Question

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 8 young adults were randomly selected and their preferences recorded. (Round your answers to three decimal places.)

(a) What is the probability that more than 5 preferred McDonald's?

(b) What is the probability that between 3 and 5 (inclusive) preferred McDonald's?

(c) What is the probability that between 3 and 5 (inclusive) preferred Burger King?

Answer 1

A)probability that people prefer mcdonald = 0.5

therefore people prefer burger king = 1- 0.5 = 0.5

a)

PROBABILITY(X>5) = P(6)+P(7)=P(8)

P(5) = 8C5*(0.5)^8 = 0.2187

P(6) = 8C6*(0.5)^8 = 0.1094

P(7) = 8C75*(0.5)^8 = 0.0312

P(8) = 8C8*(0.5)^8 = 0.003

HENCE P(X>5) = 0.3633

b) we nee to find the probability that out of 8 people 3to5 people prefer mcdonald

= p(3)+p(4)+p(5)

P(3) = 8C3*(0.5)^8 = 0.2187

P(4) = 8C45*(0.5)^8 =0.2734

P(5) = 8C5*(0.5)^8 =0.2187

THEREFORE PROBABILITY BETWEEN 3 AND 5= 0.2187+0.2734+0.2187 = 0.7109

B) AS THE PREFERENCE OVER MCDONALD AND PREFERENCE OVER BURGER KING PROBABILITY ARE SAME = 0.5

AND IN BOTH THE PARTS WE NEED TO FIND THE PROBABILITY THAT BETWWEN 3 AND 5

THEREFORE THE ANSWER AND PROCEDURE FOR SECOND PART WILL ALSO BE SAME AND THE ANSWER WILL ALSO BE THE SAM = 0.7109