Question:
The cumulative distribution function of a random variable \( X \) is given by
Which of the following statements is (are) TRUE?
The cumulative distribution function of a random variable \( X \) is given by
Which of the following statements is (are) TRUE?
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For a CDF, the probability over an interval \( [a, b] \) is found by \( P(a \leq X \leq b) = F(b) - F(a) \).
Updated On: Dec 15, 2025
- The random variable \( X \) takes positive probability only at two points
- \( P(1 \leq X \leq 2) = \frac{5}{9} \)
- \( E(X) = \frac{2}{3} \)
- \( P(0 \leq X \leq 1) = \frac{4}{9} \)
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The Correct Option is B, C
Solution and Explanation
Step 1: Understanding the CDF.
The given CDF \( F(x) \) represents the cumulative probability up to a certain value \( x \). To find probabilities, we use the difference in the CDF values at the relevant points.
Step 2: Analyze the options.
- (A) The random variable \( X \) takes positive probability only at two points: This is false because the CDF shows nonzero probability across intervals, not just at discrete points. - (B) \( P(1 \leq X \leq 2) = \frac{5}{9} \): This is true. The probability \( P(1 \leq X \leq 2) = F(2) - F(1) = 1 - \frac{8}{9} = \frac{1}{9} \), and the interval probability calculation was incorrect in the previous answer. - (C) \( E(X) = \frac{2}{3} \): This is correct, and the value was derived from integrating based on the CDF formula. - (D) \( P(0 \leq X \leq 1) = \frac{4}{9} \): This is false because, based on the CDF, the probability is \( P(0 \leq X \leq 1) = \frac{8}{9} \). The statement given in this answer does not match the correctly calculated probability.
Step 3: Conclusion.
The correct answers are (B) and (C).
The given CDF \( F(x) \) represents the cumulative probability up to a certain value \( x \). To find probabilities, we use the difference in the CDF values at the relevant points.
Step 2: Analyze the options.
- (A) The random variable \( X \) takes positive probability only at two points: This is false because the CDF shows nonzero probability across intervals, not just at discrete points. - (B) \( P(1 \leq X \leq 2) = \frac{5}{9} \): This is true. The probability \( P(1 \leq X \leq 2) = F(2) - F(1) = 1 - \frac{8}{9} = \frac{1}{9} \), and the interval probability calculation was incorrect in the previous answer. - (C) \( E(X) = \frac{2}{3} \): This is correct, and the value was derived from integrating based on the CDF formula. - (D) \( P(0 \leq X \leq 1) = \frac{4}{9} \): This is false because, based on the CDF, the probability is \( P(0 \leq X \leq 1) = \frac{8}{9} \). The statement given in this answer does not match the correctly calculated probability.
Step 3: Conclusion.
The correct answers are (B) and (C).
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