After solving;
(a) P(40.5<Y<48.9) = 0.3721
(b) P(40.5<Y<48.9|X=68.6) = 0.6075
(c) P(40.5<Y<48.9∣X>=68) = 0.3925
(d) In b and c we use conditional probability in bi-variate case. In a it is uni-variate normal distribution.
Given that;
μ(x) = 60.6
σ(x) = 11.2
μ(y) = 46.8
σ(y) = 8.4
ρ = 0.94
That is (X,Y) follows BVN(60.6,46.8,11.2,8.4,0.94)
(a) We have to find P(40.5<Y<48.9).
P(40.5<Y<48.9) = P[(40.5-μ(y))/σ(y) < (Y-μ)/σ(y) < (48.9-μ(y))/σ(y)]
Here Z = (Y-μ)/σ(y) follows N(0,1)
P(40.5<Y<48.9) = P{(40.5-46.8)/8.4 < Z < (48.9-46.8)/8.4}
P(40.5<Y<48.9) = P( -6.3/8.4 < Z < 2.1/8.4)
P(40.5<Y<48.9) = P(-0.75 < Z < 0.25)
P(40.5<Y<48.9) = p(0<Z<0.75)+p(0<Z<0.25)
P(40.5<Y<48.9) = 0.2734+0.0987
P(40.5<Y<48.9) = 0.3721
b) We have to find P(40.5<Y<48.9∣X=68.6).
We know that Y|X follows N(μ(y)+ρ σ(y)/σ(x) (x-μ(x)),σ^2(y)(1-ρ^2))
μ(y)+ρ σ(y)/σ(x) (x-μ(x)) = 46.8+0.94* 8.4/11.2 (X-60.6)
μ(y)+ρ σ(y)/σ(x) (x-μ(x)) = 46.8+0.705(X-60.6)
σ^2(y)(1-ρ^2) = 8.4^2(1-0.94^2)
σ^2(y)(1-ρ^2) = 70.56(1-0.8836)
σ^2(y)(1-ρ^2) = 70.56*0.1164
σ^2(y)(1-ρ^2) = 8.2132
Y|X follows N(46.8+0.705(X-60.6),8.2132)
X = 68.6
Y|X follows N(46.8+0.705(68.6-60.6),8.2132)
N(46.8+5.64,8.2132)
N(52.44,8.2132)
P(40.5<Y<48.9|X=68.6)=p{(40.5-52.44)/√(8.2132)<Z<√(48.9-52.44)/√(8.2132)
P(40.5<Y<48.9|X=68.6) = P(-11.94/2.8658 < z < -3.54/2.8658)
P(40.5<Y<48.9|X=68.6) = p(-4.17 < z < -1.24)
P(40.5<Y<48.9|X=68.6) = 1-0.3925
P(40.5<Y<48.9|X=68.6) = 0.6075
c) Now we have to find the P(40.5<Y<48.9∣X>=68)
P(40.5<Y<48.9∣X>=68) = 1-0.6075
P(40.5<Y<48.9∣X>=68) = 0.3925
d) In b and c we use conditional probability in bi-variate case. In a it is uni-variate normal.
To learn more about normal distribution link is here
brainly.com/question/15103234
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