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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鈥擷 for the right tire and Y for the left tire, with joint pdf fsx, yd 5 5 Ksx2 1 y2 d 20 # x # 30, 20 # y # 30 0 otherwise

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{K(}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{)}}}&{,{\rm{20}} \le {\rm{x}} \le {\rm{30,20}} \le {\rm{y}} \le {\rm{30}}}\\{\rm{0}}&{,{\rm{ otherwise }}}\end{array}} \right.\)

a. What is the value of K?

b. What is the probability that both tires are underfilled?

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

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

e. Are X and Y independent rv鈥檚?z

Short Answer

Expert verified

a. The value of \({\rm{K = }}\frac{{\rm{3}}}{{{\rm{380,000}}}}\);

b. The probability is \({\rm{P(X > 26}}\)and \({\rm{Y < 26) = 0}}{\rm{.3024;}}\)

c. The probability is \({\rm{P(|X - Y|}} \le 2) = 0.3594\);

d. The (marginal) distribution of air pressure\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{10{\rm{ \times K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}}&{,20 \le x \le 30}\\0&{,{\rm{ otherwise }}}\end{array}} \right.\)鈥

e. The random variables are not independent

Step by step solution

01

Definition of Probability

Probability denotes the possibility of something happening. It is a field of mathematics that studies the probability of a random event occurring.From\({\rm{0}}\)to\({\rm{1}}\), the value is expressed. Probability is a mathematical concept that predicts how likely occurrences are to occur.

02

Find the value of K?

Let \({\rm{X}}\)and \(Y\)be continuous random variables.

The joint probability density function

\({\rm{f(x,y)}}\)is a function that is non-negative and for which

\(\int_{ - \infty }^\infty {\int_{ - \infty }^\infty f } {\rm{(x,y)dxdy = 1}}\)

For every adequate set \({\rm{A}}\)the following holds

(a):

From relation

\(\int_{ - \infty }^\infty {\int_{ - \infty }^\infty f } {\rm{(x,y)dxdy = 1}}\)

we can find value of \({\rm{K}}\).

The area \(20 \le x \le {\rm{30,20}} \le {\rm{y}} \le 30\) is a simple square, therefore the following holds

\(\begin{array}{l}\int_{ - \infty }^\infty {\int_{ - \infty }^\infty f } {\rm{(x,y)dxdy = }}\int_{{\rm{20}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{30}}} {\rm{K}} } \left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dxdy = K}}\int_{{\rm{20}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{2}}}} } {\rm{dydx + K}}\int_{{\rm{20}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{y}}^{\rm{2}}}} } {\rm{dxdy}}\\{\rm{ = K}}\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{2}}}} \left( {\left. {\rm{y}} \right|_{{\rm{20}}}^{{\rm{30}}}} \right){\rm{dx + K}}\int_{{\rm{20}}}^{{\rm{30}}} {{{\rm{y}}^{\rm{2}}}} \left( {\left. {\rm{x}} \right|_{{\rm{20}}}^{{\rm{30}}}} \right){\rm{dy}}\\{\rm{ = }}\left. {{\rm{10K \times }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{30}}}{\rm{ + }}\left. {{\rm{10K \times }}\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{30}}}\\{\rm{ = 2 \times 10K}}\left( {\frac{{{\rm{3}}{{\rm{0}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - }}\frac{{{\rm{2}}{{\rm{0}}^{\rm{3}}}}}{{\rm{3}}}} \right)\\{\rm{ = 20K \times }}\frac{{{\rm{19,000}}}}{{\rm{3}}}\\{\rm{ = }}\frac{{{\rm{380,000}}}}{{\rm{3}}}{\rm{ \times K}}\end{array}\)

Hence, we have

\(\frac{{{\rm{380,000}}}}{{\rm{3}}}{\rm{ \times K = 1}}\)

or equally

Therefore,The value of \({\rm{K = }}\frac{{\rm{3}}}{{{\rm{380,000}}}}\)

03

Step 3:  Find the probability that both tires are underfilled?

(b):

The underfill limit is 26 psi, therefore we need the following probability

Therefore, The probability is \({\rm{P(X > 26}}\)and \({\rm{Y < 26) = 0}}{\rm{.3024;}}\)

04

Step 4:  Find the probability that the difference in air pressure between the two tires?

(c):

We need to find subset of

\(20 \le x \le 30,\;\;\;20 \le y \le 30\)

for which the difference of \({\rm{X}}\)and \({\rm{Y}}\) is at most \({\rm{2}}\). We can represent the distance between two points \({\rm{x}}\) and \({\rm{y}}\)as \({\rm{|x - y|}}\).

Therefore, we need to find probability of event

\(\{ |{\rm{X - Y|}} \le 2\} \)

Look at the picture. We need to integrate over region \({\rm{II}}\). The following holds

We have that

\(\int_{{\rm{20}}}^{{\rm{28}}} {\int_{{\rm{x + 2}}}^{{\rm{30}}} {\rm{f}} } {\rm{(x,y)dydx = }}\int_{{\rm{22}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{x - 2}}} {\rm{f}} } {\rm{(x,y)dydx}}\)

however, see the following algebra behind the integrals:

\(\begin{array}{l}{{\rm{I}}_{\rm{1}}}{\rm{ = }}\int_{{\rm{20}}}^{{\rm{28}}} {\int_{{\rm{x + 2}}}^{{\rm{30}}} {\rm{K}} } \left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dydx}}\\{\rm{ = K}}\int_{{\rm{20}}}^{{\rm{28}}} {\int_{{\rm{x + 2}}}^{{\rm{30}}} {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)} } {\rm{dydx}}\\{\rm{ = K}}\int_{{\rm{20}}}^{{\rm{28}}} {\int_{{\rm{x + 2}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{2}}}} } {\rm{dydx + K}}\int_{{\rm{20}}}^{{\rm{28}}} {\int_{{\rm{x + 2}}}^{{\rm{30}}} {{{\rm{y}}^{\rm{2}}}} } {\rm{dydx}}\\{\rm{ = K}}\int_{{\rm{28}}}^{{\rm{28}}} {{{\rm{x}}^{\rm{2}}}\left( {\left. {\rm{y}} \right|_{{\rm{x + 2}}}^{{\rm{30}}}} \right)} {\rm{dx + K}}\int_{{\rm{20}}}^{{\rm{28}}} {\left( {\left. {\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{x + 2}}}^{{\rm{30}}}} \right)} {\rm{dx}}\\{\rm{ = K}}\underbrace {\int_{{\rm{20}}}^{{\rm{28}}} {{{\rm{x}}^{\rm{2}}}} {\rm{(30 - x - 2)dx}}}_{{{\rm{I}}_{\rm{3}}}}{\rm{ + }}\frac{{\rm{K}}}{{\rm{3}}}\underbrace {\int_{{\rm{20}}}^{{\rm{28}}} {\left( {{\rm{3}}{{\rm{0}}^{\rm{3}}}{\rm{ - (x + 2}}{{\rm{)}}^{\rm{3}}}} \right)} {\rm{dx}}}_{{{\rm{I}}_{\rm{4}}}}\end{array}\)

where we have

\(\begin{array}{l}{{\rm{I}}_{\rm{3}}}{\rm{ = 28}}\int_{{\rm{20}}}^{{\rm{28}}} {{{\rm{x}}^{\rm{2}}}} {\rm{dx - }}\int_{{\rm{20}}}^{{\rm{28}}} {{{\rm{x}}^{\rm{3}}}} {\rm{dx}}\\{\rm{ = }}\left. {{\rm{28 \times }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{28}}}{\rm{ - }}\left. {\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{20}}}^{{\rm{28}}}\\{\rm{ = 16,554}}{\rm{.67,}}\end{array}\)

similarly we get

\(\begin{array}{l}{{\rm{I}}_{\rm{4}}}{\rm{ = }}\int_{{\rm{20}}}^{{\rm{28}}} {\rm{3}} {{\rm{0}}^{\rm{3}}}{\rm{dx - }}\int_{{\rm{20}}}^{{\rm{28}}} {{{\rm{f}}_{\rm{1}}}} {\rm{(x)dx}}\\{\rm{ = 3}}{{\rm{0}}^{\rm{3}}}{\rm{ \times 8 - }}\left. {\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{20}}}^{{\rm{28}}}{\rm{ - }}\left. {{\rm{6 \times }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{28}}}{\rm{ - 12 \times }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\left. {{\rm{8 \times x}}} \right|_{{\rm{20}}}^{{\rm{28}}}\\{\rm{ = 72,064}}{\rm{.}}\end{array}\)

