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each rear tire on an experimental airplane issupposed to be filled with a pressure of 40 pound per squareinch(psi).Let X denote the actual air pressure for the right tireand Y denote the actual air pressure for the left tire.Suppose that Xand Y are random varibles with the jointdensity
f(x,y)= k(x^2+y^2, 30<-x<50;
30<-y<50
0, elsewhere
a.) find k
b.) find P(30<-x<-40 and 40<-Y<50)
c.) Find the probability that both tires areunderfilled.
show all steps to solve...thx

Respuesta :

MrRoyal

Answer:

a. K = 3/3920000

b. 0.26

c. 0.189

Step-by-step explanation:

f(x,y) is a join distribution, so the following condition must be satisfied

∫∫ f(x,y) dydx {-∞ to ∞} {-∞ to ∞} = 1

A. Finding K

∫∫ k(x² + y²) {30 to 50}{30 to 50} dydx = 1

k ∫∫ (x² + y²) {30 to 50}{30 to 50} dydx= 1

k ∫ (x²y + y³/3) {30 to 50}{30 to 50} dx= 1

k ∫ (50x² + 50³/3 - 30x² - 30³/3) dx {30 to 50} = 1

k ∫ (20x² + (50³ - 30³)/3) dx {30 to 50} = 1

k ∫ (20x² + 98000/3) dx {30 to 50} = 1

k(20x³/3 + 98000x/3) {30,50} = 1

k(20 * 50³/3 + 98000 * 50/3 - 20 * 30³/3 - 98000 * 30/3) = @

k(3920000)/3 = 1

K = 3/3920000

b.

P(30<-x<-40 and 40<-Y<50) = ∫∫ k(x² + y²) {30 to 40}{40 to 50} dydx

k ∫∫ (x² + y²) {30 to 40}{40 to 50} dydx

k ∫ (x²y + y³/3) {30 to 40}{40 to 50} dx=

k ∫ (50x² + 50³/3 - 40x² - 40³/3) dx {30 to 40}

k ∫ (10x² + (50³ - 40³)/3) dx {30 to 40}

k ∫ (10x² + 61000/3) dx {30 to 40}

k(10x³/3 + 61000x/3) {30,40}

k(10 * 40³/3 + 61000 * 40/3 - 10 * 30³/3 - 61000 * 30/3)

k(980000/3)

But k = 3/3920000

So

P(30<-x<-40 and 40<-Y<50) = 3/3920000 * 980000/3

P(30<-x<-40 and 40<-Y<50) = 0.25

c. Probability that both tries are undefined.

This means that both pressure are between 30 and 40.

So, we'll solve for P(30<-x<-40 and 30<-Y<40)

P(30<-x<-40 and 30<-Y<40) = ∫∫ k(x² + y²) {30 to 40}{30 to 40} dydx

k ∫∫ (x² + y²) {30 to 40}{30 to 40} dydx

k ∫ (x²y + y³/3) {30 to 40}{30 to 40} dx=

k ∫ (40x² + 40³/3 - 30x² - 30³/3) dx {30 to 40}

k ∫ (10x² + (40³ - 30³)/3) dx {30 to 40}

k ∫ (10x² + 37000/3) dx {30 to 40}

k(10x³/3 + 37000x/3) {30,40}

k(10 * 40³/3 + 37000 * 40/3 - 10 * 30³/3 - 37000 * 30/3)

k(740000/3)

But k = 3/3920000

So

P(30<-x<-40 and 30<-Y<40) = 3/3920000 * 740000/3

P(30<-x<-40 and 30<-Y<40) = 0.188775510204081

P(30<-x<-40 and 30<-Y<40) = 0.189

The value of k and respective probabilities are; k = 7.653 * 10⁻⁷ and P(30 < -x < -40 and 40 < -Y < 50) = 0.25

What is the Probability Density Function?

A) We are given the joint density as;

f(x,y)= k(x² + y², 30 < -x < 50; 30 < -y < 50

To find K, we will use double integral;

∫₅₀³⁰∫³⁰₅₀ k(x² + y²) dydx = 1

k∫₅₀³⁰[(x²y + y³/3)]₅₀³⁰ dx= 1

k∫₅₀³⁰(50x² + 50³/3 - 30x² - 30³/3) dx  = 1

k∫₅₀³⁰(20x² + 98000/3) dx = 1

k[(20x³/3 + 98000x/3)]₅₀³⁰ = 1

k((20 × 50³/3) + (98000 × 50/3) - (20 × 30³/3) - (98000 × 30/3) = 1

k(3920000)/3 = 1

K = 3/3920000 = 7.653 * 10⁻⁷

B. We want to find P(30 < -x < -40 and 40 < -Y < 50) . Thus, we can express the probability using double integral as;

k∫₄₀³⁰∫₅₀⁴⁰ (x² + y²) dydx

k∫₄₀³⁰[(x²y + y³/3)]₅₀⁴⁰ dx

k∫₄₀³⁰(50x² + 50³/3 - 40x² - 40³/3) dx

k∫₄₀³⁰(10x² + 61000/3) dx

k[(10x³/3 + 61000x/3)]₄₀³⁰

k((10 * 40³/3) + (61000 * 40/3) - (10 * 30³/3) - (61000 * 30/3))

⇒ k(980000/3)

But k = 7.653 * 10⁻⁷

Thus;

P(30 < -x < -40 and 40 < -Y < 50) = 7.653 * 10⁻⁷ * 980000/3

P(30 < -x < -40 and 40 < -Y < 50) = 0.25

C) Probability that both tries are undefined means that both pressures are between 30 and 40.

Thus, this is; P(30 < -x < -40 and 30 < -Y < 40)

= ∫₄₀³⁰∫₄₀³⁰ k(x² + y²) dydx

k∫₄₀³⁰∫₄₀³⁰ (x² + y²) dydx

k∫₄₀³⁰](x²y + y³/3)]₄₀³⁰ dx

k∫₄₀³⁰(40x² + 40³/3 - 30x² - 30³/3) dx

k∫₄₀³⁰(10x² + 37000/3)

k[(10x³/3 + 37000x/3)]₄₀³⁰

k((10 * 40³/3) + (37000 * 40/3) - (10 * 30³/3) - (37000 * 30/3))

= k(740000/3)

But k = 7.653 * 10⁻⁷

Thus;

P(30 < -x < -40 and 30 < -Y < 40) = 7.653 * 10⁻⁷ * 740000/3

P(30 < -x < -40 and 30 < -Y < 40) = 0.189

Read more about probability density at; https://brainly.com/question/2500166

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