Question

In the US, about 17% of luxury car owners have a Mercedes-Benz. Assume you randomly select two luxury car owners. What's the probability distribution for the number of the two that own a Mercedes-Benz? Round to four decimal places. xP(x) 2 What is the mean(expected value) of this probability distribution?Round to two decimal places as needed. What is the standard deviation of this probability distribution? Round to two decimal places as needed.

Answer 1

Answer:

Binomial probability distribution is used for the number of the two that own a Mercedes-Benz

Mean: 0.34

Standard deviation: 0.53

Step-by-step explanation:

For each luxury car owner, there are only two possible outcomes. Either they own a Mercedes-Benz, or they do not. The probability of each of them owning a Mercedes-Benz is independent from each other. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In the US, about 17% of luxury car owners have a Mercedes-Benz. This means that $p = 0.17$

Two car owners, so $n = 2$

What's the probability distribution for the number of the two that own a Mercedes-Benz?

Binomial

What is the mean(expected value) of this probability distribution?

$E(X) = np = 2*0.17 = 0.34$

What is the standard deviation of this probability distribution?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2*0.17*0.83} = 0.53$