"n=10,p=0.1,(1-p)=0.9\\\\\ni)\\ p(x\\geq 2)=1-p(x<2)\\\\\np(x\\geq 2)=1-[p(x=0)+p(x=1)]\\\\\n1-[\\binom{10}{0}(0.1)^{0}(0.9)^{10}+\\binom{10}{1}(0.1)^{0}(0.9)^{9}]\\\\\n1-[0.3487+0.3874]\\\\\n1-0.7361\\\\\n0.2639\\\\\nii)\\ E(x)=np=10*0.1=1\\\\\nExpected\\ value\\ is\\ 1\\\\\niii)\\ Var(x)=npq\\\\\nstandard\\ dev. =\\sqrt{npq}=\\sqrt{10 \\times 0.1 \\times0.9}\\\\\n0.9487\\\\\nstandard\\ deviation\\ is\\ 0.9487"
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