Problem | Back to Problems Page |
The range of a random variable X is (0, 1, 2). If C is a constant such that P(X) = 3C3,
P(X = 1) = 4C − 10C2, P(x = 2) = 5C − 1, then C = Net Answers :
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Solution | |
Let "x" represent the value in the range of the random variable "X".
P(X = 0) = 3C3 ⇒ f(0) = 3C3
Since f(x) represents the probability mass function, the discrete probability distribution of "x" would be
Since f(x) is a probability mass function,
C = 1 and C = 2 are ignored as they would give a value greater than 1 for 3C3. Since 3C3 represents probability it cannot be greater than 1. The discrete probability distribution of "x" with the values of "C" replaced The probability distribution replacing the values of of "k" would be
Calculations for Mean and Standard Deviations
Expectation/Mean of the distribution
Variance of the distribution
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Credit : Vijayalakshmi Desu |