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A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 40% of this population prefers the color brown. If 18 buyers are randomly selected, what is the probability that exactly a third of the buyers would prefer brown? Round your answer to four decimal places.

User Martina
by
7.7k points

2 Answers

3 votes

Answer:

0.1655

Explanation:

User Kevin Junghans
by
9.2k points
2 votes

Answer:

0.1655

Explanation:

This is a binomial probability.

THe formula is:


nCr*p^r*(1-p)^(n-r)

Where n is the total number of people

r is how many we want (we want 1/3rd of 18 which is 6, r= 6)

p is probability of success (0.4)

Putting the information, we get:


nCr*p^r*(1-p)^(n-r)\\=18C6*(0.4)^(6)(0.6)^(12)\\=0.1655

Thus, the probability that exactly a third of the buyers would prefer brown is 0.1655

User Thomas Hupkens
by
7.7k points
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