A random variable X has the following probability distribution
then F(4) =
Show Hint
In these distribution equations, the linear coefficients often add up close to 10. Spotting that $10k \approx 1$ gives a quick hint that $k = 0.1$. Summing up to 4 gives $8k$, and $8 \times 0.1 = 0.8 = \frac{4}{5}$ instantly!
Step 1: Understanding the Question:
We are given a discrete probability distribution of a random variable $X$ expressed in terms of an unknown constant $k$. We need to determine the cumulative distribution function value at $X = 4$, denoted as $F(4) = P(X \le 4)$.
Step 2: Key Formula or Approach:
For any valid probability distribution, the sum of all individual probabilities must equal exactly 1:
$$ \sum P(X = x_i) = 1 $$
The cumulative distribution function $F(4)$ is computed by summing all the individual probabilities from the lowest value up to 4:
$$ F(4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) $$
Step 3: Detailed Explanation:
Let's sum the row of probabilities from the distribution table to solve for $k$:
$$ 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1 $$
Grouping the like terms together:
$$ (k^2 + 2k^2 + 7k^2) + (k + 2k + 2k + 3k + k) = 1 $$
$$ 10k^2 + 9k = 1 \implies 10k^2 + 9k - 1 = 0 $$
Solving this quadratic equation by splitting the middle term:
$$ 10k^2 + 10k - k - 1 = 0 $$
$$ 10k(k + 1) - 1(k + 1) = 0 \implies (10k - 1)(k + 1) = 0 $$
This yields two possible values: $k = \frac{1}{10}$ or $k = -1$. Since individual probability values can never be negative, $k$ must be positive, so we select $k = \frac{1}{10} = 0.1$.
Now, we calculate the cumulative value $F(4)$:
$$ F(4) = P(X \le 4) = 0 + k + 2k + 2k + 3k = 8k $$
Substituting our value of $k = \frac{1}{10}$:
$$ F(4) = 8 \times \left(\frac{1}{10}\right) = \frac{8}{10} = \frac{4}{5} $$
Step 4: Final Answer:
The value of $F(4)$ is $\frac{4}{5}$, which corresponds to option (D).
