According to the Center for Disease Control (CDC), 1 in 88 children in the U.S. have been diagnosed with autism. Suppose you randomly select 50 children. Based on this information, answer questions 19 - 20. Question 19 1 Point HW 6 Question 19: Find the probability that no more than one child in the sample has been diagnosed with autism. Add your answer Question 20 1 Point HW 6 Question 20: Find the probability that more than one child in the sample has been diagnosed with autism. Add your answer

To calculate the probabilities, we can model the situation with a binomial distribution, where the probability of success (a child being diagnosed with autism) is \( p = \frac{1}{88} \) and the number of trials (children selected) is \( n = 50 \). For Question 19: To find the probability of no more than one child diagnosed with autism, we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Calculating for \( k = 0 \) and \( k = 1 \): 1. For \( k = 0 \): \[ P(X = 0) = \binom{50}{0} \left(\frac{1}{88}\right)^0 \left(\frac{87}{88}\right)^{50} \approx 0.32 \] 2. For \( k = 1 \): \[ P(X = 1) = \binom{50}{1} \left(\frac{1}{88}\right)^1 \left(\frac{87}{88}\right)^{49} \approx 0.18 \] Adding these probabilities gives: \[ P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.32 + 0.18 = 0.50 \] So for Question 19, \( P(X \leq 1) \approx 0.50 \). For Question 20: To find the probability of more than one child diagnosed, we use: \[ P(X > 1) = 1 - P(X \leq 1) \approx 1 - 0.50 = 0.50 \] So, for Question 20, \( P(X > 1) \approx 0.50 \). In summary, you can expect about a 50% chance that no more than one child is diagnosed and a 50% chance that more than one child is diagnosed. Statistically speaking, the world of probabilities is filled with surprises!