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 need to understand the binomial distribution. In this case, each free throw attempt is a Bernoulli trial, with a probability of success (making the shot) being 0.7. The number of successful attempts out of 120 will follow a binomial distribution.
(A) The true probability distribution of X follows a binomial distribution with parameters n = 120 (number of trials) and p = 0.7 (probability of success on each trial).
(B) The mean of a binomial distribution is given by μ = np, where n is the number of trials and p is the probability of success. In this case, μ = 120 * 0.7 = 84.
The variance of a binomial distribution is given by σ^2 = np(1-p). In this case, σ^2 = 120 * 0.7 * (1 - 0.7) = 25.2. Therefore, the mean of the number of successful throws is 84 and the variance is 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 values of n, the binomial distribution can be well approximated by a normal distribution with mean np and variance np(1-p).
In our case, n = 120, p = 0.7, μ = np = 120 * 0.7 = 84, and σ^2 = np(1-p) = 120 * 0.7 * (1 - 0.7) = 25.2. The standard deviation is then σ = sqrt(σ^2) = sqrt(25.2) ≈ 5.02.
Now, we convert the problem to a normal approximation. We want to find the probability of making at least 90 successful free throws, which is equivalent to finding the probability that X is greater than or equal to 90. We can standardize this value using the z-score formula: z = (x - μ) / σ, where x is the value we are interested in.
In this case, the z-score for x = 90 would be z = (90 - 84) / 5.02 ≈ 1.20. We can use a z-table or a calculator to find the area under the standard normal curve to the right of z = 1.20. This will give us the approximate 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). We want to find the probability that 80 ≤ X ≤ 100.
First, we find the z-scores for x = 80 and x = 100:
z1 = (80 - 84) / 5.02 ≈ -0.80
z2 = (100 - 84) / 5.02 ≈ 3.19
We can then use a z-table or a calculator to find the area under the standard normal curve between z1 and z2 to approximate the probability.