Question

The variance of the random variable \[x\] whose probability distribution is given by \[X={{x}_{i}}:-1,0,1\] ,\[p\left( X={{x}_{i}} \right):0.4,0.2,0.4\] is

Answer 1

Hint: We are given to find the variance of the given random variable \[x\], in order to find the variance, firstly, we have to find the mean of the given distribution. After finding the mean, then we will be calculating the variance by applying the formula of variance and this would be our required answer.

Complete step by step answer:
Now let us have a brief regarding the mean and the variance of random variables. A random variable can be defined as the set of possible values from a random experiment. The mean of the random variable is nothing but the average value of the values that random variable can take. The mean is also called as the expected value and is denoted by \[\mu \]. On the other hand, variance is the value how far is the value spread from the mean. The variance is denoted by \[\sigma \].
Now let is find the variance of the given distribution.
Firstly, we will be finding the mean of the distribution.
\[\overset{-}{\mathop{x}}\,=\dfrac{-1+0+1}{3}=0\]
We obtain the mean of the given distribution as \[0\].
Now let us calculate the value of variance.
The formula for calculating variance is \[\sum {{\left( {{x}_{i}}-\overset{-}{\mathop{x}}\, \right)}^{2}}P\left( {{x}_{i}} \right)=\sum {{\left( {{x}_{i}} \right)}^{2}}P\left( {{x}_{i}} \right)\]
Now let us substitute the obtained values and compute the variance.
We get,
\[{{\left( -1 \right)}^{2}}\left( 0.4 \right)+0\left( 0.2 \right)+{{1}^{2}}\left( 0.4 \right)=0.8\]
\[\therefore \] The variance of the given distribution is \[0.8\]

Note: In order to calculate the variance, it is necessary that we have to find or know the value of mean as it depends upon the mean. We must note that if the set of possible values range is big, then the variance will be small.