Question

A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 30% of this population prefers the color brown. If 18 buyers are randomly selected, what is the probability that exactly 8 buyers would prefer brown? Round your answer to four decimal places.

Answer 1

Answer:

0.0811 is the probability that exactly 8 buyers would prefer brown.

Step-by-step explanation:

We are given the following information:

We treat people preferring color brown as a success.

P(prefer color brown) = 30% = 0.30

Then the number of buyers follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 18

We have to evaluate:

$P(x = 8)\\= \binom{18}{8}(0.3)^8(1-0.3)^7\\= 0.121$

0.0811 is the probability that exactly 8 buyers would prefer brown.