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The probability distribution of a discrete random variable X is:X=x12345P(X=x)k2k3k4k5kFind P(X ≤ 4) |
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Answer» Given, \(\displaystyle\sum_{i=1}^{n}\) pi = 1 ∴ k + 2k + 3k + 4k + 5k = 1 ∴ 15k = 1 ∴ K = \(\frac{1}{15}\)
∴ P(x ≤ 4) = P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = \(\frac{1}{15}\) + \(\frac{2}{15}\) + \(\frac{3}{15}\) + \(\frac{4}{15}\) = \(\frac{10}{15}\) = \(\frac{2}{3}\) |
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