Question

(12 pts) Suppose a basketball player practicing shooting makes 45 percent of his shots. If the basketball player would like to know the likelihood of making x out of the next 25 shots, what distribution might he? Justify yours by checking conditions for the distribution you propose.

(9 pts) Using the distribution you proposed, calculate the following:

P(making 15 out of the next 25 shots)

P(making at least more than of the next 25)

P(making at most 11 than of the next 25)

(12 pts) Now calculate the:

P(making at least 1200 shots in 2500 shots attempted)

What distribution did you use? Why? Explain. (Hint: You must check conditions for the distribution you use and show all calculation steps.)

Answer 1

Solution:-

p = 0.45

The distribution is binomial distribution.

The conditions for binomial distribution are:

Each event has only one outcome.

Each outcome is independent.

There are finite number of outcomes.

np > 10 and np(1-p) > 10.

a) P(making 15 out of the next 25 shots) = 0.0520

x = 15

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x = 15) = 0.0520

c) P(making at most 11 than of the next 25) = 0.5426

x = 11

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x < 11) = 0.5426

d) P(making at least 1200 shots in 2500 shots attempted) = 0.00139

n = 2500

x = 1200

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 1200) = 0.00139

The distribution is binomial distribution.

The conditions for binomial distribution are:

Each event has only one outcome.

Each outcome is independent.

There are finite number of outcomes.