Question

according to a recent poll, half of all pet owners would get their next dog or cat from a shelter. consider a random sample of 25 pet owners and define x as the number of pet owners who would acquire their next dog or cat from a shelter. assume that x is a binomial random variable. complete parts a through f below.
a. for this binomial experiment, define a success
oa. a success is a pet owner who did not acquire their dog or cat from a shelter
ob. a success is a pet owner who would not acquire their next dog or cat from a shelter
oc. a success is a pet owner who acquired their dog or cat from a shelter
od. a success is a pet owner who would acquire their next dog or cat from a shelter
b. for this binomial experiment, what is n?
n = 25
c. for this binomial experiment, what is p?
p = 0.5 (round to one decimal place as needed.)
d. find p(x = 11).
p(x = 11) = 0.133 (round to three decimal places as needed.)
e. find p(x ≥ 14).
p(x ≥ 14) = (round to three decimal places as needed.)

Explanation:

Step1: Recall binomial probability formula

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 \(P(X\geq14)\), we need to calculate \(P(X = 14)+P(X = 15)+\cdots+P(X = 25)\)

Step2: Calculate \(P(X = k)\) for \(k = 14\) to \(k = 25\)

• When \(n = 25\), \(p=0.5\), \(k = 14\)

\(C(25,14)=\frac{25!}{14!(25 - 14)!}=\frac{25!}{14!×11!}=\frac{25\times24\times\cdots\times12\times11!}{14\times13\times\cdots\times1\times11!}=4457400\)
\(P(X = 14)=C(25,14)\times(0.5)^{14}\times(0.5)^{25 - 14}=4457400\times(0.5)^{25}\approx0.161\)

• When \(k = 15\)

\(C(25,15)=\frac{25!}{15!(25 - 15)!}=\frac{25!}{15!×10!}=\frac{25\times24\times\cdots\times16\times10!}{15\times14\times\cdots\times1\times10!}=3268760\)
\(P(X = 15)=C(25,15)\times(0.5)^{15}\times(0.5)^{25 - 15}=3268760\times(0.5)^{25}\approx0.161\)

• When \(k = 16\)

\(C(25,16)=C(25,9)=\frac{25!}{16!(25 - 16)!}=\frac{25!}{16!×9!}=\frac{25\times24\times\cdots\times17\times9!}{16\times15\times\cdots\times1\times9!}=2042975\)
\(P(X = 16)=C(25,16)\times(0.5)^{16}\times(0.5)^{25 - 16}=2042975\times(0.5)^{25}\approx0.120\)

• When \(k = 17\)

\(C(25,17)=C(25,8)=\frac{25!}{17!(25 - 17)!}=\frac{25!}{17!×8!}=\frac{25\times24\times\cdots\times18\times8!}{17\times16\times\cdots\times1\times8!}=1081575\)
\(P(X = 17)=C(25,17)\times(0.5)^{17}\times(0.5)^{25 - 17}=1081575\times(0.5)^{25}\approx0.064\)

• When \(k = 18\)

\(C(25,18)=C(25,7)=\frac{25!}{18!(25 - 18)!}=\frac{25!}{18!×7!}=\frac{25\times24\times\cdots\times19\times7!}{18\times17\times\cdots\times1\times7!}=480700\)
\(P(X = 18)=C(25,18)\times(0.5)^{18}\times(0.5)^{25 - 18}=480700\times(0.5)^{25}\approx0.028\)

• When \(k = 19\)

\(C(25,19)=C(25,6)=\frac{25!}{19!(25 - 19)!}=\frac{25!}{19!×6!}=\frac{25\times24\times\cdots\times20\times6!}{19\times18\times\cdots\times1\times6!}=177100\)
\(P(X = 19)=C(25,19)\times(0.5)^{19}\times(0.5)^{25 - 19}=177100\times(0.5)^{25}\approx0.010\)

• When \(k = 20\)

\(C(25,20)=C(25,5)=\frac{25!}{20!(25 - 20)!}=\frac{25!}{20!×5!}=\frac{25\times24\times23\times22\times21\times5!}{20\times19\times18\times17\times16\times5!}=53130\)
\(P(X = 20)=C(25,20)\times(0.5)^{20}\times(0.5)^{25 - 20}=53130\times(0.5)^{25}\approx0.003\)

• When \(k = 21\)

\(C(25,21)=C(25,4)=\frac{25!}{21!(25 - 21)!}=\frac{25!}{21!×4!}=\frac{25\times24\times23\times22\times4!}{21\times20\times19\times18\times4!}=12650\)
\(P(X = 21)=C(25,21)\times(0.5)^{21}\times(0.5)^{25 - 21}=12650\times(0.5)^{25}\approx0.001\)

• When \(k = 22\)

\(C(25,22)=C(25,3)=\frac{25!}{22!(25 - 22)!}=\frac{25!}{22!×3!}=\frac{25\times24\times23\times3!}{22\times21\times20\times3!}=2300\)
\(P(X = 22)=C(25,22)\times(0.5)^{22}\times(0.5)^{25 - 22}=2300\times(0.5)^{25}\approx0.000\)

• When \(k = 23\)

\(C(25,23)=C(25,2)=\frac{25!}{23!(25 - 23)!}=\frac{25!}{23!×2!}=\frac{25\times24\times2!}{23\times22\times2!}=300\)
\(P(X = 23)=C(25,23)\times(0.5)^{23}\times(0.5)^{25 - 23}=300\times(0.5)^{25}\approx0.000\)

• When \(k = 24\)

\(C(25,24)=C(25,1)=\frac{25!}{24!(25 - 24)!}=\frac{25\times24!}{24!\times1!}=25\)
\(P(X = 24)=C(25,24)\times(0.5)^{24}\times(0.5)^{25 - 24}=25\times(0.5)^{25}\approx0.000\)

• When \(k = 25\)

\(C(25,25)=C(25,0)=1\)
\(P(X = 25)=C(25,25)\times(0.5)^{25}\times(0.5)^{25 - 25}=1\times(0.5)^{25}\approx0.000\)

Step3: Sum up the probabilities

\(P(X\geq14)=P(X = 14)+P(X = 15)+\cdots+P(X = 25)\)
\(P(X\geq14)\approx0.161 + 0.161+0.120 + 0.064+0.028+0.010+0.003+0.001+0.000+0.000+0.000+0.000\approx0.548\)

Answer:

\(P(x\geq14)\approx0.548\)