according to the federal communications commission, 70% of all u.s. households have vcrs. in a random sample…

according to the federal communications commission, 70% of all u.s. households have vcrs. in a random sample of 15 households, what is the probability that the number of households with vcrs is between 10 and 12, inclusive? select one: a. 0.2061 b. 0.7 c. 0.5947 d. 0.4053
Answer
Explanation:
Step1: Identify the binomial distribution parameters
This is a binomial distribution problem with $n = 15$ (number of trials/sample size), $p=0.7$ (probability of success). The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.
Step2: Calculate $P(X = 10)$
$C(15,10)=\frac{15!}{10!(15 - 10)!}=\frac{15!}{10!5!}=\frac{15\times14\times13\times12\times11}{5\times4\times3\times2\times1}=3003$ $P(X = 10)=C(15,10)\times(0.7)^{10}\times(0.3)^{5}=3003\times0.0282475249\times0.00243\approx0.2061$
Step3: Calculate $P(X = 11)$
$C(15,11)=\frac{15!}{11!(15 - 11)!}=\frac{15!}{11!4!}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365$ $P(X = 11)=C(15,11)\times(0.7)^{11}\times(0.3)^{4}=1365\times0.0197732674\times0.0081\approx0.2186$
Step4: Calculate $P(X = 12)$
$C(15,12)=\frac{15!}{12!(15 - 12)!}=\frac{15!}{12!3!}=\frac{15\times14\times13}{3\times2\times1}=455$ $P(X = 12)=C(15,12)\times(0.7)^{12}\times(0.3)^{3}=455\times0.0138412872\times0.027\approx0.1700$
Step5: Calculate $P(10\leq X\leq12)$
$P(10\leq X\leq12)=P(X = 10)+P(X = 11)+P(X = 12)\approx0.2061 + 0.2186+0.1700=0.5947$
Answer:
C. 0.5947