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

0.5947

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)=(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)=(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)=(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