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:
We are given a discrete cumulative distribution function table $F(X)$, where $F(x) = P(X \le x)$. We need to calculate the sum of the individual point probabilities $P(X=4) + P(X=5)$.

Step 2: Key Formula or Approach:
For any discrete random variable, the individual probability mass function $P(X=x)$ can be obtained from the cumulative distribution function $F(X)$ using the subtraction formula: $$P(X=x) = F(x) - F(x-1)$$ Alternatively, the sum of contiguous point probabilities can be simplified directly using: $$P(X=4) + P(X=5) = P(3 \lt X \le 5) = F(5) - F(3)$$

Step 3: Detailed Explanation:
Let's find the required sum using our direct cumulative step difference method: $$\text{Sum} = F(5) - F(3)$$ From the given distribution table, read off the matching cumulative values: $$F(5) = 0.85$$ $$F(3) = 0.48$$ Now subtract these values: $$\text{Sum} = 0.85 - 0.48 = 0.37$$ Let's verify this by computing the individual point values separately: $$P(X=4) = F(4) - F(3) = 0.62 - 0.48 = 0.14$$ $$P(X=5) = F(5) - F(4) = 0.85 - 0.62 = 0.23$$ $$P(X=4) + P(X=5) = 0.14 + 0.23 = 0.37$$ Both methods provide the exact same result.

Step 4: Final Answer:
The value of $P[X=4] + P[X=5]$ is $0.37$, which corresponds to option (C).