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) = 0.562 (round to the nearest thousandth as needed.) (c) p(2 ≤ x ≤ 5) = 0.558 (round to the nearest thousandth as needed.)

Explanation:

Step1: Identify the problem type

This is a probability - related problem for a binomial distribution. Here, \(n = 10\) (number of trials, i.e., number of adults selected) and \(p=0.28\) (probability of an adult saying they are more likely to make purchases during a sales - tax holiday).

Step2: For part (a)

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). For \(n = 10\), \(k = 2\), \(p = 0.28\), and \(1-p=0.72\).

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Step3: For part (b)

\(P(X>2)=1 - P(X\leq2)=1-(P(X = 0)+P(X = 1)+P(X = 2))\).

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Step4: For part (c)

\(P(2\leq X\leq5)=P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)\).

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Answer:

(a) \(0.255\)
(b) \(0.562\)
(c) \(0.558\)