Suppose a basketball player successfully makes 70% of the free throw shots they attempt. Let X

represent the number of successful attempts in 120 independent attempted free throws.
(A) What is the true probability distribution of X?
(B) Find the mean and variance of the number of successful throws.
(C) Approximate probability that the player will successfully make at least 90 of the attempted free
throws. Justify the approximation you use.
(D) Using the same approximation method from part (C), approximate the probability that the player
will successfully make between 80 and 100 of the attempted free throws.

To answer these questions, we will use the binomial probability distribution formula. The binomial distribution is appropriate here because each free throw attempt is independent and either a success or a failure.

(A) The true probability distribution of X can be given by the binomial distribution, which is defined as:

P(X=k) = C(n, k) * p^k * (1-p)^(n-k)

Where:
- P(X=k) is the probability of having exactly k successful throws (k successes out of n attempts).
- C(n, k) is the binomial coefficient, also known as "n choose k" or the number of ways to choose k successes out of n trials.
- p is the probability of a single success (in this case, the probability of making a free throw).
- n is the number of independent attempts (in this case, 120 free throws).

(B) The mean (expected value) and variance of the number of successful throws can be calculated using the following formulas:

Mean (μ) = n * p

Variance (σ^2) = n * p * (1-p)

Using the given information, the mean is μ = 120 * 0.7 = 84 successful throws, and the variance is σ^2 = 120 * 0.7 * (1-0.7) = 25.2.

(C) To approximate the probability that the player will successfully make at least 90 of the attempted free throws, we can use the normal approximation to the binomial distribution. For large n and a moderate value of p (not close to 0 or 1), we can use the normal distribution to approximate the binomial distribution.

First, we calculate the mean and standard deviation of the binomial distribution:

Mean (μ) = n * p = 120 * 0.7 = 84

Standard Deviation (σ) = √(n * p * (1-p)) = √(120 * 0.7 * (1-0.7)) = √(25.2) ≈ 5.02

Next, we use the normal distribution to approximate the probability:

P(X ≥ 90) ≈ 1 - P(z < (90 - μ) / σ)

P(z < (90 - μ) / σ) can be looked up in the standard normal distribution table or using a calculator to find the corresponding z-score and then use the table or calculator to find the probability.

(D) To approximate the probability that the player will successfully make between 80 and 100 of the attempted free throws, we can use the same normal approximation method as in part (C).

P(80 ≤ X ≤ 100) ≈ P((80 - μ) / σ ≤ z ≤ (100 - μ) / σ)

Again, we can use the standard normal distribution table or a calculator to find the corresponding z-scores, then use the table or calculator to find the probability.