Answer:
P ( A & B ) = 0.08751
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