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berliedecius4051
26.02.2020 •
Mathematics
Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 27 psi. Suppose the actual air pressure in each tire is a random variable—X for the right tire and Y for the left tire, with joint pdf f(x,y) = K(x2 + y2) 20 ≤ x ≤ 30, 20 ≤ y ≤ 30 0 otherwise (a) What is the value of K? (Enter your answer as a fraction.) K = 3/380000 Correct: Your answer is correct. (b) What is the probability that both tires are underfilled? (Round your answer to four decimal places.) .430426 Correct: Your answer is correct. (c) What is the probability that the difference in air pressure between the two tires is at most 2 psi? (Round your answer to four decimal places.) .3593 Correct: Your answer is correct. (d) Determine the (marginal) distribution of air pressure in the right tire alone.
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Ответ:
Part a: The value of K is found as 3/380000.
Part b: The probability that both the tires will be underfilled is 0.4304.
Part c: The probability that the difference of the air pressure between the two tyres is 2psi is 0.3593.
Part d: The marginal distribution of air pressure in the right tyre is![\dfrac{1}{{20}} + \dfrac{{3{x^2}}}{{38000}}](/tpl/images/0524/6516/d32c7.png)
Step-by-step explanation:
Part a
As the distribution is given as
Now by the definition of the pdf
So the value of K is found as 3/380000.
Part b
The probability of both tyres being underfilled is given such that the X≤27 psi thus
So the probability that both the tires will be underfilled is 0.4304.
Part c
The probability that the difference between the two tyres is more than 2 psi. For this the cases are as
For these cases the probability is given as![P\left( {\left| {X - Y} \right| \le 2} \right) = \left\{ {\int\limits_{28}^{30} {\int\limits_{x - 2}^{30} {K\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{20}^{22} {\int\limits_{20}^{x + 2} {K\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{22}^{28} {\int\limits_{x - 2}^{x + 2} {K\left( {{x^2} + {y^2}} \right)dydx} } } \right\}](/tpl/images/0524/6516/8ad7b.png)
![\\ = K\left[ {\left\{ {\int\limits_{28}^{30} {\int\limits_{x - 2}^{30} {\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{20}^{22} {\int\limits_{20}^{x + 2} {\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{22}^{28} {\int\limits_{x - 2}^{x + 2} {\left( {{x^2} + {y^2}} \right)dydx} } } \right\}} \right]\\ = K\left[ {\left\{ {\dfrac{{14804}}{{15}}} \right\} + \left\{ {\dfrac{{8204}}{{15}}} \right\} + \left\{ {\dfrac{{46864}}{{15}}} \right\}} \right]](/tpl/images/0524/6516/9b2df.png)
![= \dfrac{3}{{38000}}\left[ {\left\{ {\dfrac{{14804}}{{15}}} \right\} + \left\{ {\dfrac{{8204}}{{15}}} \right\} + \left\{ {\dfrac{{46864}}{{15}}} \right\}} \right]\\ = \dfrac{{4267}}{{11875}}\\ = 0.3593](/tpl/images/0524/6516/98b48.png)
So the probability that the difference of the air pressure between the two tyres is 2psi is 0.3593.
Part d
The marginal distribution of air pressure in right tire
So the marginal distribution of air pressure in the right tyre is![\dfrac{1}{{20}} + \dfrac{{3{x^2}}}{{38000}}](/tpl/images/0524/6516/d32c7.png)
Ответ: