The probability distribution of a random variable X is given below:

If $$ E(X)=\frac{263}{15} $$. then $$ P(X<20)$$ is equal to:

$$E(X) = \sum X \cdot P(X)$$:

$$E(X) = (4k)\left(\frac{2}{15}\right) + \left(\frac{30}{7}k\right)\left(\frac{1}{15}\right) + \left(\frac{32}{7}k\right)\left(\frac{2}{15}\right) + \left(\frac{34}{7}k\right)\left(\frac{3}{15}\right) + \left(\frac{36}{7}k\right)\left(\frac{1}{15}\right) + \left(\frac{38}{7}k\right)\left(\frac{2}{15}\right) + \left(\frac{40}{7}k\right)\left(\frac{3}{15}\right) + (6k)\left(\frac{1}{15}\right)$$

$$E(X) = \frac{k}{15} \left[ 14 + \frac{428}{7} \right] = \frac{k}{15} \left[ \frac{98 + 428}{7} \right] = \frac{526k}{105}$$

We are given $$E(X) = \frac{263}{15}$$:

$$\frac{263}{15} = \frac{526k}{105} \implies 263 = \frac{526k}{7} \implies 1 = \frac{2k}{7} \implies k = \frac{7}{2} = 3.5$$

Now substitute $$k = 3.5$$ back into each $$X$$ term to find which are strictly less than 20:

• $$4(3.5) = 14$$

• $$\frac{30}{7}(3.5) = 15$$

• $$\frac{32}{7}(3.5) = 16$$

• $$\frac{34}{7}(3.5) = 17$$

• $$\frac{36}{7}(3.5) = 18$$

• $$\frac{38}{7}(3.5) = 19$$

• $$\frac{40}{7}(3.5) = 20$$ (Not less than 20)

• $$6(3.5) = 21$$ (Not less than 20)

The values meeting the condition $$X < 20$$ are the first 6 entries. Summing their respective probabilities:

$$P(X < 20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} + \frac{2}{15} = \frac{11}{15}$$