54​% 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.

Respuesta :

Answer:

[tex]P(5)=0.238[/tex]

[tex]P(x\geq 6)=0.478[/tex]

[tex]P(x<4)=0.114[/tex]

Step-by-step explanation:

In this case we can calculate the probability using the binomial probability formula

[tex]P(X=x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}[/tex]

Where p is the probability of obtaining a "favorable outcome " x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly​ five

This is:

[tex]x=5,\ n=10,\ p=0.54.[/tex]

So:

[tex]P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}[/tex]

[tex]P(5)=0.238[/tex]

(b) at least​ six

This is: [tex]x\geq 6,\ n=10,\ p=0.54.[/tex]

[tex]P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)[/tex]

[tex]P(x\geq 6)=0.478[/tex]

(c) less than four

This is: [tex]x< 4,\ n=10,\ p=0.54.[/tex]

[tex]P(x<4)=P(3) + P(2)+P(1)+P(0)[/tex]

[tex]P(x<4)=0.114[/tex]

This question is based on the probability. Therefore, the required probabilities  are :  (a) [tex]P(5) = 0.238[/tex],  (b)[tex]P(x \geq 6) = 0.478[/tex]  and (c) [tex]P(x <4) = 0.114[/tex].

Given:

54​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults.

We have to 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.

According to the question,

[tex]P(5) = 0.238\\P(x \geq 6) = 0.478\\P(x <4) = 0.114[/tex]

In this we have to calculate the probability using the binomial probability formula,

[tex]P(X=x) = \dfrac{n!}{x!(n-x)!} \times p^{x} \times (1-p)^{n-x}[/tex]

Where, p is the probability of obtaining a "favorable outcome ", x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly​ five  

x=5, n= 10, p = 0.54

[tex]P(X=5)= \dfrac{10!}{5!(10-5)!} \times 0.5^{x} \times (1-0.54)^{10-5}[/tex]

P(X=5) = 0.238

(b) at least​ six

[tex]x\geq 6, n=10, p=0.54\\P(x\geq 6) = P(6) + P(7) + P(8) + P(9)+P(10)\\P(x\geq 6) = 0.478[/tex]

(c) less than four

[tex]x< 6, n=10, p=0.54\\P(x< 4) = P(3) + P(2) + P(1) + P(0)\\P(x< 4) = 0.114[/tex]

Therefore, the answers are :  (a) [tex]P(5) = 0.238[/tex],  (b)[tex]P(x \geq 6) = 0.478[/tex]

and (c) [tex]P(x <4) = 0.114[/tex].

For more details, please refer this link:

https://brainly.com/question/795909