaining piece of the stick is called the scrap.

Therefore, the dividend is 40in., the divisor is 7 in. The quotient of this is the number of pieces and the remainder is the length of the scrap. When we divide 40 by 7, there are fivet 7’s in 40 (5*7= 35) and the remainder is 5.

Hence, the number of pieces of 7 in = 5(quotient)

The length of the scrap = 5 (remainder)

Division with remainders word problem 2:

There are 11 apples and 4 kids . We have to give equal number of apples to each kids. How many apples will each kids get and find the amount of apple left over?

Solution:

Here we have to divide number of apples  10 by number of kids  4.  Hence ,

11 / 4 = 2 with remainder 3

Number of apples each kid will get = 2

The apples left over = 3

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Division with remainders word problem 3:

Consider there are 10 boys in a class. Divide them into 3 groups. How many boys each group will consists and how many boys are left over?

Solution:

Here the dividend is total number of boys 10 and the divisor is 3. We have to divide total number of boys by number of groups to find the number boys each group having .

10 / 3 = 3

Remainder = 1

Hence each group gets 3 boys and 1 boy is left over.

Basic Geometrical Concepts Solve Online

Geometrical is a study of structural mathematics .Basic geometrical has the concept of shapes and structure of an object. It contains two dimensional and three dimensional objects. In this article basic geometrical concepts solve online we are going to see about some basic geometric problems through online.
Generally online help means that here tutor and student should communicate directly with help of online chatting. Student can get more geometrical help and idea from tutor through online with help of chatting. It may be in audio or non audio.


Basic Geometrical Concepts Shapes Solve Online :

Basic geometrical concepts are plane, point, line, angle, ray, two dimensional objects, and three dimensional objects. Main purpose of using geometry is finding area, volume and perimeter of the given shape.

Basic geometrical concepts:

Two dimensional objects:

Square
Rectangle
Triangle
Circle
Parallelogram
Trapezoid

Three dimensional objects:

Cylinder
Cone
Cube
Sphere
Hemisphere
Prism
Pyramid
3d geometry shapes

Solve Example Problems in Basic Geometrical Concepts:

Solve Example problems in basic geometrical concepts:

Example problem 1:

Find the angle of triangle from the given figure?

Geometry-triangle

Solution:

Given data:

Two angles of triangle is 550 and 380

Solve the third angle of triangle

Total angle of triangle=1800

55+38+x=180

93+x=180

X=180-93

X=870

Third angle of triangle is 870

Example problem 2:

Which one is 3D shape of geometrical object?

Geometry- shapes

Solution:

Option (a)-Two dimensional object(square)

Option (b)-Two dimensional objects(Circle)

Option (c)-Three dimensional object(Parallelogram)

Option(d)-Three dimensional object(Cone)

Example problem 3:

How many edges and vertices in the cube?

Cube

Solution:

Cube having 12 edges

Cube having 8 vertices

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So the cube having top portion of four edges

Side portion of four edges

Bottom portion of four edges

So totally have 12 edges

Vertices mean nothing but corner point of the given shapes. When the two edges or lines are meeting It will produce the vertices

Solve Proving Proofs

Mathematical statement considered the sequence of statements by proofs, every statement being adjusted with some definition or a proposition or an axiom that is before launched by the method of deduction using only the permitted logical rules. Repeatedly we prove a proposition straightly from what is given in the proposition. But so many times it is easier to prove a same proposition not proving the proposition by itself,

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Solve Proving Proofs - Types of Proofs:

Solve Proving Proofs Indirect Method:

Solve proving proofs - Indirect method:

Prove the given function f : R  =>  R defined by f(x)  =  2x  +  5 is one – one.

Sol :  A function is one – one if  f(x1)  =  f(x2)  => x1  =  x2.

Using this we have to show that “ 2x1 + 5  = 2x2  +  5” => “  x1  =  x2”.This is of the form p  =>  q, where , p is 2x1  +  5 =  2x2  +  5 and q : x1  =  x2. so this is the direct method proofing.

We can also prove the same by using contrapositive of the statement. Now contrapositive of this statement is ~ q=>  ~ p, is, contrapositive of “if f(x1)  =  f(x2), then  x1  =  x2” is “if x1  =  x2, then f(x1) =  f(x2)”.

Now                 x1  !=   x2

=>                   2x1  !=   2x2

=>                   2x1  +  5  !=   2x2  + 5

=>                   f(x1)  !=   f(x2).
Solve Proving Proofs of Direct Approach:

Solve Proving Proofs -direct Proof:


It  is the proof of  a proposition in which we directly start the proof  with what is given in the proposition.

1)Straight forward approach.

2) Mathematical Induction approach.

3)Proofs By cases or by   exhaustion.

Solve Proving Proofs -Indirect Proof:

Instead of proving the proposition directly, we establish the proof of the proposition through  proving a proposition which is equivalent to the given proposition.

1)Proofs  by contradiction.

2)Proofs by using contrapositive statement.

3)Proofs by a by a counter example.


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Solving Proving Proofs - some Examples for Proving Proofs:

Solving proving proofs - Direct Method:

Prove the given function f:  R  =>  R

Defined by f(x) = 2x + 5 is one – one.

Sol :  Note that a function f is one – one if

f(x1)  =  f(x2)    x1  =  x2 (definition of one – one function)

Now given that   f(x1)  =  f(x2),

=> 2 x1 +  5 = 2 x2 +  5

=>                 2 x1  +  5  –  5  = 2 x2  +  5  –  5(adding the same quantity on both sides)

=>                 2 x1  +  0  =  2 x2  +  0

=>                 2 x1  =  2 x2 (using additive identity of real number)

=>                 (2)/(2) x1  =  (2)/(2) x2 (dividing by the same non zero quantity)

=>                 x1  =  x2.