Aresearcher 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.

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)=

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)==0.00003

P(x=1)==0.00046

P(x=2)==0.00320

P(x=3)==0.01389

P(x=4)==0.04165

P(x=5)==0.09164

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

do i do it on a graph?

step-by-step explanation:

the correct answer is d

hope this