Question

59% of u.s. adults have very little confidence in newspapers. you randomly select 10 u.s. adults. find the probability that the number of u.s. adults who have very little confidence in newspapers is (a) exactly five than four. (a) p(5)=0.209 (round to three decimal places as needed.) (b) p(x≥6)=□ (round to three decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(x)=C(n,x)\times p^{x}\times(1 - p)^{n - x}\), where \(n = 10\), \(p=0.59\), \(1-p = 0.41\).

Step2: Calculate \(P(x\geq6)\)

\(P(x\geq6)=P(x = 6)+P(x = 7)+P(x = 8)+P(x = 9)+P(x = 10)\)
For \(P(x = k)=C(10,k)\times(0.59)^{k}\times(0.41)^{10 - k}\), \(C(n,k)=\frac{n!}{k!(n - k)!}\)

• \(P(x = 6)=\frac{10!}{6!(10 - 6)!}\times(0.59)^{6}\times(0.41)^{4}\)

\(C(10,6)=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\)
\(P(x = 6)=210\times(0.59)^{6}\times(0.41)^{4}\approx210\times0.0422\times0.0283\approx0.252\)

• \(P(x = 7)=\frac{10!}{7!(10 - 7)!}\times(0.59)^{7}\times(0.41)^{3}\)

\(C(10,7)=\frac{10\times9\times8}{3\times2\times1}=120\)
\(P(x = 7)=120\times(0.59)^{7}\times(0.41)^{3}\approx120\times0.0249\times0.0689\approx0.206\)

• \(P(x = 8)=\frac{10!}{8!(10 - 8)!}\times(0.59)^{8}\times(0.41)^{2}\)

\(C(10,8)=\frac{10\times9}{2\times1}=45\)
\(P(x = 8)=45\times(0.59)^{8}\times(0.41)^{2}\approx45\times0.0147\times0.1681\approx0.111\)

• \(P(x = 9)=\frac{10!}{9!(10 - 9)!}\times(0.59)^{9}\times(0.41)^{1}\)

\(C(10,9)=10\)
\(P(x = 9)=10\times(0.59)^{9}\times(0.41)^{1}\approx10\times0.0087\times0.41\approx0.036\)

• \(P(x = 10)=\frac{10!}{10!(10 - 10)!}\times(0.59)^{10}\times(0.41)^{0}\)

\(C(10,10)=1\)
\(P(x = 10)=(0.59)^{10}\approx0.006\)

\(P(x\geq6)=0.252 + 0.206+0.111+0.036 + 0.006=0.611\)

Answer:

\(0.611\)