Find the probability distribution of the random variable X, which can take only the values 1,2,3 given that P(1)=10/30,P(2)=1/3 and P(3)=12/33.
The Probability Distribution is as follows:
"\\begin{matrix} x & 1 & 2 & 3\\\\ P(x) & \\frac{10}{33} & \\frac{1}{3} & \\frac{12}{33}. \\end{matrix}"
The mean of a discrete random variable X is the number given by
"\\mu =E(x)=\\sum x P(x)=1\\left(\\frac{10}{33}\\right)+2\\left(\\frac{1}{3}\\right)+3\\left(\\frac{12}{33}\\right),\\\\ =\\frac{10}{33}+\\frac{2}{3}+\\frac{36}{33}=\\frac{10}{33}+\\frac{22}{33}+\\frac{36}{33}=\\frac{68}{33}=2\\frac{2}{33}"
Therefore, mean of the random variable x is
"=2\\frac{2}{33}"
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