Answer to Question #222339 in Statistics and Probability for maximila

Question #222339

Suppose X and Y are 2 independent normal random variable such that x~N(65,28) and Y~N(85,36).Find p(134<X+Y<166) and p(Y>x)


1
Expert's answer
2021-08-19T15:20:32-0400

The distribution of a sum of two normally distributed independent variables XN(μ1,σ12)X\sim N(\mu_1, \sigma_1^2) and YN(μ2,σ22)Y\sim N(\mu_2, \sigma_2^2) is another normal distribution (X+Y)N(μ1+μ2,σ12+σ22).(X+Y)\sim N(\mu_1+\mu_2, \sigma_1^2+\sigma_2^2).

Given XN(65,28),YN(85,36).X\sim N(65, 28), Y\sim N(85, 36). Then (X+Y)N(65+85,28+36).(X+Y)\sim N(65+85, 28+36).


P(134<X+Y<166)P(134<X+Y<166)

=P(X+Y<166)P(X+Y134)=P(X+Y<166)-P(X+Y\leq134)

=P(Z<16615064)P(Z13415064)=P(Z<\dfrac{166-150}{\sqrt{64}})-P(Z\leq\dfrac{134-150}{\sqrt{64}})

=P(Z<2)P(Z2)=P(Z<2)-P(Z\leq-2)

0.977250.02275=0.9545\approx0.97725-0.02275=0.9545

The distribution of a difference of two normally distributed independent variables YN(μ2,σ22)Y\sim N(\mu_2, \sigma_2^2) and XN(μ1,σ12)X\sim N(\mu_1, \sigma_1^2) is another normal distribution (YX)N(μ2μ1,σ12+σ22).(Y-X)\sim N(\mu_2-\mu_1, \sigma_1^2+\sigma_2^2).

Given XN(65,28),YN(85,36).X\sim N(65, 28), Y\sim N(85, 36). Then (YX)N(8565,28+36).(Y-X)\sim N(85-65, 28+36).


P(Y>X)=P(YX>0)=1P(YX0)P(Y>X)=P(Y-X>0)=1-P(Y-X\leq0)




=1P(Z02064)=1P(Z2.5)=1-P(Z\leq\dfrac{0-20}{\sqrt{64}})=1-P(Z\leq -2.5)

0.9938\approx0.9938


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