Question

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?

Answer 1

ANSWER:

0.5947

STEP-BY-STEP EXPLANATION:

Given:

p = 0.7

n = 15

Calculate the probability with the following formula:

$P=nCx\cdot p^x\cdot(1-p)^{n-x}$

In this case 10 ≤ x ≤ 12

$\begin{gathered} P(10\le x\le12)=P(x=10)+P(x=11)+P(x=12) \\ P(x=10)=15C10\cdot0.7^{10}\cdot(1-0.7)^{15-10}=\frac{15!}{10!\cdot(15-10)!}\cdot0.7^{10}\cdot(0.3)^5=0.2061 \\ P(x=11)=15C11\cdot0.7^{11}\cdot(1-0.7)^{15-11}=\frac{15!}{11!\cdot(15-11)!}\cdot0.7^{11}\cdot(0.3)^4=0.2186 \\ P(x=12)=15C12\cdot0.7^{12}\cdot(1-0.7)^{15-12}=\frac{15!}{12!\cdot(15-12)!}\cdot0.7^{12}\cdot(0.3)^3=0.1700 \\ P(10\le x\le12)=0.2061+0.2186+0.1700 \\ P(10\le x\le12)=0.5947 \end{gathered}$

The probability is 0.5947