Question

A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 50% of this population prefers the color brown. If 15 buyers are randomly selected, what is the probability that more than a third of the buyers would prefer brown? Round your answer to four decimal places.

Answer 1

Answer:

The probability that more than a third of the buyers would prefer brown is 0.8491.

Step-by-step explanation:

This is binomial problem with n=15 and p=0.50.

The binomial formula of probaility is:

P(X=x)=$\frac{n!}{x!(n-x)!} p^{x} (1-p)^{n-x}$

More than a third of the buyers: 15÷3=5

We have to find P(x>5).

P(x>5) = 1 - P(x≤5) = 1 - P(x=0) - P(x=1) - P(x=2) - P(x=3) - P(x=4) - P(x=5)

P(x=0)=$\frac{15!}{0!15!} (0.5)^{0} (1-0.5)^{15}$=0.00003

P(x=1)=$\frac{15!}{1!14!} (0.5)^{1} (1-0.5)^{14}$=0.00046

P(x=2)=$\frac{15!}{2!13!} (0.5)^{2} (1-0.5)^{13}$=0.00320

P(x=3)=$\frac{15!}{3!12!} (0.5)^{3} (1-0.5)^{12}$=0.01389

P(x=4)=$\frac{15!}{4!11!} (0.5)^{4} (1-0.5)^{11}$=0.04165

P(x=5)=$\frac{15!}{5!10!} (0.5)^{5} (1-0.5)^{10}$=0.09164

P(x>5) = 1 - P(x≤5)= 1- 0.00003 - 0.00046 - 0.00320 - 0.01389 - 0.04165 - 0.09164 = 0.8491