Answer:
1) K = 7.895 × 10⁻⁶
2) 0.3024
3) 3.6775 × 10⁻²
4)
5) X and Y are not independent variables
6)
![h(x\mid y) = \frac{38000x^2+38000y^2}{3y^2+19000}](https://tex.z-dn.net/?f=h%28x%5Cmid%20y%29%20%20%3D%20%5Cfrac%7B38000x%5E2%2B38000y%5E2%7D%7B3y%5E2%2B19000%7D)
7) 0.54967
8) 25.33 psi
σ = 2.875
Step-by-step explanation:
1) Here we have
![f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}](https://tex.z-dn.net/?f=f%28x%2C%20y%29%20%3D%5Cbegin%7Bcases%7D%20%26%20%5Ctext%20%28x%5E%7B2%7D%2By%5E%7B2%7D%29%20%5Cright.%2020%5Cleq%20x%5Cleq%2030%20%26%20%5C%200%20%5C%2C%20Otherwise%5Cend%7Bcases%7D)
![\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1](https://tex.z-dn.net/?f=%5Cint_%7Bx%7D%28%20%5Cright%20%29%5Cint_%7By%7D%20f%28x%2C%20y%29dy%29dx%20%3D%201)
![K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1](https://tex.z-dn.net/?f=K%5Cint_%7Bx%7D%28%20%5Cright%20%29%5Cint_%7By%7D%28x%5E%7B2%7D%20%2By%5E%7B2%7D%29dy%29dx%20%3D%201)
![K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1](https://tex.z-dn.net/?f=K%5Cint_%7Bx%7D%28%20%28x%5E%7B2%7Dy%20%2B%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%29_%7B20%7D%5E%7B30%7D%29dx%20%3D%201)
![K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1](https://tex.z-dn.net/?f=K%5Cint_%7Bx%7D%28%20%28x%5E%7B2%7D%2830-20%29%29%20%2B%5Cfrac%7B30%5E%7B3%7D-20%5E%7B3%7D%7D%7B3%7D%29_%7B20%7D%5E%7B30%7D%29dx%20%3D%201)
![K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1](https://tex.z-dn.net/?f=K%5Cint_%7Bx%7D%28%20%2810x%5E%7B2%7D%29%2B%5Cfrac%7B19000%7D%7B3%7D%29_%7B20%7D%5E%7B30%7D%29dx%20%3D%201)
![K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1](https://tex.z-dn.net/?f=K%28%20%2810%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D%29%2B%5Cfrac%7B19000%7D%7B3%7Dx%29_%7B20%7D%5E%7B30%7D%29%3D%201)
![K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1](https://tex.z-dn.net/?f=K%28%20%2810%5Cfrac%7B30%5E%7B3%7D-20%5E%7B3%7D%7D%7B3%7D%29%2B%5Cfrac%7B19000%7D%7B3%7D%2830-20%29%29%29_%7B20%7D%5E%7B30%7D%29%20%3D%201)
![K =\frac{3}{380000}](https://tex.z-dn.net/?f=K%20%3D%5Cfrac%7B3%7D%7B380000%7D)
2) The probability that both tires are underfilled
P(X≤26,Y≤26) =
![\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx](https://tex.z-dn.net/?f=%5Cint_%7B20%7D%5E%7B26%7D%20%5Cint_%7B20%7D%5E%7B26%7DK%28x%5E%7B2%7D%2By%5E%7B2%7D%29dydx)
![=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx](https://tex.z-dn.net/?f=%3DK%5Cint_%7Bx%7D%28%20%5Cright%20%29%5Cint_%7By%7D%28x%5E%7B2%7D%20%2By%5E%7B2%7D%29dy%29dx)
![= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx](https://tex.z-dn.net/?f=%3D%20K%5Cint_%7Bx%7D%28%20%28x%5E%7B2%7Dy%20%2B%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%29_%7B20%7D%5E%7B26%7D%29dx)
![K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx](https://tex.z-dn.net/?f=K%5Cint_%7Bx%7D%28%20%28x%5E%7B2%7D%2826-20%29%29%20%2B%5Cfrac%7B26%5E%7B3%7D-20%5E%7B3%7D%7D%7B3%7D%29_%7B20%7D%5E%7B26%7D%29dx)
![K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx](https://tex.z-dn.net/?f=K%5Cint_%7Bx%7D%28%20%286x%5E%7B2%7D%29%2B%5Cfrac%7B9576%7D%7B3%7D%29_%7B20%7D%5E%7B26%7D%29dx)
![K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})](https://tex.z-dn.net/?f=K%28%20%286%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D%29%2B%5Cfrac%7B9576%7D%7B3%7Dx%29_%7B20%7D%5E%7B26%7D%29)
![K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})](https://tex.z-dn.net/?f=K%28%20%286%5Cfrac%7B26%5E%7B3%7D-20%5E%7B3%7D%7D%7B3%7D%29%2B%5Cfrac%7B9576%7D%7B3%7D%2826-20%29%29%29_%7B20%7D%5E%7B26%7D%29)
