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# Examples On Continuous Variables Expected Value

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Examples On Continuous Variables Expected Value
Examples On Continuous Variables Expected Value

We have already looked at Variance and Standard deviation as measures of dispersion under
the section on Averages.

We can also measure the dispersion of Random variables across a given distribution using
Variance and Standard deviation. This allows us to better understand whatever the distribution
represents.

The Variance of a random variable X is also denoted by σ;2 but when sometimes can be
written as Var(X).

Variance of a random variable can be defined as the expected value of the square of the
difference between the random variable and the mean.

Given that the random variable X has a mean of μ, then the variance is expressed as:

In the previous section on Expected value of a random variable, we saw that the
method/formula for calculating the expected value varied depending on whether the random
variable was discrete or continuous.
Know More About Applications Of Integration

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As a consequence, we have two different methods for calculating the variance of a random
variable depending on whether the random variable is discrete or continuous.

For a Discrete random variable, the variance σ2 is calculated as:
For a Continuous random variable, the variance σ2 is calculated as:

In both cases f(x) is the probability density function.

The Standard Deviation σ in both cases can be found by taking the square root of the
variance.

Example 1

A software engineering company tested a new product of theirs and found that the number of
errors per 100 CDs of the new software had the following probability distribution:

x   f(x)
2 0.01
3 0.25
4 0.4
5 0.3
6 0.04
Find the Variance of X

Solution

The probability distribution given is discrete and so we can find the variance from the
following:

We need to find the mean μ first:

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The possible outcomes for one coin toss can be described by the state space \Omega =
{heads, tails}. We can introduce a real-valued random variable Y as follows:

Y(\omega) = \begin{cases} 1, & \text{if} \ \ \omega = \text{heads} ,\\ 0, & \text{if} \ \ \omega =
\text{tails} . \end{cases}

If the coin is equally likely to land on either side then it has a probability mass function given
by:

\rho_Y(y) = \begin{cases}\frac{1}{2},& \text{if }y=1,\\
\frac{1}{2},& \text{if }y=0.\end{cases}

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