Mathematics: 47% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four

The two quantites are defined as follows:So, a child with mass kg sits at a height m has potential energySimilarly, if the same child moves with speed m/s, he has kinetic energy...

47% of U.S. adults have very little confidence

Ответ:

foxesforlife45

22.01.2022

Mathematics

$a. \ P(X=5)=0.2417\\\\b. \ P(X\geq 6)=0.3056\\\\c. \ P(X$

Step-by-step explanation:

a. This is a binomial probability distribution problem expressed as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}$

Where:

is the probability a successful event.

the number of successful events.

is the total number of events.

Given n=10, p=0.47 the probability of exactly 5 is calculated as:

$P(X=5)={n\choose x}p^x(1-p)^{n-x}\\\\={10\choose 5}0.47^5(1-0.47)^5\\\\=0.2417$

Hence, the probability that exactly 5 adults have very little confidence is 0.2417

b. Given n=10, p=0.47,

-We substitute our values in formula and the probability of at least 6 is calculated as:

$P(X\geq 6)={n\choose x}p^x(1-p)^{n-x}\\\\\\\ \ \ \ \ \ ={10\choose 6}0.47^6(0.53)^4+{10\choose 7}0.47^7(0.53)^3+{10\choose 8}0.47^8(0.53)^2+{10\choose 9}0.47^9(0.53)^1+{10\choose 10}0.47^{10}(0.53)^0\\\\\\=0.1786+0.0905+0.0301+0.0059+0.0005\\\\\\=0.3056$