wat is standard deviation???....need it for my gre prep....plzzzz help!!!?

Question

sum problems on standard deviation from d quantitative section....which r likely to be expected in gre....would be of gr8 help!!!

Answer 1

Standard deviation is the square root of the variance, which is the average (mean) squared value of the deviation from the mean (average). It is a measure of how spread out (how broad) a probability distribution is. If you have a sample of size n {x_i|i=1,2, ..., n} from some population, to get the mean you just sum up all the x_i and then divide by the number n of points. Let m stand for the mean. Now, to get the variance, you compute (x_i - m) and square it for all n points and sum up all of those squared "deviations". Now divide that sum by n (or by n-1 to get an "unbiased" estimate) and that gives you the variance. Take the square root of that and you have the standard deviation.

If you run a web search with the query: "standard deviation" (in quotes), you will get a ton of good hits. I did that and put links to three that look pretty good under "Sources" below.

Answer 2

In probability and statistics, the standard deviation of a probability distribution, random variable, or population or multiset of values is defined as the square root of the variance.

The standard deviation is measured in the same units as the values of the population. For a population of distances in meters, the standard deviation is also measured in meters, whereas the variance is measured in square meters.

At least 75% of the values in any population are at most two standard deviations away from the mean (see Chebyshev's inequality). The actual percentage depends on the distribution, for example it is approximately 95% if the population has a normal distribution.

The term standard deviation was introduced to statistics by Karl Pearson On the dissection of asymmetrical frequency curves, 1894.

The standard deviation is the root mean square (RMS) deviation of the values from their arithmetic mean. For example, in the population {4, 8}, the mean is 6 and the standard deviation is 2. This may be written: {4, 8} ≈ 6±2. In this case 100% of the values in the population are within one standard deviation of the mean.

Standard deviation is the most common measure of statistical dispersion, measuring how spread out the values in a data set are. If the data points are all close to the mean, then the standard deviation is close to zero. If many data points are far from the mean, then the standard deviation is far from zero. If all the data values are equal, then the standard deviation is zero.

The standard deviation (σ) of a population can be estimated by a modified standard deviation (s) of a sample. The formulae are given below.

Standard deviation of a random variable

The standard deviation of a random variable X is defined as:

\sigma = \sqrt{\operatorname{E}((X-\operatorname{E}(X))^2)} = \sqrt{\operatorname{E}(X^2) - (\operatorname{E}(X))^2}

where E(X) is the expected value of X.

Not all random variables have a standard deviation, since these expected values need not exist. For example, the standard deviation of a random variable which follows a Cauchy distribution is undefined.

If the random variable X takes on the values x1,...,xN (which are real numbers) with equal probability, then its standard deviation can be computed as follows. First, the mean of X, \overline{x}, is defined as:

\overline{x} = \frac{1}{N}\sum_{i=1}^N x_i = \frac{x_1+x_2+\cdots+x_N}{N}

(see sigma notation). Next, the standard deviation simplifies to:

\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}

In other words, the standard deviation of a discrete uniform random variable X can be calculated as follows:

1. For each value xi calculate the difference x_i - \overline{x} between xi and the average value \overline{x}.
2. Calculate the squares of these differences.
3. Find the average of the squared differences. This quantity is the variance σ2.
4. Take the square root of the variance.

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Estimating population standard deviation from sample standard deviation

In the real world, finding the standard deviation of an entire population is unrealistic except in certain cases, such as standardized testing, where every member of a population is sampled. In most cases, sample standard deviation (s) is used to estimate population standard deviation (σ). Given only a sample of values x1,...,xN from some larger population, many authors define the sample (or estimated) standard deviation by

s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}.

The reason for this definition is that s2 is an unbiased estimator for the variance σ2 of the underlying population, if it is uncorrelated and has uniform variance of σ2. However, s is not an unbiased estimator for the standard deviation σ; it tends to underestimate the population standard deviation. Although an unbiased estimator for "σ" is known when the random variable is normally distributed, the formula is complicated and amounts to a minor correction. Moreover, unbiasedness, in this sense of the word, is not always desirable; see statistical bias. Some have argued[citation needed] that even the difference between N and N − 1 in the denominator is overly complex and insignificant. Without that term, what is left is the simpler expression:

s = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}.

This form has a uniformly smaller mean squared error than does the unbiased estimator, and is the maximum-likelihood estimate when the population (or the random variable X) is normally distributed.
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Example

We will show how to calculate the standard deviation of a population. Our example will use the ages of four young children: { 5, 6, 8, 9 }.

Step 1. Calculate the mean average, \overline{x}:

\overline{x}=\frac{1}{N}\sum_{i=1}^N x_i

We have N = 4 because there are four data points:

x_1 = 5\,\!
x_2 = 6\,\!
x_3 = 8\,\!
x_4 = 9\,\!

\overline{x}=\frac{1}{4}\sum_{i=1}^4 x_i Replacing N with 4

\overline{x}=\frac{1}{4} \left ( x_1 + x_2 + x_3 +x_4 \right )

\overline{x}=\frac{1}{4} \left ( 5 + 6 + 8 + 9 \right )

\overline{x}= 7 This is the mean.

Step 2. Calculate the standard deviation \sigma\,\!:

\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}

\sigma = \sqrt{\frac{1}{4} \sum_{i=1}^4 (x_i - \overline{x})^2} Replacing N with 4

\sigma = \sqrt{\frac{1}{4} \sum_{i=1}^4 (x_i - 7)^2} Replacing \overline{x} with 7

\sigma = \sqrt{\frac{1}{4} \left [ (x_1 - 7)^2 + (x_2 - 7)^2 + (x_3 - 7)^2 + (x_4 - 7)^2 \right ] }

\sigma = \sqrt{\frac{1}{4} \left [ (5 - 7)^2 + (6 - 7)^2 + (8 - 7)^2 + (9 - 7)^2 \right ] }

\sigma = \sqrt{\frac{1}{4} \left ( (-2)^2 + (-1)^2 + 1^2 + 2^2 \right ) }

\sigma = \sqrt{\frac{1}{4} \left ( 4 + 1 + 1 + 4 \right ) }

\sigma = \sqrt{\frac{10}{4}}

\sigma = 1.5811\,\! This is the standard deviation.


Were this set a sample drawn from a larger population of children, and the question at hand was the standard deviation of the population, convention would replace the N (or 4) here with N−1 (or 3).


get this correct in this site
http://en.wikipedia.org/wiki/Standard_deviation

sigma is a summation sign
get examples here

Answer 3

Standard deviation is a measure of average deviation from the mean in a data set. If the data points are all close to the mean, then the standard deviation is close to zero. If many data points are far from the mean, then the standard deviation is far from zero. If all the data values are equal, then the standard deviation is zero.

Answer 4

by defination, SD is sqroot of Variance. in words, SD is a unit measurement to judge how much a data is deviated. so the ans to such qn will be :1 SD = a no. (no unit), and this data is how many SD deviated from the mean.... to the left of mean is negative SD.. and the opp is positive.. if u still have qn email me.