Question

In response to concerns about nutritional contents of fast foods, McDonald's has announced that it will use a new cooking oil for its french fries that will decrease substantially trans fatty acid levels and increase the amount of more beneficial polyunsaturated fat. The company claims that 97 out of 100 people cannot detect a difference in taste between the new and old oils. Assuming that this figure is correct (as a long-run proportion), what is the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald's,

(a) At least 39 can taste the difference between the two oils? (Round your answer to four decimal places.)


(b) At most 6% can taste the difference between the two oils? (Round your answer to four decimal places.)

Answer 1

Answer:

Given,

sample n = 1000

q = 97/100

= 0.97

p = 1 - 0.97

= 0.03

standard deviation = sqrt(p(1-p)/n)

substitute values

= sqrt(0.03(1-0.03)/1000)

= 0.0054

a)

P(X > 39) = P((x-u)/s > (0.039 - 0.03)/0.0054)

= P(z > 1.67)

= 0.0474597 [since from z table]

= 0.0475

b)

P(X < 0.06) = P((x-u)/s < (0.06 - 0.03)/0.0054)

= P(z < 5.56)

= 1 [since from z table]