Question

28% 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. (a) p(2)=0.255 (round to the nearest thousandth as needed.) (b) p(x > 2)=□ (round to the nearest thousandth as needed.)

Explanation:

Step1: Recall probability - complement rule

We know that the sum of the probabilities of all possible outcomes is 1. Let \(X\) be the number of adults who say they are more likely to make purchases during a sales - tax holiday. If we know \(P(X = k)\) follows a binomial distribution \(P(X=k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 10\) and \(p=0.28\). We want to find \(P(X>2)=1 - P(X\leq2)\). And \(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\).

Step2: Calculate \(P(X = 0)\)

The binomial coefficient \(C(n,k)=\frac{n!}{k!(n - k)!}\), so \(C(10,0)=\frac{10!}{0!(10 - 0)!}=1\), and \(P(X = 0)=C(10,0)\times(0.28)^{0}\times(1 - 0.28)^{10-0}=1\times1\times(0.72)^{10}\approx0.0374\).

Step3: Calculate \(P(X = 1)\)

\(C(10,1)=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10\), \(P(X = 1)=C(10,1)\times(0.28)^{1}\times(0.72)^{9}=10\times0.28\times(0.72)^{9}\approx0.1478\).

Step4: Calculate \(P(X = 2)\)

\(C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45\), \(P(X = 2)=C(10,2)\times(0.28)^{2}\times(0.72)^{8}=45\times0.0784\times(0.72)^{8}\approx0.2428\).

Step5: Calculate \(P(X\leq2)\)

\(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.0374 + 0.1478+0.2428=0.428\).

Step6: Calculate \(P(X>2)\)

\(P(X>2)=1 - P(X\leq2)=1 - 0.428 = 0.572\).

Answer:

0.572