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: Understanding the Concept:
The probabilities sum to 1, and the c.d.f. is \(F(x) = P(X \le x)\).

Step 2: Find k:
\[ k + 2k + 3k + 4k = 10k = 1 \Rightarrow k = \frac1{10} \]
So the probabilities are \(\dfrac1{10}, \dfrac2{10}, \dfrac3{10}, \dfrac4{10}\).

Step 3: Cumulative sums:
\(F(1) = \dfrac1{10}\). \(F(2) = \dfrac1{10} + \dfrac2{10} = \dfrac3{10}\). \(F(3) = \dfrac3{10} + \dfrac3{10} = \dfrac6{10}\). \(F(4) = \dfrac6{10} + \dfrac4{10} = 1\).

Step 4: Match:
The table with \(\dfrac1{10}, \dfrac3{10}, \dfrac6{10}, 1\) is option (A). Option (B) has a decreasing step, and (C) and (D) do not end in 1 or increase steadily.

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