Question

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. Calculate the probability that the total useful life of two randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator.

Answer 1

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.

$\mu=10\text{ years}\\\sigma=3\text{ years}$

Let $X_1$ and $X_2$ are two randomly selected refrigerator's life whose sum will exceed third selected refrigerator $X_3$ .

So that, $X =X_1+X_2-1.9\times X_3$

Mean $=E(X_1)+E(X_2)-E(X_3)=10+10-1.9\times10 =1$

Standard deviation =$\sqrt{(Var(X_1)+Var(X_1)+Var(X_1))}$

$=\sqrt{(3^2+3^2+(1.9\times3)^2)}$

$=7.1056315694$

Z-score : $z=\dfrac{X-E[x]}{\sqrt{Var[x]}}=\dfrac{0-1}{7.0156315694}\approx-0.14$

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 :-

$P(Z>-0.14)=P(Z$

Hence, the required probability is 0.55567.