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.
(C) What is the probability that at least 330 nesting attempts are successful?

• A. $0.02$
• B. $0.005$
• C. $0.98$
• D. $0.03$

Answer: (C)

Verified step by step guidance

1. Step 1: Recognize that this is a binomial probability problem. The number of trials (n) is 500, the probability of success (p) for each trial is 0.7, and we are interested in the probability of at least 330 successes.
2. Step 2: To simplify the calculation, approximate the binomial distribution using a normal distribution. The mean (μ) and standard deviation (σ) of the binomial distribution can be calculated as follows: μ = n * p and σ = sqrt(n * p * (1 - p)).
3. Step 3: Convert the problem to a normal distribution problem. Use the continuity correction by adjusting the value of 330 to 329.5 (since we are looking for 'at least 330').
4. Step 4: Standardize the value 329.5 to a z-score using the formula: z = (X - μ) / σ, where X is the value of interest (329.5), μ is the mean, and σ is the standard deviation.
5. Step 5: Use the standard normal distribution table or a statistical software to find the probability corresponding to the calculated z-score. Subtract this probability from 1 to find the probability of at least 330 successes.