"p(x)=\\frac {e^{-0.2}(0.2)^x}{x!} ,x=0,1,2,...\\\\\nquestions\\; 8\\\\\np(x=0)=\\frac {e^{-0.2}(0.2)^0}{0!} =0.8187\\\\\n\\text {The answer is (5) none of the above}\\\\\nquestions\\; 9\\\\\n(1)p(x=1)=\\frac {e^{-0.2}(0.2)^1}{1!} =0.1637\\\\\n(2)p(x=2)=\\frac {e^{-0.2}(0.2)^2}{2!} =0.0164\\\\\n(3)p(x\\geq 3)=1-p(x\\leq 2)\\\\\n=1-[p(0)+p(1)+p(2)] =0.0012\\\\\n(4)p(x\\geq 5)=1-p(x\\leq 4)\\\\\n=1-[p(0)+p(1)+p(2)+p(3)+p(4)] \\approx 0.0000\\\\\n\\text {The answer is (5) none of the above}\\\\\nquestion 10\\\\\n\\text {The answer is (3) } \\mu=0.2, \\sigma ^2=0.2\\\\"
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