![38304\times K =\frac{3\times38304}{380000}](https://tex.z-dn.net/?f=38304%5Ctimes%20K%20%3D%5Cfrac%7B3%5Ctimes38304%7D%7B380000%7D)
= 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,
≤ 2}
{20 ≤ x ≤ 30, 20 ≤ y ≤ 30,
≤ 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(
≤2) =
+
+ ![\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx](https://tex.z-dn.net/?f=%5Cint_%7B22%7D%5E%7B28%7D%20%5Cint_%7Bx-2%7D%5E%7Bx%2B2%7DK%28x%5E%7B2%7D%2By%5E%7B2%7D%29dydx)
= K (
+
+
)
=
= 3.6775 × 10⁻²
4) The marginal pressure distribution in the right tire is
![f_{x}\left ( x \right )=\int_{y} f(x ,y)dy](https://tex.z-dn.net/?f=f_%7Bx%7D%5Cleft%20%28%20x%20%5Cright%20%29%3D%5Cint_%7By%7D%20f%28x%20%2Cy%29dy)
![=K( \right )\int_{y}(x^{2} +y^{2})dy)](https://tex.z-dn.net/?f=%3DK%28%20%5Cright%20%29%5Cint_%7By%7D%28x%5E%7B2%7D%20%2By%5E%7B2%7D%29dy%29)
![= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})](https://tex.z-dn.net/?f=%3D%20K%28%20%28x%5E%7B2%7Dy%20%2B%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%29_%7B20%7D%5E%7B30%7D%29)
![K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})](https://tex.z-dn.net/?f=K%28%20%28x%5E%7B2%7D%2830-20%29%29%20%2B%5Cfrac%7B30%5E%7B3%7D-20%5E%7B3%7D%7D%7B3%7D%29_%7B20%7D%5E%7B30%7D%29)
![K(10x^{2}+\frac{19000}{3})}](https://tex.z-dn.net/?f=K%2810x%5E%7B2%7D%2B%5Cfrac%7B19000%7D%7B3%7D%29%7D)
![\frac{3}{38000} (10x^{2}+\frac{19000}{3})}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B38000%7D%20%2810x%5E%7B2%7D%2B%5Cfrac%7B19000%7D%7B3%7D%29%7D)
![= \frac{1}{20} +\frac{3x^{2} }{38000}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B20%7D%20%2B%5Cfrac%7B3x%5E%7B2%7D%20%7D%7B38000%7D)
5) Here we have
The product of marginal distribution given by
=![\frac{(3x^2+1900)(3y^2+1900)}{1444000000}](https://tex.z-dn.net/?f=%5Cfrac%7B%283x%5E2%2B1900%29%283y%5E2%2B1900%29%7D%7B1444000000%7D)
≠ 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
= Here we have
![h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}](https://tex.z-dn.net/?f=h%28y%5Cmid%20x%29%20%3D%5Cfrac%7Bx%5E2%2By%5E2%7D%7B%5Cfrac%7B1%7D%7B20%7D%20%2B%5Cfrac%7B3x%5E2%7D%7B38000%7D%20%7D%20%3D%20%5Cfrac%7B38000x%5E2%2B38000y%5E2%7D%7B3x%5E2%2B19000%7D)
Similarly, the the conditional probability of X given that Y = y is given by
![h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}](https://tex.z-dn.net/?f=h%28x%5Cmid%20y%29%20%3D%5Cfrac%7Bx%5E2%2By%5E2%7D%7B%5Cfrac%7B1%7D%7B20%7D%20%2B%5Cfrac%7B3y%5E2%7D%7B38000%7D%20%7D%20%3D%20%5Cfrac%7B38000x%5E2%2B38000y%5E2%7D%7B3y%5E2%2B19000%7D)
7) Here we have
When the pressure in the left tire is at least 25 psi gives
![K\int\limits^{25}_{20} \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx](https://tex.z-dn.net/?f=K%5Cint%5Climits%5E%7B25%7D_%7B20%7D%20%20%5Cfrac%7B38000x%5E2%2B38000y%5E2%7D%7B3x%5E2%2B19000%7D%20%7B%7D%20%5C%2C%20dx)
Since x = 22 psi, we have
= 0.45033
For P(Y≥25) we have
= 0.54967
8) The expected pressure is the conditional mean given by
= 25.33 psi
The standard deviation is given by
![Standard \, deviation =\sqrt{Variance}](https://tex.z-dn.net/?f=Standard%20%5C%2C%20deviation%20%3D%5Csqrt%7BVariance%7D)
Variance = ![K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy](https://tex.z-dn.net/?f=K%5Cint%5Climits%5E%7B30%7D_%7B20%7D%20%5By-E%28Y%5Cmid%20x%29%20%5D%5E2h%28y%20%5Cmid%20x%29%5C%2C%20dy)
![=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy](https://tex.z-dn.net/?f=%3DK%5Cint%5Climits%5E%7B30%7D_%7B20%7D%20%5By-25.33%5D%5E2%2810.066y%5E2%2B6291.39%29%5C%2C%20dy)
=
= 8.268
The standard deviation = √8.268 = 2.875.