Statistics Help

Question

An LG Dishwasher, which costs $700, has a 17% chance of needing to be replaced in the first 2 years of purchase. A two-year extended warrantee costs $112.10 on a dishwasher. Assume the dishwasher is replaced in the first 2 years.

a) Write out the probability distribution for the value of the extended warranty.

   Make sure you enter your X values from smallest to largest.

X	P(X)

b) What is the expected value of the extended warranty?

  Round final answer to 2 decimal places and write the units in the second box.

Answer 1

basically we have two scenarios here, replacement or no replacement

P(replacement) = 0.17 and the value in that case is 700 - 112.10 = 587.90

P(no replacement) = 1 - 0.17 = 0.83, and the value in that case is just the -112.10

then to find the expected value you can take a weighted average of those terms

-112.10(0.83) + 587.90(0.17)

=-93.04 + 99.94

=6.90

meaning that the expected value of the extended warranty in this situation is $6.90

Answer 2

Let X represent the value of the extended warranty for the dishwasher. The two outcomes are:

1. Dishwasher replacement needed: The warranty pays $700 (the cost of replacement).
2. No replacement needed: The warranty pays $0.

From the given information:

1. The probability of needing a replacement, P(Replacement), is 17%, or 0.17.
2. The probability of not needing a replacement, P(No Replacement), is 1 - 0.17 = 0.83.

So the probability distribution of X is:

X = 700; P(X) = 0.17

X = 0; P(X) = 0.83

The expected value can be calculated using this formula:

E(X) = ∑X ⋅ P(X)

Plugging in the values:

E(X) = (700 ⋅ 0.17) + (0 ⋅ 0.83) = 119

Hope this was helpful.