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)

7) 0.54967
8) 25.33 psi
σ = 2.875
Step-by-step explanation:
1) Here we have









2) The probability that both tires are underfilled
P(X≤26,Y≤26) =








= 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) =
+
+ 
= K (
+
+
)
=
= 3.6775 × 10⁻²
4) The marginal pressure distribution in the right tire is







5) Here we have
The product of marginal distribution given by
=
≠ 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

Similarly, the the conditional probability of X given that Y = y is given by

7) Here we have
When the pressure in the left tire is at least 25 psi gives

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

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.