05

Explain function

Where the function \({{\rm{f}}_{\rm{1}}}\) is

\(\begin{array}{l}{{\rm{f}}_{\rm{1}}}{\rm{(x) = (x + 2}}{{\rm{)}}^{\rm{3}}}\\{\rm{ = }}{{\rm{x}}^{\rm{3}}}{\rm{ + 3 \times }}{{\rm{x}}^{\rm{2}}}{\rm{ \times 2 + 3 \times x \times }}{{\rm{2}}^{\rm{2}}}{\rm{ + }}{{\rm{2}}^{\rm{3}}}\\{\rm{ = }}{{\rm{x}}^{\rm{3}}}{\rm{ + 6}}{{\rm{x}}^{\rm{2}}}{\rm{ + 12x + 8}}\end{array}\)

The first integral is

\(\begin{array}{l}{{\rm{I}}_{\rm{1}}}{\rm{ = K \times 16,554}}{\rm{.67 + }}\frac{{\rm{K}}}{{\rm{3}}}{\rm{ \times 72,064}}\\{\rm{ = 0}}{\rm{.3203}}\end{array}\)

For the second integral, we have

\(\begin{array}{l}{{\rm{I}}_{\rm{2}}}{\rm{ = }}\int_{{\rm{22}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{x - 2}}} {\rm{K}} } \left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dydx}}\\{\rm{ = K}}\int_{{\rm{22}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{x - 2}}} {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)} } {\rm{dydx}}\\{\rm{ = K}}\int_{{\rm{22}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{x - 2}}} {{{\rm{x}}^{\rm{2}}}} } {\rm{dydx + K}}\int_{{\rm{22}}}^{{\rm{30}}} {\int_{{\rm{20}}}^{{\rm{x - 2}}} {{{\rm{y}}^{\rm{2}}}} } {\rm{dydx}}\\{\rm{ = K}}\int_{{\rm{20}}}^{{\rm{30}}} {\left( {\left. {\rm{y}} \right|_{{\rm{20}}}^{{\rm{x - 2}}}} \right)} {\rm{dx + K}}\int_{{\rm{22}}}^{{\rm{30}}} {\left( {\left. {\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{x - 2}}}} \right)} {\rm{dx}}\\{\rm{ = K}}\underbrace {\int_{{\rm{22}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{2}}}} {\rm{(x - 2 - 20)dx}}}_{{{\rm{I}}_{\rm{5}}}}{\rm{ + }}\frac{{\rm{K}}}{{\rm{3}}}\underbrace {\int_{{\rm{22}}}^{{\rm{30}}} {\left( {{{{\rm{(x - 2)}}}^{\rm{3}}}{\rm{ - 2}}{{\rm{0}}^{\rm{3}}}} \right)} {\rm{dx}}}_{{{\rm{I}}_{\rm{6}}}}\end{array}\)

where

\(\begin{array}{l}{{\rm{I}}_{\rm{5}}}{\rm{ = }}\int_{{\rm{22}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{3}}}} {\rm{dx - 22}}\int_{{\rm{22}}}^{{\rm{30}}} {{{\rm{x}}^{\rm{2}}}} {\rm{dx}}\\{\rm{ = }}\left. {\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{22}}}^{{\rm{30}}}{\rm{ - }}\left. {{\rm{22 \times }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{22}}}^{{\rm{30}}}\\{\rm{ = 24,021}}{\rm{.33,}}\end{array}\)

and also

\(\begin{array}{l}{{\rm{I}}_{\rm{6}}}{\rm{ = }}\int_{{\rm{22}}}^{{\rm{30}}} {\underbrace {{{{\rm{(x - 2)}}}^{\rm{3}}}}_{{{\rm{f}}_{\rm{2}}}{\rm{(x)}}}} {\rm{dx - 2}}{{\rm{0}}^{\rm{3}}}{\rm{ \times }}\int_{{\rm{22}}}^{{\rm{30}}} {\rm{d}} {\rm{x}}\\{\rm{ = }}\left. {\frac{{{{\rm{x}}^{\rm{4}}}}}{{\rm{4}}}} \right|_{{\rm{22}}}^{{\rm{30}}}{\rm{ - }}\left. {{\rm{6 \times }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{22}}}^{{\rm{30}}}{\rm{ + }}\left. {{\rm{12 \times }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right|_{{\rm{22}}}^{{\rm{30}}}{\rm{ - }}\left. {{\rm{8 \times x}}} \right|_{{\rm{22}}}^{{\rm{30}}}{\rm{ - }}\left. {{\rm{2}}{{\rm{0}}^{\rm{3}}}{\rm{ \times x}}} \right|_{{\rm{22}}}^{{\rm{30}}}\\{\rm{ = 49,664}}\end{array}\)

