Let's clarify and solve the problem step by step. **Given:** - n = 1,000 (number of trials) - p = 0.9 (probability of success in each trial) - We are asked: **Can the binomial distribution with these parameters be approximated by a normal distribution?** --- ## Step 1: State the Binomial Distribution A binomial distribution is defined by: - Number of trials: **n** - Probability of success: **p** The mean (μ) and variance (σ²) are: - Mean: μ = n·p - Variance: σ² = n·p·(1-p) --- ## Step 2: Conditions for Normal Approximation A binomial distribution \( B(n, p) \) can be approximated by a normal distribution when **both** \( n·p \) and \( n·(1-p) \) are **greater than or equal to 5** (some sources use 10, but 5 is commonly accepted). This ensures the distribution is not too skewed and has enough data points for the normal approximation to be reasonable. --- ## Step 3: Check the Conditions Let's compute: - \( n·p = 1,000 × 0.9 = 900 \) - \( n·(1-p) = 1,000 × 0.1 = 100 \) Both 900 and 100 are much greater than 5. --- ## Step 4: Conclusion **Since both \( n·p \) and \( n·(1-p) \) are much greater than 5, the binomial distribution with n = 1,000 and p = 0.9 can be very well approximated by a normal distribution.** --- ## **Final Answer** **Yes, the binomial distribution with \( n = 1,000 \) and \( p = 0.9 \) can be approximated by the normal distribution, because \( n·p = 900 \) and \( n·(1-p) = 100 \) are both much greater than 5.**

Certainly! Here's the step-by-step solution divided into three parts, each with its own explanation: --- ### **Part 1: Understanding the Binomial Distribution** **Step:** A binomial distribution is characterized by two parameters: the number of trials \( n \) and the probability of success in each trial \( p \). The key measures are the mean \( \mu = n \times p \) and the variance \( \sigma^2 = n \times p \times (1 - p) \). **Explanation:** This foundational step helps us understand the shape and characteristics of the binomial distribution based on the given parameters. Knowing the mean and variance is essential for assessing whether an approximation to the normal distribution is appropriate. --- ### **Part 2: Conditions for Normal Approximation** **Step:** The general rule for approximating a binomial distribution with a normal distribution is that both \( n \times p \) and \( n \times (1 - p) \) should be greater than or equal to 5 (some use 10 for more accuracy). **Explanation:** This condition ensures the distribution isn't too skewed and that the normal approximation is valid. When these values are sufficiently large, the binomial distribution resembles a bell curve, making the normal approximation appropriate. --- ### **Part 3: Applying the Conditions and Final Conclusion** **Step:** Calculate: - \( n \times p = 1,000 \times 0.9 = 900 \) - \( n \times (1 - p) = 1,000 \times 0.1 = 100 \) Both are much greater than 5, satisfying the condition. **Explanation:** Because both values are large, the binomial distribution with these parameters can be effectively approximated by a normal distribution. This simplifies calculations and analysis for large \( n \). --- ### **Final Statement:** **Yes, the binomial distribution with \( n = 1,000 \) and \( p = 0.9 \) can be approximated by a normal distribution because both \( n \times p \) and \( n \times (1 - p) \) are significantly greater than 5.**

Yes, your understanding and the step-by-step explanation are correct. You correctly identified the key conditions for the normal approximation to the binomial distribution and verified that they are satisfied with the given parameters. Therefore, the binomial distribution with \( n = 1000 \) and \( p = 0.9 \) can indeed be approximated by a normal distribution.