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lidiya [134]
3 years ago
15

(b) what is the probability that more than eleven loads occur during a 4-year period

Mathematics
1 answer:
sergey [27]3 years ago
7 0
Given that an article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. 
Suppose the mean time between occurrences of loads is 0.4 year. 
a). How many loads can be expected to occur during a 4-year period? 
b). What is the probability that more than 11 loads occur during a 4-year period? 
c). How long must a time period be so that the probability of no loads occurring during that period is at most 0.3?



Part A:
The number of loads that can be expected to occur during a 4-year period is given by:
\frac{4}{0.4} =10 \ loads


Part B:

The expected value of the number of loads to occur during the 4-year period is 10 loads.
This means that the mean is 10.

The probability of a poisson distribution is given by

P(k \ events \ in \ an \ interval)= \frac{\lambda^ke^{-\lambda}}{k!}

where: k = 0, 1, 2, . . ., 11 and λ = 10.

P(k=0)= \frac{10^0e^{-10}}{0!} =0.000045 \\  \\ P(k=1)= \frac{10^1e^{-10}}{1!} =0.000454 \\  \\ P(k=2)= \frac{10^2e^{-10}}{2!} =0.002270 \\  \\ P(k=3)= \frac{10^3e^{-10}}{3!} =0.007567 \\  \\ P(k=4)= \frac{10^4e^{-10}}{4!} =0.018917 \\  \\ P(k=5)= \frac{10^5e^{-10}}{5!} =0.037833 \\  \\ P(k=6)= \frac{10^6e^{-10}}{6!} =0.063055 \\  \\ P(k=7)= \frac{10^7e^{-10}}{7!} =0.090079 \\  \\ P(k=8)= \frac{10^8e^{-10}}{8!} =0.112599 \\  \\ P(k=9)= \frac{10^9e^{-10}}{9!} =0.125110

P(k=10)= \frac{10^{10}e^{-10}}{10!} =0.125110 \\  \\ P(k=11)= \frac{10^{11}e^{-10}}{11!} =0.113736

The probability that more than 11 loads occur during a 4-year period is given by:

1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]

= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]

= 1 - 0.571665 = 0.428335

Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283




Part C:

The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:

\frac{\lambda^0e^{-\lambda}}{0!} \leq0.3 \\  \\ \Rightarrow e^{-\lambda}\leq0.3 \\  \\ \Rightarrow-\lambda\leq\ln{0.3} \\  \\ \Rightarrow -\lambda\leq-1.204 \\  \\ \Rightarrow\lambda\geq1.204

Therefore, the time period that must be so that the probability of no loads occurring during that period is at most 0.3 is given by:

\frac{4}{1.204} =3.3 \ years
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When consumers apply for credit, their credit is rated using FICO (Fair, Isaac, and Company) scores. Credit ratings are given be
Flura [38]

Answer:

a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.  

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n}  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}  

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is \bar X= 660.313

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. The degrees of freedom are given by:

df=n-1=16-1=15

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for t_{\alpha/2}=-2.95

"=T.INV(1-0.005,15)" for t_{1-\alpha/2}=2.95

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}  

And if we find the limits we got:

660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588  

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]  

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

4 0
2 years ago
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