The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Preve

Question

The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number resulting in a defect. Assume the births are independent.
1. Identify an appropriate probability model for X.
a. Uniform distribution with mean 2.5.
b. Poisson distribution with mean 5/33.
c. binomial distribution with n = 5 and p = 1/33.
d. binomial distribution with n = 5 and p = 32/33.
e. Normal distribution with mean 5 and variance 1/33.

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Parker 21 hours 2021-09-13T12:10:14+00:00 1 Answer 0

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    2021-09-13T12:12:09+00:00

    Answer:

    c. binomial distribution with n = 5 and p = 1/33.

    Step-by-step explanation:

    For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability of a birth resulting in a defect is independent of other births. So we use the binomial probability distrbution 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 proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).

    This means that the probability of a birth resulting in a defect is p = \frac{1}{33}

    A local hospital randomly selects five births.

    This means that n = 5

    So the correct answer is:

    c. binomial distribution with n = 5 and p = 1/33.

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