Question #261244

34​% 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.


1
Expert's answer
2021-11-09T09:41:50-0500

Let X=X= the number of adults who say they are more likely to make purchases during a sales tax holiday: XBin(n,p).X\sim Bin(n, p).

Given n=10,p=0.34,q=1p=0.66n=10, p=0.34, q=1-p=0.66

(a)


P(X=2)=(102)(0.34)2(0.66)1020.187293P(X=2)=\dbinom{10}{2}(0.34)^2(0.66)^{10-2}\approx0.187293

(b)


P(X>2)=1P(X=0)P(X=1)P(X>2)=1-P(X=0)-P(X=1)

P(X=2)=1(100)(0.34)0(0.66)100-P(X=2)=1-\dbinom{10}{0}(0.34)^0(0.66)^{10-0}

(101)(0.34)1(0.66)101(102)(0.34)2(0.66)102-\dbinom{10}{1}(0.34)^1(0.66)^{10-1}-\dbinom{10}{2}(0.34)^2(0.66)^{10-2}

=10.015683368810.08079311205=1-0.01568336881-0.08079311205

0.18729312338=0.71623039576-0.18729312338=0.71623039576

0.716230\approx0.716230

(c)


P(2X5)=P(X=2)+P(X=3)P(2\leq X\leq 5)=P(X=2)+P(X=3)

+P(X=4)+P(X=5)+P(X=4)+P(X=5)

=(102)(0.34)2(0.66)102=\dbinom{10}{2}(0.34)^2(0.66)^{10-2}

+(103)(0.34)3(0.66)103+\dbinom{10}{3}(0.34)^3(0.66)^{10-3}

+(104)(0.34)4(0.66)104+\dbinom{10}{4}(0.34)^4(0.66)^{10-4}

+(105)(0.34)5(0.66)105+\dbinom{10}{5}(0.34)^5(0.66)^{10-5}

0.18729312+0.25729156\approx0.18729312+0.25729156

+0.23195224+0.14338866+0.23195224+0.14338866

0.819926\approx0.819926


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