From the sampling distribution of "x_a-x_b," we know that the distribution is approximately normal with mean
"\\mu _{x_a-x_b}=\\mu _a-\\mu _b=0" and variance
"\\sigma^2_{x_a-x_b}={\\sigma _a^2 \\over n_a}+{\\sigma_b^2 \\over n_b}={1 \\over 18}+{1 \\over 18}={1 \\over 9}" Corresponding to the value "x_a-x_b=1.0," we have
"z={1-(\\mu_a-\\mu _b) \\over \\sqrt{1\/9}}={1-0 \\over \\sqrt{1\/9}}=3.0" So
"P(x_a-x_b>1.0)=P(z>3.0)=1-P(z<3.0)="
"=1-0.998650=0.001350"
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