The best player on a basketball team makes 80% of all free throws. The second-best player makes 75% of all free throws. The third-best player makes 70% of all free throws. Based on their experimental probabilities, estimate the number of free throws each player will make in his or her next 60 attempts.

Expected number of free throws in 60 attempts:

        Best player = 48

        2nd best player = 45

        3rd best player = 42

Step-by-step explanation:

Solution:-

- The probability that best player makes free throw, p1 = 0.8

- The probability that second-best player makes free throw, p2 = 0.75

- The probability that third-best player makes free throw, p3 = 0.70

- Total number of attempts made in free throws, n = 60.

- The estimated number of free throws that any player makes is defined by:

                      E ( Xi ) = n*pi

Where,           Xi = Player rank

                      pi = Player rank probability

- Expected value for best player making the free throws would be:

                    E (X1) = n*p1

                              = 60*0.8

                              = 48 free throws

- Expected value for second-best player making the free throws would be:

                    E (X2) = n*p2

                              = 60*0.75

                              = 45 free throws

- Expected value for third-best player making the free throws would be:

                    E (X3) = n*p3

                              = 60*0.70

                              = 42 free throws

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

happy to

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