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 =
= = 8.268
The standard deviation = √8.268 = 2.875.