Let $$X$$ be a random variable with distribution.

If the mean of $$X$$ is 2.3 and variance of $$X$$ is $$\sigma^2$$, then $$100\sigma^2$$ is equal to _________.

We have a discrete random variable $$X$$ that can take the five values $$-2,\,-1,\,3,\,4,\,6$$ with respective probabilities $$\dfrac15,\,a,\,\dfrac13,\,\dfrac15,\,b$$. Because these probabilities must add up to $$1$$, we first write

$$\dfrac15 + a + \dfrac13 + \dfrac15 + b = 1.$$

The two fractions on the left that are already known combine as

$$\dfrac15 + \dfrac15 = \dfrac25,$$

and using a common denominator of $$15$$ we get

$$\dfrac25 + \dfrac13 = \dfrac{6}{15} + \dfrac{5}{15} = \dfrac{11}{15}.$$

So the probability-sum equation becomes

$$a + b + \dfrac{11}{15} = 1 \quad\Longrightarrow\quad a + b = 1 - \dfrac{11}{15} = \dfrac{4}{15}. \quad -(1)$$

Next, the mean (expected value) is given to be $$2.3$$. The formula for the mean of a discrete variable is

$$E[X] = \sum x_i P(X = x_i).$$

Substituting all values and probabilities we obtain

$$E[X] = (-2)\!\left(\dfrac15\right) + (-1)\,a + 3\!\left(\dfrac13\right) + 4\!\left(\dfrac15\right) + 6b.$$

We now simplify each product:

$$(-2)\!\left(\dfrac15\right) = -\dfrac25,\qquad 3\!\left(\dfrac13\right) = 1,\qquad 4\!\left(\dfrac15\right) = \dfrac45.$$

Hence

$$E[X] = -\dfrac25 - a + 1 + \dfrac45 + 6b.$$

Combining the three constant fractions gives

$$-\dfrac25 + 1 + \dfrac45 = \dfrac{-2 + 5 + 4}{5} = \dfrac75 = 1.4.$$

Therefore

$$1.4 - a + 6b = 2.3.$$

Writing $$2.3$$ as the fraction $$\dfrac{23}{10}$$ and $$1.4$$ as $$\dfrac{14}{10},$$ we get

$$-\!a + 6b = \dfrac{23}{10} - \dfrac{14}{10} = \dfrac{9}{10}. \quad -(2)$$

We now solve the simultaneous linear equations (1) and (2):

From (1)  $$a = \dfrac{4}{15} - b.$$

Substituting this into (2), we have

$$-\!\left(\dfrac{4}{15} - b\right) + 6b = \dfrac{9}{10}.$$

Simplifying inside the brackets and then combining terms:

$$-\dfrac{4}{15} + b + 6b = \dfrac{9}{10}$$

$$-\dfrac{4}{15} + 7b = \dfrac{9}{10}$$

$$7b = \dfrac{9}{10} + \dfrac{4}{15}.$$

Adding the two fractions on the right using the common denominator $$30$$, we get

$$\dfrac{9}{10} = \dfrac{27}{30},\qquad \dfrac{4}{15} = \dfrac{8}{30},\qquad \dfrac{27}{30} + \dfrac{8}{30} = \dfrac{35}{30} = \dfrac76.$$

Thus

$$7b = \dfrac76 \quad\Longrightarrow\quad b = \dfrac16.$$

Using $$a + b = \dfrac{4}{15},$$ we now find

$$a = \dfrac{4}{15} - \dfrac16 = \dfrac{8}{30} - \dfrac{5}{30} = \dfrac{3}{30} = \dfrac1{10}.$$

With both unknown probabilities known, we can compute the variance. First, we need $$E[X^2].$$ The required formula is

$$E[X^2] = \sum x_i^2 P(X = x_i).$$

We list each term:

$$(-2)^2\!\left(\dfrac15\right) = 4\!\left(\dfrac15\right) = \dfrac45,$$

$$( -1)^2 a = 1\!\left(\dfrac1{10}\right) = \dfrac1{10},$$

$$3^2\!\left(\dfrac13\right) = 9\!\left(\dfrac13\right) = 3,$$

$$4^2\!\left(\dfrac15\right) = 16\!\left(\dfrac15\right) = \dfrac{16}{5},$$

$$6^2 b = 36\!\left(\dfrac16\right) = 6.$$

Adding them together:

$$E[X^2] = \dfrac45 + \dfrac1{10} + 3 + \dfrac{16}{5} + 6.$$

Converting each to the denominator $$10$$ for ease of addition, we have

$$\dfrac45 = \dfrac8{10},\quad \dfrac{16}{5} = \dfrac{32}{10},\quad 3 = \dfrac{30}{10},\quad 6 = \dfrac{60}{10}.$$

So

$$E[X^2] = \dfrac8{10} + \dfrac1{10} + \dfrac{30}{10} + \dfrac{32}{10} + \dfrac{60}{10} = \dfrac{131}{10}.$$

Now, the variance formula is

$$\sigma^2 = E[X^2] - \bigl(E[X]\bigr)^2.$$

We know $$E[X] = 2.3 = \dfrac{23}{10}.$$ Therefore

$$\sigma^2 = \dfrac{131}{10} - \left(\dfrac{23}{10}\right)^2 = \dfrac{131}{10} - \dfrac{529}{100}.$$

Writing $$\dfrac{131}{10} = \dfrac{1310}{100},$$ we obtain

$$\sigma^2 = \dfrac{1310}{100} - \dfrac{529}{100} = \dfrac{781}{100}.$$

Finally, multiplying by $$100$$ as required, we get

$$100\sigma^2 = 100 \times \dfrac{781}{100} = 781.$$

So, the answer is $$781$$.