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.
(B) What is the probability that less than 330 attempts are successful?

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

Answer: (A)

Verified step by step guidance

1. Step 1: Identify the type of distribution. Since the problem involves a fixed number of independent trials (500 nesting attempts) with a constant probability of success (70%), this is a binomial distribution. The parameters are n = 500 (number of trials) and p = 0.7 (probability of success).
2. Step 2: Define the random variable. Let X represent the number of successful nesting attempts. X follows a binomial distribution, X ~ Binomial(n=500, p=0.7). The goal is to find P(X < 330), the probability that fewer than 330 attempts are successful.
3. Step 3: Approximate the binomial distribution using a normal 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 μ = 500 * 0.7 and σ = sqrt(500 * 0.7 * 0.3).
4. Step 4: Apply the continuity correction. Since the binomial distribution is discrete and the normal distribution is continuous, apply a continuity correction by finding P(X < 330) as P(X ≤ 329.5) in the normal distribution.
5. Step 5: Standardize the value for the normal distribution. Use the z-score formula z = (X - μ) / σ to convert X = 329.5 into a z-score. Then, use standard normal distribution tables or software to find the cumulative probability corresponding to this z-score, which gives P(X < 330).