Answer: 0.55567
Step-by-step explanation:
Given : A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent.
i.e.
[tex]\mu=10\text{ years}\\\sigma=3\text{ years}[/tex]
Let [tex]X_1[/tex] and [tex]X_2[/tex] are two randomly selected refrigerator's life whose sum will exceed third selected refrigerator [tex]X_3[/tex] .
So that, [tex]X =X_1+X_2-1.9\times X_3[/tex]
Mean [tex]=E(X_1)+E(X_2)-E(X_3)=10+10-1.9\times10 =1[/tex]
Standard deviation =[tex]\sqrt{(Var(X_1)+Var(X_1)+Var(X_1))}[/tex]
[tex]=\sqrt{(3^2+3^2+(1.9\times3)^2)}[/tex]
[tex]=7.1056315694[/tex]
Z-score : [tex]z=\dfrac{X-E[x]}{\sqrt{Var[x]}}=\dfrac{0-1}{7.0156315694}\approx-0.14[/tex]
Now , The probability that the total useful life of two(i..e n=2) randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator would be :-
[tex]P(Z>-0.14)=P(Z<0.14)=0.55567[/tex]
Hence, the required probability is 0.55567.
