This problem requires us to determine the variance of a discrete random variable, given its probability distribution and mean. We will systematically apply the fundamental definitions and formulas for probability distributions, expected value (mean), and variance.

1. Key Concepts and Formulas

   For a discrete random variable $X$ with possible values $x_1, x_2, \ldots, x_n$ and corresponding probabilities $P(X=x_1), P(X=x_2), \ldots, P(X=x_n)$:

   • Sum of Probabilities: The sum of probabilities for all possible outcomes must be equal to 1. $\sum_{i=1}^{n} P(X=x_i) = 1$
   • Mean (Expected Value): The mean, denoted as $E(X)$ or $\mu$, is the weighted average of all possible values of $X$, where the weights are their respective probabilities. $E(X) = \sum_{i=1}^{n} x_i \cdot P(X=x_i)$
   • Variance: The variance, denoted as $\sigma^2$ or $Var(X)$, measures the spread or dispersion of the distribution around its mean. It is calculated using the formula: $Var(X) = E(X^2) - [E(X)]^2$ where $E(X^2) = \sum_{i=1}^{n} x_i^2 \cdot P(X=x_i)$.

2. Step-by-Step Solution

   The distribution of the random variable $X$ is given by:

   $x$	-2	-1	3	4	6
   $P(X=x)$	$1/5$	$a$	$1/3$	$1/5$	$b$

   The mean of $X$, $E(X) = 2.3$.

   Step 1: Formulating the First Equation using the Sum of Probabilities

   Explanation: A fundamental property of any probability distribution is that the sum of all probabilities for all possible outcomes must equal 1. We use this to establish a relationship between the unknown probabilities $a$ and $b$.

   $P(X=-2) + P(X=-1) + P(X=3) + P(X=4) + P(X=6) = 1$ Substituting the given probabilities: $\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1$ Combine the known fractional probabilities: $\left(\frac{1}{5} + \frac{1}{5}\right) + \frac{1}{3} + a + b = 1$ $\frac{2}{5} + \frac{1}{3} + a + b = 1$ To sum the fractions, find a common denominator (15): $\frac{2 \times 3}{5 \times 3} + \frac{1 \times 5}{3 \times 5} + a + b = 1$ $\frac{6}{15} + \frac{5}{15} + a + b = 1$ $\frac{11}{15} + a + b = 1$ Isolate $a + b$: $a + b = 1 - \frac{11}{15}$ $a + b = \frac{15 - 11}{15}$ $\mathbf{a + b = \frac{4}{15}} \quad \text{(Equation 1)}$

   Step 2: Formulating the Second Equation using the Mean of X

   Explanation: The mean (or expected value) $E(X)$ is calculated by multiplying each possible value of $X$ by its corresponding probability and then summing these products. We are given $E(X) = 2.3$, which allows us to form a second equation involving $a$ and $b$.

   $E(X) = \sum x \cdot P(X=x)$ Substituting the values from the table and the given mean: $(-2)\left(\frac{1}{5}\right) + (-1)(a) + (3)\left(\frac{1}{3}\right) + (4)\left(\frac{1}{5}\right) + (6)(b) = 2.3$ Perform the multiplications: $-\frac{2}{5} - a + 1 + \frac{4}{5} + 6b = 2.3$ Combine the known numerical terms: $\left(-\frac{2}{5} + \frac{4}{5}\right) + 1 - a + 6b = 2.3$ $\frac{2}{5} + 1 - a + 6b = 2.3$ Convert fractions to decimals for easier calculation with 2.3: $0.4 + 1 - a + 6b = 2.3$ $1.4 - a + 6b = 2.3$ Isolate the terms with $a$ and $b$: $-a + 6b = 2.3 - 1.4$ $\mathbf{-a + 6b = 0.9} \quad \text{(Equation 2)}$

   Step 3: Solving for Unknown Probabilities 'a' and 'b'

   Explanation: We now have a system of two linear equations with two unknowns ($a$ and $b$). We will solve this system using the elimination method.

