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.

Question

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.

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Yến Oanh 5 years 2021-08-26T18:32:43+00:00 1 Answers 543 views 0

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  1. Nem
    -1
    2021-08-26T18:34:36+00:00
    Reply

    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

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

    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, P(X = 5 ) = \frac{10!}{5!((10-5)! } * 0.52^{5} * (1-0.52)^{10-5}

    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) = \frac{10!}{6!((10-6)! } * 0.52^{6} * (1-0.52)^{10-6} + \frac{10!}{7!((10-7)! } * 0.52^{7} * (1-0.52)^{10-7} +\frac{10!}{8!((10-8)! } * 0.52^{8} * (1-0.52)^{10-8} + \frac{10!}{9!((10-9)! } * 0.52^{9} * (1-0.52)^{10-9}+\frac{10!}{10!((10-10)! } * 0.52^{10} * (1-0.52)^{10-10}

    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) = \frac{10!}{3!((10-3)! } * 0.52^{3} * (1-0.52)^{10-3} + \frac{10!}{2!((10-2)! } * 0.52^{2} * (1-0.52)^{10-2}+ \frac{10!}{1!((10-1)! } * 0.52^{1} * (1-0.52)^{10-1} + \frac{10!}{0!((10-0)! } * 0.52^{0} * (1-0.52)^{10-0}

    On solving we get,

    P(x < 4)= 0.1410 ≈ 0.14  

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