Mathematics, 13.03.2020 17:03 canyonmorlan

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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Answer from: vege

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

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.

Answer from: Quest

Idon’t know actually i’m in 5th grade and don’t know

Answer from: Quest

answer: $201

step-by-step explanation: to get the answer, you first get 3 nieces that he sold and multiply it by 67 and get $201, so he earned $201 selling his nieces.