1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
asambeis [7]
3 years ago
9

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

an time between occurrences of loads is 0.5 year. (a) how many loads can be expected to occur during a 2-year period? 1 4 loads (b) what is the probability that more than six loads occur during a 2-year period? (round your answer to three decimal places.) 2 your answer cannot be understood or graded. more information (c) how long must a time period be so that the probability of no loads occurring during that period is at most 0.2? (round your answer to four decimal places.) 3 yr
Mathematics
1 answer:
kirill115 [55]3 years ago
3 0

Answer:

a) \lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

b) P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

c)  e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given \mu = 0.5 and we can solve for the parameter \lambda like this:

\frac{1}{\lambda} = 0.5

\lambda =2

So then X(t) \sim Poi (\lambda t)

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

\lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

P(X(2) >6)

And we can use the complement rule like this

P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

And we can solve this like this using the masss function:

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

T \sim Exp(\lambda=2)

The probability of no arrival during a period of duration t is given by:

f(T) = e^{-\lambda t}

And we want to find a value of t who satisfy this:

e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

You might be interested in
Explain how to solve 3^x-4=6 using the change of base formula log base b of y equals log y over log b. include the solution for
Alexxandr [17]

Answer:

x = log 10/log 3

Step-by-step explanation:

3^x - 4 = 6

3^x = 10

We take log base 3 of both sides since log_3 3^x is simply x.

log_3 3^x = log_3 10

x = log_3 10

We have an answer for x, but it is a log base 3. We want log base 10.

Now we use the change of base formula.

log_b y = log y/log b

x = log 10/log 3

7 0
3 years ago
What is the name of the line that the function approaches?
Kitty [74]
D. asymptote

In the attached graph, the function has two asymptotes, one horizontal (the x-axis) and one vertical (the line x = 2).


7 0
3 years ago
Which expression gives the measure of
AnnyKZ [126]
What??????????????? lol.
3 0
3 years ago
8x^3-6x^2+13x-4<br> Please simplify this expression and tell me how u simplified it.
jeka94
This expression is already simplified. 

5 0
2 years ago
Lisa started to drive from Boston to Washington, DC, which is 440 miles away, at 8:00 AM. For the first four hours of her trip,
sukhopar [10]

Answer:

60 mph

Step-by-step explanation:

Given;

Total distance covered d = 440 miles

Average speed in the first 4 hours v1 = 50 mph

Total time taken for the whole trip t = 8:00 to 4:00pm

t = 16:00-8:00 = 8 hours

Firstly we need to calculate the distance covered during the first 4 hours;

d1 = average speed × time = v1 × t1

t1 = 4

d1 = 50 × 4 = 200 miles

Then we need to calculate the distance covered in the second period of the journey;

d = d1 + d2

d2 = d - d1

Substituting the values;

d2 = 440 - 200

d2 = 240 miles

The time taken for the second period of travel t2;

t2 = t -t1 = 8-4

t2 = 4 hours

Average speed = distance travelled ÷ time taken

The average speed for the second part of the trip v2 is;

v2 = d2 ÷ t2

Substituting t2 and d2;

v2 = 240 miles ÷ 4 hours

v2 = 60 mph

The average speed during the second part of the trip is 60 mph

3 0
2 years ago
Other questions:
  • Help please with b. and d.
    14·1 answer
  • Rachel rounded 16,473 to the nearest hundred then she rounded it to the nearest thousand what is the final number
    10·2 answers
  • Amelia bought 9/12 pound of sliced salami at the deli counter. Which of the following decimals did the scale show,
    6·2 answers
  • my dad is 40 miles away at work...if he leaves work at 6:24pm...and travels at an average of 43 miles per hour...what time is he
    11·2 answers
  • Answer now pleaseee
    7·2 answers
  • Please help with this
    11·1 answer
  • Which angle pairs are congruent angle pairs? Select all that apply.
    12·1 answer
  • I will give brainliest to whoever helps me with this question. Also, if you could please explain to me how you solved it I would
    8·1 answer
  • Please help me it would mean a lot thanks!
    8·2 answers
  • Which sequences are geometric? <br><br> Check all that apply.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!