Suppose that X ∼ N(-1.4,3.1), Y ∼ N(2.9,1.7), and Z ∼ N(1.1, 0.4) are independent random variables.
Find the probability that 4.8X + 3Y + 4Z ≥ 9.6.
Round your answer to the nearest thousandth.
Ans:
E(4.8X + 3Y + 4Z)=-1.4*4.8+3*2.9+4*1.1=6.38
SD(4.8X + 3Y + 4Z)=SQRT(4.8^2*3.1+3^2*1.7+4^2*0.4)=9.65
z=(9.6-6.38)/9.65
z=0.334
P(z>=0.334)=0.369
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