Suppose the weight and the circumference are independent. Find the probability that a randomly selected baseball will have a weight between 5.11 and 5.13 ounces and a circumference between 9.04 and 9.07 inches. (Round your answer to four decimal places.) .0571 Incorrect: Your answer is incorrect.

P ( A & B ) = 0.08751

Step-by-step explanation:

- the complete question is as follows:

Given:

" Sport and Leisure According to Major League Baseball rules, a baseball should weigh between 5 and 5.25 ounces and have a circumference of between 9 and 9.25 inches. Suppose the weight of a baseball (in ounces) has a uniform distribution with a = 5.085 and b = 5.155, and the circumference (in inches) has a uniform distribution with a = 9.0 and b = 9.1.

Find:

Suppose the weight and the circumference are independent. Find the probability that a randomly selected baseball will have a weight between 5.11 and 5.13 ounces and a circumference between 9.04 and 9.06 inches. "

Solution:

- The cumulative density function CDF for a uniform distribution is given as:

                                     F(x) = ( x - a ) / ( b - a )

- We will denote an RV for the weight of the ball as X. Its CDF would be given as:

                                     F(X) = ( X - a ) / ( b - a )

                                     a = 5.085 , b = 5.155

                                     F(X) = ( X - 5.085 ) / ( 0.07 )

- The probability of weight of a random ball that lies between 5.11 and 5.13 is:

                           F(5.13) - F(5.11) = [( 5.13 - 5.085 ) - ( 5.11 - 5.085 )] / ( 0.07 )

                                                    = [ 0.02 / 0.07 ]

                                                    = 2 / 7

- We will denote an RV for the circumference of the ball as Y. Its CDF would be given as:

                                     F(Y) = ( Y - a ) / ( b - a )

                                     a = 9.0 , b = 9.1

                                     F(Y) = ( Y - 9 ) / ( 0.1 )

- The probability of circumference of a random ball that lies between 9.04 and 9.07 is:

                           F(9.07) - F(9.04) = [( 9.07 - 9 ) - ( 9.04 - 9 )] / ( 0.1 )

                                                    = [ 0.03 / 0.1 ]

                                                    = 3 / 10

- We are asked that the randomly selected ball will have a weight that lies between 5.13 and 5.11 ounces and circumference between 9.04 and 9.07. While the two events A and B are independent we have:

                          P ( A & B ) = P ( A ) * P ( B )

Where,

                          P ( A ) = F(5.13) - F(5.11) = 2 / 7

                          P ( B ) = F(9.07) - F(9.04) = 3 / 10

Hence,

                          P ( A & B ) = ( 2 * 3 ) / ( 7 * 10 )

                                            = 0.08571

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5