Question

5.1 basics of probability distributions. compute the mean and standard deviation of a discrete random variable. an lg dishwasher, which costs $750, has a 10% chance of needing to be replaced in the first two years of purchase. a two - year extended warranty costs $112.10 on a dishwasher. a) write out the probability distribution for the customers value of the extended warranty. make sure you enter your x values from smallest to largest. b) what is the customers expected value of the extended warranty? round final answer to 2 decimal places and write the units in the second box. c) which of the following is the correct interpretation of the expected value of the warranty? select an answer question help: video message instructor post to forum

Explanation:

Step1: Define the random - variable values

Let $X$ be the value of the extended - warranty to the customer. If the dishwasher does not need to be replaced (with probability $P(\text{no replacement})=0.9$), the customer has paid $\$112.10$ for the warranty and received no benefit, so $X=- 112.10$. If the dishwasher needs to be replaced (with probability $P(\text{replacement}) = 0.1$), the customer has paid $\$112.10$ for the warranty but saved $\$750$, so $X=750 - 112.10=637.90$.

Step2: Write the probability distribution

The probability distribution is:

$X$	$P(X)$
637.90	0.1

Step3: Calculate the expected value

The formula for the expected value $E(X)$ of a discrete random variable is $E(X)=\sum_{i}x_{i}P(x_{i})$.
$E(X)=(-112.10)\times0.9 + 637.90\times0.1$
$E(X)=-100.89+63.79$
$E(X)=-37.10$

Step4: Interpret the expected value

The expected value of the warranty represents the average amount of money the customer can expect to gain or lose by purchasing the extended - warranty. A negative expected value means that, on average, the customer loses money when purchasing the extended - warranty.

Answer:

a)

$X$	$P(X)$
637.90	0.1

b) - 37.10 dollars
c) On average, the customer loses $37.10$ dollars by purchasing the extended - warranty.