Suppose X is a binomial random variable with n 5 25 and p 5 0.80. a. Find the mean, variance, and standard deviation of X. b. Find the proba

Question

Suppose X is a binomial random variable with n 5 25 and p 5 0.80. a. Find the mean, variance, and standard deviation of X. b. Find the probability X is within one standard deviation of the mean. c. Find the probability X is more than two standard deviations from the mean

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Arianna 2 hours 2021-09-13T22:35:51+00:00 1 Answer 0

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    2021-09-13T22:37:30+00:00

    Answer:

    (a) The mean, variance and standard deviation of random variable X are 20, 4 and 2 respectively.

    (b) The probability that X is within one standard deviation of the mean is 0.7926.

    (c) The probability that X is more than two standard deviations from the mean is 0.0038.

    Step-by-step explanation:

    The random variable X follows a Binomial distribution with parameter n = 25 and p = 0.80.

    The probability mass function of X is:

    P(X=x)={25\choose x}0.80^{x}(1-0.80)^{25-x};\ x=0,1,2,3...

    The mean, variance and standard deviation of a Binomial distribution is given by:

    \mu=np\\\sigma^{2}=np(1-p)\\\sigma=\sqrt{np(1-p)}

    (a)

    Compute the mean, variance and standard deviation of random variable X as follows:

    \mu=np=25\times0.80=20\\\sigma^{2}=np(1-p)=25\times0.80\times(1-0.80)=4\\\sigma=\sqrt{np(1-p)}=\sqrt{4}=2

    Thus, the mean, variance and standard deviation of random variable X are 20, 4 and 2 respectively.

    (b)

    Compute the probability that X is within one standard deviation of the mean as follows:

    P (μ – σ ≤ X ≤ μ + σ) = P (20 – 2 ≤ X ≤ 20 + 2)

                                   = P (18 ≤ X ≤ 22)

                                   = P (X ≤ 22) – P (X ≤ 18)

                                  = P (X = 18) + P (X = 19) + P (X = 20) + P (X = 21) + P (X = 22)

                                  ={25\choose 18}0.80^{18}(1-0.80)^{25-18}+{25\choose 19}0.80^{19}(1-0.80)^{25-19}\\+{25\choose 20}0.80^{20}(1-0.80)^{25-20}+{25\choose 21}0.80^{21}(1-0.80)^{25-21}\\+{25\choose 22}0.80^{22}(1-0.80)^{25-22}\\=0.1108+0.1633+0.1960+0.1867+0.1358\\=0.7926

    Thus, the probability that X is within one standard deviation of the mean is 0.7926.

    (c)

    Compute the probability that X is more than two standard deviations from the mean as follows:

    P (X > μ + 2σ) = P (X > 20 + (2×2))

                          = P (X > 24)

                          = P (X = 25)

                          ={25\choose 25}0.80^{25}(1-0.80)^{25-25}\\=1\times 0.00378\times1\\=0.0038

    Thus, the probability that X is more than two standard deviations from the mean is 0.0038.

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