Question

The distribution function \(F(X)\) of a discrete random variable \(X\) is given by: \[ \begin{array}{c|cccccc} X & 1 & 2 & 3 & 4 & 5 & 6 \\ F(X) & 0.20 & 0.37 & 0.48 & 0.62 & 0.85 & 1 \end{array} \] Then \(P[X = 4] + P[X = 5] =\)

• A. 0.14
• B. 0.85
• C. 0.37
• D. 0.23

Hint:

To calculate the sum of consecutive probabilities $P(X=k) + P(X=k+1) + \dots + P(X=n)$, you don't need to compute each piece. The sum is simply equal to the higher cumulative index minus the one right before the start: $F(n) - F(k-1)$. Here, $F(5) - F(3) = 0.85 - 0.48 = 0.37$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
From a cumulative distribution table F(X), find P(X=4) + P(X=5).

Step 2: Key Formula or Approach:
P(X=x) = F(x) – F(x–1); sum = F(5) – F(3).

Step 3: Detailed Explanation:
Sum = F(5) – F(3) = 0.85 – 0.48 = 0.37. Verification: P(4)=0.62–0.48=0.14; P(5)=0.85–0.62=0.23; sum=0.37.

Step 4: Final Answer:
The sum is 0.37, matching option (C).