Where the function \({{\rm{f}}_{\rm{2}}}\)can be written as

\(\begin{array}{l}{{\rm{f}}_{\rm{2}}}{\rm{(x) = (x - 2}}{{\rm{)}}^{\rm{3}}}\\{\rm{ = }}{{\rm{x}}^{\rm{3}}}{\rm{ - 3 \times }}{{\rm{x}}^{\rm{2}}}{\rm{ \times 2 + 3 \times x \times }}{{\rm{2}}^{\rm{2}}}{\rm{ - }}{{\rm{2}}^{\rm{3}}}\\{\rm{ = }}{{\rm{x}}^{\rm{3}}}{\rm{ - 6 \times }}{{\rm{x}}^{\rm{2}}}{\rm{ + 12 \times x - 8}}\end{array}\)

The second integral is

\(\begin{array}{l}{{\rm{I}}_{\rm{2}}}{\rm{ = K \times 24,021}}{\rm{.33 + }}\frac{{\rm{K}}}{{\rm{3}}}{\rm{ \times 49,664}}\\{\rm{ = 0}}{\rm{.3203}}\end{array}\)

The probability is \({\rm{P(|X - Y|}} \le 2) = 0.3594\);

06

Step 6: Determine the (marginal) distribution of air pressure in the right tire alone

(d):

The continuous random variable's marginal probability mass function \({\rm{X}}\)is

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\int_{ - \infty }^\infty f {\rm{(x,y)dy,}}\;\;\;{\rm{ for }} - \infty < {\rm{x}} < \infty \)

The continuous random variable's marginal probability mass function \({\rm{Y}}\)is

\({{\rm{f}}_{\rm{Y}}}{\rm{(x) = }}\int_{ - \infty }^\infty f {\rm{(x,y)dx,}}\;\;\;{\rm{ for }} - \infty < y < \infty \)

The following holds

\(\begin{array}{l}{{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\int_{{\rm{20}}}^{{\rm{30}}} {\rm{K}} \left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){\rm{dy}}\\{\rm{ = 10K}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}\left. {{\rm{K}}\frac{{{{\rm{y}}^{\rm{3}}}}}{{\rm{3}}}} \right|_{{\rm{20}}}^{{\rm{30}}}\\{\rm{ = 10 \times K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05,}}\\20 \le x \le 30,\\{f_X}(x) = 0,x \notin (20,30).\end{array}\)

The same marginal distribution would be obtained for \({\rm{Y}}\) if we substitute \({\rm{x}}\)with \({\rm{y}}\).

The distribution of air pressure is \({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\left\{ {\begin{array}{*{20}{l}}{10{\rm{ \times K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}}&{,20 \le x \le 30}\\0&{,{\rm{ otherwise }}}\end{array}} \right.\)

07

Step 7:Explain X And Y are Independent?

(e):

Two independent variables If and only if, \({\rm{X}}\) and \({\rm{Y}}\) are independent.1. \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\), for every \({\rm{(x,y)}}\) and when \({\rm{X}}\)and \({\rm{Y}}\)discrete rv's,

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\)and when \({\rm{X}}\)and \({\rm{Y}}\)continuous rv's,

otherwise they are dependent.

The random variables are obviously not independent. Because of this, functions

\(\begin{array}{c}{\rm{f(x,y) = K}}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right){{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\\{\rm{ = }}\left( {{\rm{10 \times K}}{{\rm{x}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}} \right){\rm{ \times }}\left( {{\rm{10 \times K}}{{\rm{y}}^{\rm{2}}}{\rm{ + 0}}{\rm{.05}}} \right){\rm{,}}\end{array}\)

are not equal

Therefore,The random variables are not independent

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