Part CCSIR NET June 2025the-cdf-jumps-at-0-and-at-2-so-there-are-atoms-either-side-of-the-density

The cdf jumps at 0 and at 2 so there are atoms either side of the density

Let X be a random variable with the cumulative distribution function F_X(x) = 0 if x < 0, F_X(x) = (x + 2)/5 if 0 ≤ x < 2, and F_X(x) = 1 if x ≥ 2. Then, which of the following statements are true?

A. $E (X^{2}) = 4/3$ ✓
B.P(|X − 4/5| ≥ 1) = 19/25
C.The upper bound of P(|X − 4/5| ≥ 1), using Chebyshev's inequality, is 52/75✓
D.The value of the moment generating function, M_X(t), at t = 1 is $(e^{2} - 1) /5$

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: Uncorrelated ⇒ independent

The cdf jumps at 0 and at 2 so there are atoms either side of the density

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