Probability Part 3

Concept:The expected value $E(X)$ for a discrete random variable is the sum of each value multiplied by its probability.We first find the unknown constant $k$ using the given $E(X)$, then compute $P(X<20)$.Explanation:Step 1: Apply the formula $E(X) = \sum X_i \cdot P(X_i)$.From the table, write $E(X)$ as a sum of terms:$E(X) = 4k \cdot \frac{2}{15} + \frac{30}{7}k \cdot \frac{1}{15} + \frac{32}{7}k \cdot \frac{2}{15} + \frac{34}{7}k \cdot \frac{1}{5} + \frac{36}{7}k \cdot \frac{1}{15} + \frac{38}{7}k \cdot \frac{2}{15} + \frac{40}{7}k \cdot \frac{1}{5} + 6k \cdot \frac{1}{15}$.Step 2: Factor $\frac{k}{15}$ common from all terms.$E(X) = \frac{k}{15}\left(8 + \frac{30}{7} + \frac{64}{7} + \frac{102}{7} + \frac{36}{7} + \frac{76}{7} + \frac{120}{7} + 6\right)$.Combine the constant terms: $8+6=14$.Combine the fractions: $\frac{30+64+102+36+76+120}{7} = \frac{428}{7}$.So $E(X) = \frac{k}{15}\left(14 + \frac{428}{7}\right) = \frac{k}{15} \cdot \frac{98+428}{7} = \frac{k}{15} \cdot \frac{526}{7} = \frac{526k}{105}$.Step 3: Use given $E(X) = \frac{263}{15}$ to solve for $k$.$\frac{263}{15} = \frac{526k}{105}$ simplifies to $263 = \frac{526k}{7}$.Thus $k = \frac{263 \times 7}{526} = \frac{7}{2} = 3.5$.Step 4: Substitute $k=3.5$ into each $X$ value and find which are less than 20.$X_1 = 4 \times 3.5 = 14 < 20$, $P = \frac{2}{15}$.$X_2 = \frac{30}{7} \times 3.5 = 15 < 20$, $P = \frac{1}{15}$.$X_3 = \frac{32}{7} \times 3.5 = 16 < 20$, $P = \frac{2}{15}$.$X_4 = \frac{34}{7} \times 3.5 = 17 < 20$, $P = \frac{1}{5} = \frac{3}{15}$.$X_5 = \frac{36}{7} \times 3.5 = 18 < 20$, $P = \frac{1}{15}$.$X_6 = \frac{38}{7} \times 3.5 = 19 < 20$, $P = \frac{2}{15}$.$X_7 = \frac{40}{7} \times 3.5 = 20$ and $X_8 = 6 \times 3.5 = 21$ are not less than 20.Step 5: Sum the probabilities for $X<20$.$P(X<20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} + \frac{2}{15} = \frac{11}{15}$.Answer:$P(X<20) = \frac{11}{15}$, which corresponds to option D.