n=10
p=0.1
q=0.9
X~Binomial(10,0.1)
"P(X) = C^{10}_x(0.1)^x(0.9)^{10-x}"
i) Accepted if
"P(X<2) = P(X=0)+P(X=1) \\\\\n\nP(X=0) = \\frac{10!}{0!(10-0)!} \\times (0.1)^0 \\times (0.9)^{10-0} \\\\\n\n= 1 \\times 1 \\times 0.3486 \\\\\n\n= 0.3486 \\\\\n\nP(X=1) = \\frac{10!}{1!(10-1)!} \\times (0.1)^1 \\times (0.9)^{10-1} \\\\\n\n= 10 \\times 0.1 \\times 0.3874 \\\\\n\n= 0.3874 \\\\\n\nP(X<2) = 0.3486 + 0.3874 = 0.736"
ii) Rejected if
"P(X\u22652) = 1 -P(X<2) \\\\\n\n= 1 -0.736 \\\\\n\n= 0.264"
Comments
Leave a comment