Question

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable X for the right tire and Y for the left tire, wiht joint pdf
f(x, y) = {K(x2 + y2) 0 20 le x le 30 and 20 le

1) What is the value of K?

2) What is the probability that both tires are underfilled?

3) What is the probability that the difference in air pressure between the two tires is at most 2 psi?

4) Determine the marginal distribution of air pressure in the right tire alone.

5) Are X and Y independent random variables?

6) Determine the conditional pdf of Y given that X=x and the conditional pdf of X given that Y=y.

7) If the pressure in the right tire is found to be 22 psi, what is the probability that the left tire has a pressure of at least 25 psi? Compare this to P(Y>=25)

8) If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left tire, and what is the standard deivation of pressure in this tire?

Answer 1

Answer:

1) K = 7.895 × 10⁻⁶

2) 0.3024

3)  3.6775 × 10⁻²

4) $f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) X and Y are not independent variables

6)

$h(x\mid y) = \frac{38000x^2+38000y^2}{3y^2+19000}$

7)  0.54967

8)  25.33 psi

σ = 2.875

Step-by-step explanation:

1) Here we have

$f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}$

$\int_{x}\int_{y} f(x, y)dydx = 1$    

$\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1$

$K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1$

$K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1$

$K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1$

$K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1$

$K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1$

$K =\frac{3}{380000}$

2) The probability that both tires are underfilled

P(X≤26,Y≤26) =

$\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx$

$=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx$

$= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx$

$K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx$

$K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})$

$K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})$

$38304\times K =\frac{3\times38304}{380000}$

= 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, $| x-y |$ ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, $\sqrt{(x-y)^2}$ ≤ 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( $| x-y |$ ≤2) =  $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ +  $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ +  $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$

= K (  $\int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx$ +  $\int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx$ +  $\int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx$)

= $K\left [ \left (\frac{14804}{15} \right )+\left (\frac{8204}{15} \right ) +\left (\frac{46864}{15} \right )\right ] = \frac{3}{380000}\times \frac{69872}{15} =\frac{4367}{118750}$ = 3.6775 × 10⁻²

4) The marginal pressure distribution in the right tire is

$f_{x}\left ( x \right )=\int_{y} f(x ,y)dy$

$=K( \right )\int_{y}(x^{2} +y^{2})dy)$

$= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})$

$K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})$

$K(10x^{2}+\frac{19000}{3})}$

$\frac{3}{38000} (10x^{2}+\frac{19000}{3})}$

$= \frac{1}{20} +\frac{3x^{2} }{38000}$

$f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}$

5) Here we have

The product of marginal distribution given by

$f_x(x) f_y(y) = ( \frac{1}{20} +\frac{3x^{2} }{38000})( \frac{1}{20} +\frac{3y^{2} }{38000})$ =$\frac{(3x^2+1900)(3y^2+1900)}{1444000000}$

≠ 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

$h(y\mid x) =\frac{f(x,y))}{f_{X}\left (x \right )}$=  Here we have

 

$h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}$

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

$h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}$

7) Here we have

When the pressure in the left tire is at least 25 psi gives

$K\int\limits^{25}_{20} \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx$

Since x = 22 psi, we have

$K\int\limits^{25}_{20} \frac{38000\cdot 25^2+38000y^2}{3\cdot 25^2+19000} {} \, dx = K \int\limits^{25}_{20} 10.066y^2+6291.39, dx = 57041.942\times \frac{3}{380000}$= 0.45033

For P(Y≥25) we have

$K \int\limits^{30}_{25} 10.066y^2+6291.39, dx = 69624.72\times \frac{3}{380000}$ = 0.54967

8) The expected pressure is the conditional mean given by

$E(Y\mid x) = K\int\limits^{30}_{20} yh(y \mid x)\, dy$

$E(Y\mid x) = K\int\limits^{30}_{20} 10.066y^3+6291.39y\, dy = \frac{3}{380000} \times 3208609.27153$

= 25.33 psi

The standard deviation is given by

$Standard \, deviation =\sqrt{Variance}$

Variance = $K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy$

$=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy$

= $\frac{3}{380000} \times 1047259.78$ = 8.268

The standard deviation = √8.268 = 2.875.