Question

21​% 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.

Answer 1

Answer:

$\left(a\right)\ 21.066\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(b\right)\ 66.872\\ \left(c\right)\ 81.567$

Step-by-step explanation:

$\left(a\right)\ P\left(2\right)=C^{\begin{matrix}2\\10\\\end{matrix}}\bullet(32)^2\bullet(1-32)^{10-2}\\ =C^{\begin{matrix}2\\10\\\end{matrix}}\bullet(32)^2\bullet(68)^8=0.21066=21.066$ $percent$

$cb)\ P(x > 2)=P(x\le2)=1-(P(x=0)+P(x=1)+P(x=2))\\ =1-C^{\begin{matrix}0\\10\\\end{matrix}}(0.32)degrees(0.68)^{10}-C^{\begin{matrix}1\\10\\\end{matrix}}(0.32)\prime(0.68)^9-C^{\begin{matrix}1\\0\\\end{matrix}}(008)(66)^9\\ =1-0.02114-0.09948-0.21066\\ =0.66872=66.872$

$(C)\ \ \ \ \ P(2\preccurlyeq$ $5)=P(2)+P(3)+P(4)+P(5)5)=P(2)+P(3)+P(4)+P(5)$

$=C^{\begin{matrix}2\\10\\\end{matrix}}(0.32)^{)^2}\bullet(0.68)^3+C^{\begin{matrix}3\\10\\\end{matrix}}(0.32)^3\bullet(0.68)^7+C^{\begin{matrix}4\\10\\\end{matrix}}(0.32)^4\bullet(0.68)^6+C^{\begin{matrix}5\\10\\\end{matrix}}\bullet(0.32)^5\bullet(0.65)^5)$

$=0.2186+0.26436+0.21771+0.12294\\ =0.81567=81.567 percent$

$(binomial\ distribution)$

I hope this helps you

:)