Question

Exercises 31–33 refer to the following setting. Choose an American household at random and

let the random variable X be the number of cars (including SUVs and light trucks) they own.

Here is the probability distribution if we ignore the few households that own more than 5cars:

What’s the expected number of cars in a randomly selected American household?

$$\begin{gathered} a)1.00 \\ b)1.75 \\ c)1.84 \\ d)2.00 \\ e)2.50 \end{gathered}$$

Short answer

The expected number of cars in a randomly selected American household is$1.75\left(b\right)$

Step-by-step solution

Step 1. Given information 

We have given probability distribution:

Step 2. To find the probability.  

The expected value is the sum of the product of the each possibility with its probability.

$$\begin{gathered} \mu =\sum _{}xP\left(x\right) \\ =0\times 0.09+1\times 0.36+2\times 0.35+3\times 0.13+4\times 0.05+5\times 0.02 \\ =1.75 \end{gathered}$$