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?

Answers: 2

Show answers

Answers

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

The answer should be 5.

Answer from: Quest

step-by-step explanation:

Another question on Mathematics

Mathematics, 22.06.2019 00:00

The function f(x)= x5+ ( x + 3)2 is used to create a table. if x = -1 then what is f(x)?

Answers: 1

Answer

Mathematics, 22.06.2019 03:00

Barrett is comparing the membership fees for two museums.the art museum charges a one-time fee of $8.25 plus $2.25 per month. the science museum charges a one-time fee of $10.75 plus $3.50 per month. how much does barrett save by joining the art museum instead of the science museum?

Answers: 3

Answer

Mathematics, 22.06.2019 04:10

Give the domain and range. a. domain: {0, 2, 4}, range: {2, 6, 10} b. domain: {0}, range: {2} c. domain: {2, 6, 10}, range: {0, 2, 4} d. domain: {2}, range: {0}

Answers: 3

Answer

Mathematics, 22.06.2019 06:00

Graph a sine function whose amplitude is 3, period is 4π , midline is y = 2, and y-intercept is (0, 2) . the graph is not a reflection of the parent function over the x-axis. use the sine tool to graph the function. the first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

Answers: 1

Answer

You know the right answer?

After a large scale earthquake, it is predicted that 15% of all buildings have been structurally com...

Questions

Mathematics, 03.06.2020 19:03

There are 8 blue marbles, 6 red marbles, and 4 white marbles in a jar. if a marble is selected at random from the jar, what is the probability that it...

Mathematics, 03.06.2020 19:03

10^xy=1000, where x and y are positive integers and x > y, what is one possible value of x?...