The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) Find the mean

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The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) Find the mean and standard deviation of the sample count X who are married. b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married? c.) What is the probability that we find less than 4 of the seniors are married? d.) What is the probability that we find at least 1 of the seniors are married?

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Audrey 1 month 2021-10-21T01:15:12+00:00 1 Answer 0 views 0

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    2021-10-21T01:16:32+00:00

    Answer:

    a) Mean 6, standard deviation 2.42

    b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

    c) 14.85% probability that we find less than 4 of the seniors are married.

    d) 99.77% probability that we find at least 1 of the seniors are married

    Step-by-step explanation:

    For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

    Binomial probability distribution

    The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

    C_{n,x} = \frac{n!}{x!(n-x)!}

    And p is the probability of X happening.

    The expected value of the binomial distribution is:

    E(X) = np

    The standard deviation of the binomial distribution is:

    \sqrt{V(X)} = \sqrt{np(1-p)}

    In this problem, we have that:

    n = 300, p = 0.02

    a.) Find the mean and standard deviation of the sample count X who are married.

    Mean

    E(X) = np = 300*0.02 = 6

    Standard deviation

    \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42

    b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

    This is P(X = 8).

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040

    10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

    c.) What is the probability that we find less than 4 of the seniors are married?

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

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023

    P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143

    P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436

    P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883

    P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485

    14.85% probability that we find less than 4 of the seniors are married.

    d.) What is the probability that we find at least 1 of the seniors are married?

    Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

    P(X = 0) + P(X \geq 1) = 1

    From c), we have that P(X = 0) = 0.0023. So

    0.0023 + P(X \geq 1) = 1

    P(X \geq 1) = 0.9977

    99.77% probability that we find at least 1 of the seniors are married

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