Question

Mo often plays basketball in his free time. Every Tuesday, he practices free throws. Mo has found out that he scores in 60 percent of the free throws, and that the throws can be considered independent.

(a) Next Tuesday, Mo plans to perform 12 free throws. What is the probability that Mo scores in more than 6 of the throws? Give the answer as a decimal number with three decimals, e.g., 0.125 and 0.053.

(b) Over the next four weeks, Mo plans to perform a total of 58 free throws. For simplicity, we assume that despite all the practice, Mo will not get any better at free throws, so the probability of Mo scoring in a throw will always be 60 percent. Use the central limit theorem to find the (approximate) probability that Mo scores in more than 29 of the 58 throws. Give the answer as a decimal number with three decimals, e.g., 0.125 and 0.053.

Answer 1

To determine the probability of Mo scoring more than 6 out of 12 free throws, binomial probabilities for 7 or more successes must be summed. For 58 free throws, the central limit theorem allows us to approximate the probability using a normal distribution after calculating the mean and standard deviation for the binomial distribution and finding the corresponding z-score.

Step-by-step explanation:

To answer (a), we must calculate the probability that Mo scores on more than 6 out of 12 free throws with a success rate of 60% for each throw. This can be solved using the binomial probability formula. However, since the question asks for more than 6, we need to find the probabilities for 7, 8, 9, 10, 11, and 12 successful throws and sum them up.

To answer (b), since we are dealing with a larger number of trials (58 free throws), we can use the central limit theorem to approximate the probability that Mo scores on more than 29 throws. The central limit theorem states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is large enough. To apply the theorem, we need to calculate the mean (μ = np) and standard deviation (σ = √npq) of the binomial distribution and then use these to find the z-score for 29.5 (using a continuity correction since we're dealing with discrete data). The probability can then be found using a standard normal distribution table.