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Question
The probability distribution of a random variable X is
X | - 2.5 | - 1.5 | -0.5 | 0.5 | 1.5 |
P(X) | 0.35 | 0.25 | 0.2 | 0.15 | 0.05 |
Then the c.d.f. of X is
Options
X -2.5 -1.5 -0.5 0.5 1.5 F(X) 0.35 0.5 0.7 0.75 0.8 X -2.5 -1.5 -0.5 0.5 1.5 F(X) 0.35 0.5 0.7 0.85 1 X -2.5 -1.5 -0.5 0.5 1.5 F(X) 0.35 0.6 0.8 0.85 1 X -2.5 -1.5 -0.5 0.5 1.5 F(X) 0.35 0.6 0.8 0.95 1
MCQ
Solution
X | -2.5 | -1.5 | -0.5 | 0.5 | 1.5 |
F(X) | 0.35 | 0.6 | 0.8 | 0.95 | 1 |
Explanation:
F(x1) = p1 = 0.35
F(x2) = p1 + p2 = 0.35 + 0.25 = 0.6
F(x3) = p1 + p2 + p3 = 0.6 + 0.2 = 0.8
F(x4) = p1 + p2 + p3 + p4 = 0.8 + 0.15 = 0.95
F(x5) = p1 + p2 + p3 + p4 + p5 = 0.95 + 0.05 = 1
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