52 %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:

a. P(x = 5) = 0.24

b. P(x ≥ 6) = 0.43

c. P(x < 4)= 0.14

Step-by-step explanation:

According to the Question,

  • Given, 52 %of U.S. Adults have very little confidence in newspapers & We have to randomly select 10 U.S. Adults.

⇒We use Binomial Probability Formula

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

Where, p=0.52 & n=10

a. The probability that the number of U.S. adults who have very little confidence in newspapers is ​exactly​ five . So, (x=5)

Thus, [tex]P(X = 5 ) = \frac{10!}{5!((10-5)! } * 0.52^{5} * (1-0.52)^{10-5}[/tex]

On Solving Above Equation we get,

P(5)=0.2441 ≈ 0.24

b. The probability that the number of U.S. adults who have very little confidence in newspapers is at least 6 . So, (x≥6)

Then,

P(x≥6) = P(6)+P(7)+P(8)+P(9)+P(10)

P(x≥6) = [tex]\frac{10!}{6!((10-6)! } * 0.52^{6} * (1-0.52)^{10-6}[/tex] + [tex]\frac{10!}{7!((10-7)! } * 0.52^{7} * (1-0.52)^{10-7}[/tex] +[tex]\frac{10!}{8!((10-8)! } * 0.52^{8} * (1-0.52)^{10-8}[/tex] + [tex]\frac{10!}{9!((10-9)! } * 0.52^{9} * (1-0.52)^{10-9}[/tex]+[tex]\frac{10!}{10!((10-10)! } * 0.52^{10} * (1-0.52)^{10-10}[/tex]

On solving above equation we get,

P(x≥6) =  0.4270 ≈ 0.43

c. The probability that the number of U.S. adults who have very little confidence in newspapers is less than 4 . So, (x<4)

Then,

P(x < 4)=P(3) + P(2)+P(1)+P(0)

P(x<4) = [tex]\frac{10!}{3!((10-3)! } * 0.52^{3} * (1-0.52)^{10-3}[/tex] + [tex]\frac{10!}{2!((10-2)! } * 0.52^{2} * (1-0.52)^{10-2}[/tex]+ [tex]\frac{10!}{1!((10-1)! } * 0.52^{1} * (1-0.52)^{10-1}[/tex] + [tex]\frac{10!}{0!((10-0)! } * 0.52^{0} * (1-0.52)^{10-0}[/tex]

On solving we get,

P(x < 4)= 0.1410 ≈ 0.14