Use the appropriate normal distribution to approximate the resulting binomial distribution. A basketball player has a 75% chance of making a free throw. What is the probability of her making 80 or more free throws in 120 trials?
A) The probability is 0.9822.
B) The probability is 0.9864.
C) The probability is 0.9744.
D) The probability is 0.9898.

Question

Quizplus · Accepted Answer

To approximate the binomial distribution with a normal distribution, we use the mean (μ) and standard deviation (σ) of the binomial distribution. For n trials with a success probability of p, μ = np and σ = sqrt(np(1-p)). Here, n = 120 and p = 0.75, so μ = 120*0.75 = 90 and σ = sqrt(120*0.75*0.25) = sqrt(22.5) ≈ 4.7434. To find the probability of making 80 or more free throws, we calculate the z-score for 79.5 (using continuity correction) and look up the corresponding probability in the z-table or use a calculator. The z-score is (79.5 - 90) / 4.7434 ≈ -2.21. The probability of z > -2.21 is approximately 0.9864, indicating the probability of making 80 or more free throws is about 0.9864.

Use the Appropriate Normal Distribution to Approximate the Resulting Binomial