Question

You plan to conduct a marketing experiment in which students are to taste one of two different brands of soft drink. Their task is to correctly identify the brand tasted. You select a random sample of 200 students and assume that the students have no ability to distinguish between the two brands. (Hint: If an individual has no ability to distinguish between the two soft drinks, then the two brands are equally likely to be selected) [15 Marks] What is the probability that the sample will have between 50% and 60% of the identifications correct? The probability is 90% that the sample percentage is contained within what symmetrical limits of the population percentage? What is the probability that the sample percentage of correct identifications is greater than 65%? Which is more likely to occur„more than 60% correct identifications in the sample of 200 or more than 55% correct identifications in a sample of 1,000? Explain.

Answer 1

Answer:

solution:-.

given that n = 180 (number of students)

p = 0.5 (probability of correct identification)

standard deviation = sqrt[p(1-p)/200] = sqrt[(0.5)(0.5)/180]

= 0.0373

(a)p(0.5 < x < 0.6) = p[(0.5 - 0.5) / 0.0373 < Z < (0.6 - 0.5) / 0.0373]

= p(0 < z < 2.6810)

= p(z < 2.6810) - p(z < 0)

= 0.9963 - 0.5

= 0.4963

Hope it helps.