Let X be a random variable whose probability distribution is given by the table

X	1	3	5	7
P(X)	1/3	1/6	1/6	1/3

Then variance of x is

The correct answer is 19/3.

To find the variance of the random variable X, we use the formula:
$\mathrm{Var}\left(X\right)=E\left[X^2\right]-{\left(E\right[X\left]\right)}^2$

First, let's calculate the expected value, E[X], which is given by:
$E\left[X\right]=\sum x_i·P(X=x_i)$

Using the values from the probability distribution table:
$E\left[X\right]=1·\frac{1}{3}+3·\frac{1}{6}+5·\frac{1}{6}+7·\frac{1}{3}$

Let's simplify each term:
$E\left[X\right]=\frac{1}{3}+\frac{3}{6}+\frac{5}{6}+\frac{7}{3}$

Grouping the terms with a common denominator:
$E\left[X\right]=(\frac{1}{3}+\frac{7}{3})+(\frac{3}{6}+\frac{5}{6})=\frac{8}{3}+\frac{8}{6}$

Since 8/6 simplifies to 4/3, we get:
$E\left[X\right]=\frac{8}{3}+\frac{4}{3}=\frac{12}{3}=4$

Next, let's calculate the expected value of X², which is given by:
$E\left[X^2\right]=\sum x_i^2·P(X=x_i)$

Substituting the values from the table:
$E\left[X^2\right]=1^2·\frac{1}{3}+3^2·\frac{1}{6}+5^2·\frac{1}{6}+7^2·\frac{1}{3}$

Calculating the squares of the X values:
$E\left[X^2\right]=1·\frac{1}{3}+9·\frac{1}{6}+25·\frac{1}{6}+49·\frac{1}{3}$

Simplifying the terms:
$E\left[X^2\right]=\frac{1}{3}+\frac{9}{6}+\frac{25}{6}+\frac{49}{3}$

Grouping the terms with common denominators:
$E\left[X^2\right]=(\frac{1}{3}+\frac{49}{3})+(\frac{9}{6}+\frac{25}{6})=\frac{50}{3}+\frac{34}{6}$

Simplifying 34/6 to 17/3:
$E\left[X^2\right]=\frac{50}{3}+\frac{17}{3}=\frac{67}{3}$

Now, we can compute the variance:
$\mathrm{Var}\left(X\right)=E\left[X^2\right]-{\left(E\right[X\left]\right)}^2$

Substituting our values:
$\mathrm{Var}\left(X\right)=\frac{67}{3}-4^2=\frac{67}{3}-16$

To subtract, find a common denominator:
$\mathrm{Var}\left(X\right)=\frac{67}{3}-\frac{48}{3}=\frac{67-48}{3}=\frac{19}{3}$