Question

The probability distribution of a random variable X is given by

\(X = x\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(X = x)\)	\(k\)	\(2k\)	\(3k\)	\(4k\)

Then the c.d.f. of X is given by

• A.

  \(X = x\)	\(1\)	\(2\)	\(3\)	\(4\)
  \(F(X = x)\)	\(\frac{1}{10}\)	\(\frac{3}{10}\)	\(\frac{6}{10}\)	\(1\)

• B.

  \(X = x\)	\(1\)	\(2\)	\(3\)	\(4\)
  \(F(X = x)\)	\(\frac{3}{10}\)	\(\frac{1}{10}\)	\(\frac{6}{10}\)	\(1\)

• C.

  \(X = x\)	\(1\)	\(2\)	\(3\)	\(4\)
  \(F(X = x)\)	\(\frac{1}{10}\)	\(\frac{3}{10}\)	\(\frac{5}{10}\)	\(\frac{1}{10}\)

• D.

  \(X = x\)	\(1\)	\(2\)	\(3\)	\(4\)
  \(F(X = x)\)	\(\frac{1}{10}\)	\(\frac{6}{10}\)	\(\frac{3}{10}\)	\(1\)

Hint:

Find k from the total probability, then add up the probabilities.

Answer 1

The Correct Option is A

Step 1: Properties of a c.d.f.:
A c.d.f. never decreases and ends at $1$.

Step 2: Check tables:
Option (B) goes $\tfrac3{10}$ then $\tfrac1{10}$, a decrease. Option (C) ends at $\tfrac1{10}$ and (D) goes $\tfrac6{10}$ then $\tfrac3{10}$.

Step 3: Verify option A:
Probabilities $\tfrac1{10},\tfrac2{10},\tfrac3{10},\tfrac4{10}$ give running totals $\tfrac1{10},\tfrac3{10},\tfrac6{10},1$, option (A).

Final Answer:
The c.d.f. is 1/10, 3/10, 6/10, 1. \[ \boxed{\text{(A) }\text{F = 1/10, 3/10, 6/10, 1}} \]