Answer:
The company is expected to make [tex]\$ 1[/tex] on every extended warranty sold.
Step-by-step explanation:
Let [tex]X[/tex] denote whether a product with the extended warranty requires replacement in that two years. ([tex]X\![/tex] would be a random variable.) Assume that [tex]X = 1[/tex] means that the product requires replacement, and [tex]X = 0[/tex] otherwise.
Assume that this product requires replacement in that two years. That is: [tex]X = 1[/tex]. The company would then bear a cost of [tex]200 \times 1 =200[/tex] for replacing this product (since [tex]X = 1\![/tex], that cost would be the same as [tex]200 = 200 \times 1 = 200\, X[/tex].)
On the other hand, assume [tex]X = 0[/tex] (that is: this product does not require replacement in that two years.) The company would not need to pay for replacing this product. Since [tex]X = 0\![/tex], the expression [tex]200\, X[/tex] would still represent the cost for the company for this warranty.
Either way, [tex]200\, X[/tex] would denote the replacement cost that the company would bear for this product. However, given that the company would charge [tex]\$ 11[/tex] for the extended warranty, the net revenue of the company on this warranty would be [tex]((-200) \, X + 11)[/tex]. (An earning of [tex]\$11 \![/tex] minus a spending of [tex]200\, X \![/tex].)
The question states that [tex]0.5\%[/tex] of the products would need replacement in this period. In other words, the expected value of [tex]X[/tex] would be [tex]\mathbb{E}(X) = 0.005[/tex].
The expected revenue of the company on this warranty would be:
[tex]\mathbb{E}(200\, X + 11)[/tex].
Apply the linearity of expected values to find the value of [tex]\mathbb{E}(200\, X + 11)[/tex]:
[tex]\begin{aligned}& \mathbb{E}((-200)\, X + 11) \\ &= (-200)\, \mathbb{E}(X) + 11 \\ &= (-200) \times 0.005 + 11 = 1\end{aligned}[/tex].
Hence, the company is expected to make [tex]\$ 1[/tex] on every extended warranty that it sold.
