Respuesta :
Answer:
1) K = 7.895 × 10⁻⁶
2) 0.3024
3) 3.6775 × 10⁻²
4) [tex]f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}[/tex]
5) X and Y are not independent variables
6)
[tex]h(x\mid y) = \frac{38000x^2+38000y^2}{3y^2+19000}[/tex]
7) 0.54967
8) 25.33 psi
σ = 2.875
Step-by-step explanation:
1) Here we have
[tex]f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}[/tex]
[tex]\int_{x}\int_{y} f(x, y)dydx = 1[/tex]
[tex]\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1[/tex]
[tex]K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1[/tex]
[tex]K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1[/tex]
[tex]K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1[/tex]
[tex]K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1[/tex]
[tex]K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1[/tex]
[tex]K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1[/tex]
[tex]K =\frac{3}{380000}[/tex]
2) The probability that both tires are underfilled
P(X≤26,Y≤26) =
[tex]\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx[/tex]
[tex]=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx[/tex]
[tex]= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx[/tex]
[tex]K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx[/tex]
[tex]K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx[/tex]
[tex]K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})[/tex]
[tex]K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})[/tex]
[tex]38304\times K =\frac{3\times38304}{380000}[/tex]
= 0.3024
That is P(X≤26,Y≤26) = 0.3024
3) The probability that the difference in air pressure between the two tires is at most 2 psi is given by
{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, [tex]| x-y |[/tex] ≤ 2}
{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, [tex]\sqrt{(x-y)^2}[/tex] ≤ 2}
{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, y ≤ x - 2}
Which gives
20 ≤ x ≤ 22 :: 20 ≤ y ≤ x + 2
22 ≤ x ≤ 28 :: x - 2 ≤ y ≤ x + 2
28 ≤ x ≤ 30 :: x - 2 ≤ y ≤ 30
From which we derive probability as
P( [tex]| x-y |[/tex] ≤2) = [tex]\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx[/tex] + [tex]\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx[/tex] + [tex]\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx[/tex]
= K ( [tex]\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx[/tex] + [tex]\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx[/tex] + [tex]\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx[/tex])
= [tex]K\left [ \left (\frac{14804}{15} \right )+\left (\frac{8204}{15} \right ) +\left (\frac{46864}{15} \right )\right ] = \frac{3}{380000}\times \frac{69872}{15} =\frac{4367}{118750}[/tex] = 3.6775 × 10⁻²
4) The marginal pressure distribution in the right tire is
[tex]f_{x}\left ( x \right )=\int_{y} f(x ,y)dy[/tex]
[tex]=K( \right )\int_{y}(x^{2} +y^{2})dy)[/tex]
[tex]= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})[/tex]
[tex]K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})[/tex]
[tex]K(10x^{2}+\frac{19000}{3})}[/tex]
[tex]\frac{3}{38000} (10x^{2}+\frac{19000}{3})}[/tex]
[tex]= \frac{1}{20} +\frac{3x^{2} }{38000}[/tex]
[tex]f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}[/tex]
5) Here we have
The product of marginal distribution given by
[tex]f_x(x) f_y(y) = ( \frac{1}{20} +\frac{3x^{2} }{38000})( \frac{1}{20} +\frac{3y^{2} }{38000})[/tex] =[tex]\frac{(3x^2+1900)(3y^2+1900)}{1444000000}[/tex]
≠ f(x,y)
X and Y are not independent variables since the product of the marginal distribution is not joint probability distribution function.
6) Here we have the conditional probability of Y given X = x and the conditional probability of X given that Y = y is given by
[tex]h(y\mid x) =\frac{f(x,y))}{f_{X}\left (x \right )}[/tex]= Here we have
[tex]h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}[/tex]
Similarly, the the conditional probability of X given that Y = y is given by
[tex]h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}[/tex]
7) Here we have
When the pressure in the left tire is at least 25 psi gives
[tex]K\int\limits^{25}_{20} \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx[/tex]
Since x = 22 psi, we have
[tex]K\int\limits^{25}_{20} \frac{38000\cdot 25^2+38000y^2}{3\cdot 25^2+19000} {} \, dx = K \int\limits^{25}_{20} 10.066y^2+6291.39, dx = 57041.942\times \frac{3}{380000}[/tex]= 0.45033
For P(Y≥25) we have
[tex]K \int\limits^{30}_{25} 10.066y^2+6291.39, dx = 69624.72\times \frac{3}{380000}[/tex] = 0.54967
8) The expected pressure is the conditional mean given by
[tex]E(Y\mid x) = K\int\limits^{30}_{20} yh(y \mid x)\, dy[/tex]
[tex]E(Y\mid x) = K\int\limits^{30}_{20} 10.066y^3+6291.39y\, dy = \frac{3}{380000} \times 3208609.27153[/tex]
= 25.33 psi
The standard deviation is given by
[tex]Standard \, deviation =\sqrt{Variance}[/tex]
Variance = [tex]K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy[/tex]
[tex]=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy[/tex]
= [tex]\frac{3}{380000} \times 1047259.78[/tex] = 8.268
The standard deviation = √8.268 = 2.875.
