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.

Question

Mathematics

Aniyah Graves

14 April, 07:02

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.

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Answers (1)

Know the Answer?

Tyree Bradley · Answer 1

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