Question

38​% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is​ (a) exactly​ two, (b) more than​ two, and​ (c) between two and​ five, inclusive. Question content area bottom Part 1 ​(a) P(2)= enter your response here ​(Round to the nearest thousandth as​ needed.) Part 2 ​(b) P(x>2)= enter your response here ​(Round to the nearest thousandth as​ needed.) Part 3 ​(c) P(2≤x≤5)= enter your response here ​(Round to the nearest thousandth as​ needed.)

Answer 1

(a) Using the binomial probability distribution with n=10 and p=0.38,
P(2) = ${10\choose 2} (0.38)^2 (1-0.38)^8$ = 0.244 (rounded to the nearest thousandth).

(b) P(x>2) = P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10), where
P(3) = ${10\choose 3} (0.38)^3 (1-0.38)^7$
P(4) = ${10\choose 4} (0.38)^4 (1-0.38)^6$
P(5) = ${10\choose 5} (0.38)^5 (1-0.38)^5$
P(6) = ${10\choose 6} (0.38)^6 (1-0.38)^4$
P(7) = ${10\choose 7} (0.38)^7 (1-0.38)^3$
P(8) = ${10\choose 8} (0.38)^8 (1-0.38)^2$
P(9) = ${10\choose 9} (0.38)^9 (1-0.38)^1$
P(10) = ${10\choose 10} (0.38)^10 (1-0.38)^0$
Calculating these probabilities gives P(x>2) = 0.550 (rounded to the nearest thousandth).

(c) P(2≤x≤5) = P(2)+P(3)+P(4)+P(5), which we have already calculated in parts (a) and (b). Thus, P(2≤x≤5) = 0.794 (rounded to the nearest thousandth).