Question

A random variable X has the following probability distribution.

| X = x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | P[X = x] | 0 | k | 2k | 2k | 3k | k^2 | 2k^2 | 7k^2 + k |

Then F(4) =

• A. $\tfrac{3}{10}$
• B. $\tfrac{1}{10}$
• C. $\tfrac{7}{10}$
• D. $\tfrac{4}{5}$

Answer 1

Answer: (D)

$\tfrac{4}{5}$

Answer 2

Step 1: Determine $k$ by using the normalization condition:

$$\sum_{x=0}^7 P(X=x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 9k + 10k^2 = 1.$$

Solve $10k^2 + 9k - 1 = 0$:

$$k = \frac{-9 + \sqrt{9^2 + 4\cdot10\cdot1}}{2\cdot10} = \frac{-9 + 11}{20} = \frac{1}{10}.$$

Step 2: Compute the cumulative distribution function at 4:

$$F(4) = P(X \le 4) = \sum_{x=0}^4 P(X=x) = 0 + k + 2k + 2k + 3k = 8k = 8 \times \tfrac{1}{10} = \tfrac{4}{5}.$$

Hence, $\displaystyle F(4) = \tfrac{4}{5}.$