After a large scale earthquake, it is predicted that 15% of all buildings have been structurally compromised.a) What is the probability that

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After a large scale earthquake, it is predicted that 15% of all buildings have been structurally compromised.a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

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Jasmine 3 days 2021-09-12T04:32:14+00:00 1 Answer 0

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    2021-09-12T04:33:38+00:00

    Answer:

    a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

    b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

    c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

    d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

    e) The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

    Step-by-step explanation:

    For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. 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.

    15% of all buildings have been structurally compromised.

    This means that p = 0.15

    20 buildings

    This means that n = 20

    a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?

    This is P(X = 1).

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

    P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368

    13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

    b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?

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

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

    P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388

    P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368

    P(X < 2) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756

    17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

    c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?

    Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So

    P(X \leq 4) + P(X > 4) = 1

    P(X > 4) = 1 - P(X \leq 4)

    In which

    P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

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

    P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388

    P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368

    P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293

    P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428

    P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821

    P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0388 + 0.1368 + 02293 + 0.2428 + 0.1821 = 0.8298

    P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8298 = 0.1702

    17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

    d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?

    P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

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

    P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293

    P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428

    P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821

    P(X = 5) = C_{20,5}.(0.15)^{5}.(0.85)^{15} = 0.1028

    P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.7570

    75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

    e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

    The expected value of the binomial distribution is:

    E(X) = np

    So

    E(X) = 20*0.15 = 3

    The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

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