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16 June, 22:06

Compute the following binomial probabilities directly from the formula for b (x; n, p) : a. b (3; 8,.6) b. b (5; 8,.6) c. P (3 X 5) when n ¼ 8 and p ¼.6 d. P (1 X) when n ¼ 12 and p ¼.1

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  1. 16 June, 23:53
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    The question is not properly typed! The complete question along with answer and explanation is provided below.

    Question:

    Compute the following binomial probabilities directly from the formula for P (x; n, p)

    a. P (3; 8, 0.6)

    b. P (5; 8, 0.6)

    c. P (3 ≤ X ≤ 5) when n = 8 and p = 0.6

    d. P (1 ≤ X) when n = 12 and p = 0.1

    Answer:

    a. P (3; 8, 0.6) = 0.1238

    b. P (5; 8, 0.6) = 0.2786

    c. P (3 ≤ X ≤ 5) = 0.6346

    d. P (1 ≤ X) = 0.6589

    Step-by-step explanation:

    We know that a binomial distribution is given by

    P (x; n, p) = nCx pˣ (1 - p) ⁿ⁻ˣ

    Where x is the variable of interest, n is the number of trials and p is the probability of success and 1 - p is the probability of failure

    a. P (3; 8, 0.6)

    Here we have x = 3, n = 8 and p = 0.6

    P (3; 8, 0.6) = 8C3*0.6³ * (1 - 0.6) ⁸⁻³

    P (3; 8, 0.6) = 56*0.00221

    P (3; 8, 0.6) = 0.1238

    b. P (5; 8, 0.6)

    Here we have x = 5, n = 8 and p = 0.6

    P (5; 8, 0.6) = 8C5*0.6⁵ * (1 - 0.6) ⁸⁻⁵

    P (5; 8, 0.6) = 56*0.004976

    P (5; 8, 0.6) = 0.2786

    c. P (3 ≤ X ≤ 5) when n = 8 and p = 0.6

    Here 3 ≤ X ≤ 5 means we need P (3) + P (4) + P (5)

    We have already calculated P (3) and P (5)

    P (4) = 8C4*0.6⁴ * (1 - 0.6) ⁸⁻⁴

    P (4) = 70*0.003317

    P (4) = 0.2322

    P (3 ≤ X ≤ 5) = 0.1238 + 0.2322 + 0.2786

    P (3 ≤ X ≤ 5) = 0.6346

    d. P (1 ≤ X) when n = 12 and p = 0.1

    Here we have n = 12 and p = 0.1

    P (1 ≤ X) means P (0) + P (1)

    P (0) = 12C0*0.1⁰ * (1 - 0.1) ¹²⁻⁰

    P (0) = 1*0.2824

    P (0) = 0.2824

    P (1) = 12C1*0.1¹ * (1 - 0.1) ¹²⁻¹

    P (1) = 12*0.03138

    P (1) = 0.3765

    P (1 ≤ X) = P (0) + P (1)

    P (1 ≤ X) = 0.2824 + 0.3765

    P (1 ≤ X) = 0.6589
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