   Our system of equations is:

   1. $a + b = \frac{4}{15}$
   2. $-a + 6b = 0.9 = \frac{9}{10}$

   Add Equation 1 and Equation 2: $(a + b) + (-a + 6b) = \frac{4}{15} + \frac{9}{10}$ $7b = \frac{4 \times 2}{15 \times 2} + \frac{9 \times 3}{10 \times 3}$ $7b = \frac{8}{30} + \frac{27}{30}$ $7b = \frac{35}{30}$ Simplify the fraction: $7b = \frac{7}{6}$ Divide by 7 to find $b$: $b = \frac{1}{6}$

   Substitute the value of $b$ into Equation 1 to find $a$: $a + \frac{1}{6} = \frac{4}{15}$ $a = \frac{4}{15} - \frac{1}{6}$ Find a common denominator (30): $a = \frac{4 \times 2}{15 \times 2} - \frac{1 \times 5}{6 \times 5}$ $a = \frac{8}{30} - \frac{5}{30}$ $a = \frac{3}{30}$ Simplify the fraction: $a = \frac{1}{10}$ So, the unknown probabilities are $\mathbf{a = \frac{1}{10}}$ and $\mathbf{b = \frac{1}{6}}$.

   Step 4: Calculating the Expected Value of X-squared, $E(X^2)$

   Explanation: To calculate the variance, we need $E(X^2)$. This is found by squaring each value of $X$, multiplying it by its corresponding probability, and then summing these products. We use the values of $a$ and $b$ we just found.

   $E(X^2) = \sum x^2 \cdot P(X=x)$ $E(X^2) = (-2)^2\left(\frac{1}{5}\right) + (-1)^2(a) + (3)^2\left(\frac{1}{3}\right) + (4)^2\left(\frac{1}{5}\right) + (6)^2(b)$ Substitute the values of $a = 1/10$ and $b = 1/6$: $E(X^2) = (4)\left(\frac{1}{5}\right) + (1)\left(\frac{1}{10}\right) + (9)\left(\frac{1}{3}\right) + (16)\left(\frac{1}{5}\right) + (36)\left(\frac{1}{6}\right)$ Perform the multiplications: $E(X^2) = \frac{4}{5} + \frac{1}{10} + 3 + \frac{16}{5} + 6$ Group terms with common denominators and combine integers: $E(X^2) = \left(\frac{4}{5} + \frac{16}{5}\right) + \frac{1}{10} + (3 + 6)$ $E(X^2) = \frac{20}{5} + \frac{1}{10} + 9$ $E(X^2) = 4 + \frac{1}{10} + 9$ $E(X^2) = 13 + \frac{1}{10}$ $E(X^2) = 13.1$

   **Step 5: Calculating the Variance, $\sigma^2$}

   Explanation: Now that we have $E(X^2)$ and the given $E(X)$, we can calculate the variance using the formula $Var(X) = E(X^2) - [E(X)]^2$.

   $\sigma^2 = E(X^2) - [E(X)]^2$ Substitute the calculated $E(X^2) = 13.1$ and the given $E(X) = 2.3$: $\sigma^2 = 13.1 - (2.3)^2$ Calculate $(2.3)^2$: $(2.3)^2 = 5.29$ Now subtract: $\sigma^2 = 13.1 - 5.29$ $\sigma^2 = 7.81$

   Step 6: Final Calculation: $100 \sigma^2$

   Explanation: The question asks for the value of $100 \sigma^2$. We simply multiply our calculated variance by 100.

   $100 \sigma^2 = 100 \times 7.81$ $\mathbf{100 \sigma^2 = 781}$

3. Common Mistakes & Tips

   • Sum of Probabilities: Always verify that the sum of all probabilities equals 1. This is a crucial first step for problems involving unknown probabilities.
   • Fractional Arithmetic: Be meticulous when performing calculations with fractions and converting them to decimals. A small error early on can propagate through the entire problem.
   • Variance Formula: Do not confuse $E(X^2)$ with $[E(X)]^2$. $E(X^2)$ is the expected value of the square of the random variable, while $[E(X)]^2$ is the square of the expected value.

4. Summary

   This problem required a systematic application of the definitions of discrete probability distributions, mean, and variance. We began by using the properties of probability (sum equals 1) and the given mean to solve for the unknown probabilities $a$ and $b$. Once all probabilities were determined, we calculated $E(X^2)$, and then used it along with the given mean to find the variance $\sigma^2$. Finally, we multiplied the variance by 100 to get the desired result.

The final answer is $\boxed{781}$.