A biologist is monitoring a large bird sanctuary where a particular bird species is known to have a 70% success rate for each nesting attempt (at least one chick fledges from the nest). This season, she observes 500 independent nesting attempts across the sanctuary.
(D) What is the probability that 330 – 370 nesting attempts are successful?

• A. $0.98$
• B. $0.96$
• C. $0.02$
• D. 0.04

Answer: (B)

Verified step by step guidance

1. Step 1: Recognize that this is a binomial probability problem. The success rate for each nesting attempt is 70% (p = 0.7), and the number of trials is 500 (n = 500). The goal is to find the probability that the number of successful nesting attempts falls between 330 and 370 (inclusive).
2. Step 2: Use the normal approximation to the binomial distribution. For large n, the binomial distribution can be approximated by a normal distribution with mean μ = n * p and standard deviation σ = sqrt(n * p * (1 - p)). Calculate these values: μ = 500 * 0.7 and σ = sqrt(500 * 0.7 * 0.3).
3. Step 3: Apply the continuity correction. Since we are looking for the probability that the number of successes is between 330 and 370, inclusive, adjust the range to 329.5 to 370.5 to account for the discrete nature of the binomial distribution.
4. Step 4: Standardize the values using the z-score formula: z = (x - μ) / σ. Compute the z-scores for 329.5 and 370.5 using the mean and standard deviation calculated in Step 2.
5. Step 5: Use the standard normal distribution table (or a statistical software) to find the probabilities corresponding to the z-scores from Step 4. Subtract the smaller probability from the larger probability to find the probability that the number of successful nesting attempts is between 330 and 370.