Given a random variable $X$, its variance $\sigma ^2 = E(X^2)-[E(X)]^2$, where $E(X)$ is the expected value of $X$. For a discrete random variable $X$, $E(X)=\sum x_iP(X=x_i)$. For a continuous variable, $E(X) = \int_{\infty}^{+\infty} xp(x)dx$.
For example, let $X$ be the outcome of rolling a perfect die, where X is a discrete random variable and $P(X=1) = P(X=2) = ... = P(X=6) = \frac{1}{6}$. Then $E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + ... + 6 \times \frac{1}{6} = 3.5$. $E(X^2) = 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6} + ... + 6^2 \times \frac{1}{6} = \frac{91}{6}$. Thus the variance $\sigma ^2 = E(X^2)-[E(X)]^2 = \frac{91}{6} - 3.5^2 = \frac{35}{12}$.
Given a random variable $X$, its variance $\sigma ^2 = E(X^2)-[E(X)]^2$, where $E(X)$ is the expected value of $X$. For a discrete random variable $X$, $E(X)=\sum x_iP(X=x_i)$. For a continuous variable, $E(X) = \int_{\infty}^{+\infty} xp(x)dx$.
For example, let $X$ be the outcome of rolling a perfect die, where X is a discrete random variable and $P(X=1) = P(X=2) = ... = P(X=6) = \frac{1}{6}$. Then $E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + ... + 6 \times \frac{1}{6} = 3.5$. $E(X^2) = 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6} + ... + 6^2 \times \frac{1}{6} = \frac{91}{6}$. Thus the variance $\sigma ^2 = E(X^2)-[E(X)]^2 = \frac{91}{6} - 3.5^2 = \frac{35}{12}$.
user
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Given a random variable $X$, its variance $\sigma ^2 = E(X^2)-[E(X)]^2$, where $E(X)$ is the expected value of $X$. For a discrete random variable $X$, $E(X)=\sum x_iP(X=x_i)$. For a continuous variable, $E(X) = \int_{\infty}^{+\infty} xp(x)dx$.
For example, let $X$ be the outcome of rolling a perfect die, where X is a discrete random variable and $P(X=1) = P(X=2) = ... = P(X=6) = \frac{1}{6}$. Then $E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + ... + 6 \times \frac{1}{6} = 3.5$. $E(X^2) = 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6} + ... + 6^2 \times \frac{1}{6} = \frac{91}{6}$. Thus the variance $\sigma ^2 = E(X^2)-[E(X)]^2 = \frac{91}{6} - 3.5^2 = \frac{35}{12}$.
user
The standard deviation $\sigma$ is the square root of the variance.

The variance is a measure of the spread of the data. The larger the variance, the more spread out the data.

The standard deviation is a measure of the spread of the data. The larger the standard deviation, the more spread out the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread of the data.

The variance and standard deviation are both measures of the spread of the data. The variance is a measure of the spread of the data, and the standard deviation is a measure of the spread
assistant
Given a random variable $X$, its variance $\sigma ^2 = E(X^2)-[E(X)]^2$, where $E(X)$ is the expected value of $X$. For a discrete random variable $X$, $E(X)=\sum x_iP(X=x_i)$. For a continuous variable, $E(X) = \int_{\infty}^{+\infty} xp(x)dx$.
For example, let $X$ be the outcome of rolling a perfect die, where X is a discrete random variable and $P(X=1) = P(X=2) = ... = P(X=6) = \frac{1}{6}$. Then $E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + ... + 6 \times \frac{1}{6} = 3.5$. $E(X^2) = 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6} + ... + 6^2 \times \frac{1}{6} = \frac{91}{6}$. Thus the variance $\sigma ^2 = E(X^2)-[E(X)]^2 = \frac{91}{6} - 3.5^2 = \frac{35}{12}$.
user
## Ad blocker interference detected!

### Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.
assistant