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.

b. a success is a pet owner who would not acquire their next dog or cat from a shelter.
c. a success is a pet owner who acquired their dog or cat from a shelter.
d. 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) = 0.345 (round to three decimal places as needed.)
f. find p(x > 10).
p(x > 10) = (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 \(P(x>10)\), we can use the property \(P(x>10)=1 - P(x\leq10)\).

Step2: Calculate \(P(x\leq10)\)

We need to calculate \(P(x = 0)+P(x = 1)+\cdots+P(x = 10)\). Using the binomial formula with \(n = 25\), \(p=0.5\):
For \(k = 0\): \(C(25,0)\times(0.5)^{0}\times(0.5)^{25}=\frac{25!}{0!(25 - 0)!}\times1\times(0.5)^{25}=(0.5)^{25}\)
For \(k = 1\): \(C(25,1)\times(0.5)^{1}\times(0.5)^{24}=\frac{25!}{1!(25 - 1)!}\times0.5\times(0.5)^{24}=25\times(0.5)^{25}\)
\(\cdots\)
For \(k = 10\): \(C(25,10)\times(0.5)^{10}\times(0.5)^{15}=\frac{25!}{10!(25 - 10)!}\times(0.5)^{25}\)
Using a binomial probability calculator or software (since calculating each term manually is time - consuming), we find that \(P(x\leq10)\approx0.388\)

Step3: Calculate \(P(x>10)\)

\(P(x>10)=1 - P(x\leq10)=1 - 0.388 = 0.612\)

Answer:

\(0.612\)