Monday, July 16, 2012

How to find Prime numbers

Introduction to prime numbers : let us first understand What is prime number in math ?A natural number greater than 1, which has no factors except 1 and itself is called a prime number.Example of prime numbers are: 2, 3, 5, 7, 11, 13, 17,…. and so on.what about 1 . Is the Number 1 a Prime Number? so , no 1 is not a prime number because 1 is neither prime nor composite. Every natural number except 1 is, either a prime number or a composite number.and now let us see Is Two a Prime Number?yes  2 is the only prime number which is even. All other prime numbers are odd.Let us now understand the concept of finding prime numbers.

How to find prime number?Eratosthenes, a Greek mathematician, gave a simple method to mark out primes. His method is known as the Sieve of Eratosthenes.We first list the numbers up to 100, except 1 which is neither prime nor composite.
1. Begin with 2 which is prime. So keep it but cross out all its multiples.
2. Next, the number is 3 is prime. Thus we keep it but cross out all its multiples. Some of these numbers have already been crossed out.
3. The next number not crossed out is 5. It is also prime. So, keep it and cross out all its multiples.
4. Continue this process keeping only the primes and striking off their multiples until we cannot strike off any more numbers.

Thus, the prime numbers from 1 to 100 are: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.Eratosthenes, probably, made holes in the paper instead of crossing out the numbers. Therefore, his paper must have looked like a sieve. That is why perhaps this method is known as sieve method.

From the above set of primes between 1 and 100, we do not see any pattern. We will not see any regular pattern even if we take numbers between 1 and 10000. Mathematicians from ages tried unsuccessfully to find a simple formula which will give all the prime numbers and only the prime numbers. It is now known that no simple formula exists, so the only way to find whether the given number is prime is to see if it has any factor other than 1 and itself.now let us know What is the largest known prime number? The largest known prime number is 243112609 - 1, It is the largest integer that is currently known to be a prime number.

Monday, July 9, 2012

Formula for Standard Deviation


In Statistics, Standard Deviation is the measure of variability or spread of scores within a set of data. It is calculated as the root mean square deviation of the values from their arithmetic mean
Standard Deviation Equation or Formula for Standard Deviation

Standard Deviation
The given data with ‘n’ data values, there are two types of data:
1. Sample data (selection from a bigger population)
2. Population data (from a group of sample)
The standard deviation formula or the standard deviation equation is given as,
Population Standard Deviation, s = square root of [sigma(xi-x(bar))^2/(n)]
Sample Standard Deviation, s = square root of [sigma(xi-x(bar))^2/(n-1)]
[ s is the standard deviation, xi = all the data items of a sample data (i is 1 to n), x(bar) = mean of the sample data, n=number of data items]

How to do Standard Deviation:
To find the Standard Deviation the steps involved are:
• First we need to count the number of data items in the given sample, this gives us ‘n’
• The mean of the data is calculated, x(bar)
• Subtract the mean from each data item and square the difference, [(x-x(bar)]2 tabulate this values
• Find the sum of the above tabulated values, sigma [(x-x(bar))^2]
• Divide the sum with the number of data items (n); sigma [(x-x(bar))^2]/n
• Once all this done, find the square root; square root {sigma [(x-x(bar))^2]/n}
Using the formula square root of {sigma [(x-x(bar))^2]/n} gives us the standard deviation of the given sample data
Consider a sample data, 5, 8, 11, 13, 18, 7, 14, 12. Let us find the standard deviation of the given sample data using a Standard Deviation Chart
There are eight scores in the given sample data, so n=8
The mean =(5+8+11+13+18+7+14+12)/8= 88/8 = 11

Data Values (x)        [x-x(bar)]2
5         (-6)^2=36
8        (-3)^2=9
11   (0)^2=0
13 (2)^2=4
18 (7)^2=49
7 (-4)^2=16
14 (3)^2=9
12 (1)^2=1

Sigma  [x-x(bar)]^2= 36+9+0+4+49+16+9+1= 124

Sigma  [x-x(bar)]2/n = 124/8 = 15.5

Standard Deviation =Square root{ Sigma  [x-x(bar)]2/n} = square root {15.5} = 3.94
So, the standard deviation of the given sample data is 3.94

Standard Deviation Graph
The standard deviation is a statistic which tells us how the various samples or data items of a sample data are distributed around the mean in a set of data. In a normal distribution, the graph of the standard deviation is in the shape of a bell curve. When the samples are mostly together then the graph we get is a steep bell curve; which shows that the standard deviation is small. When the samples are spread apart then the bell curve we get is relatively flat; which shows that the standard deviation is relatively large.