A population of values has a normal distribution with \( \mu=237.1 \) and \( \sigma=6.9 \). You intend to draw a random sample of size \( n=206 \). Find the probability that a sample of size \( n=206 \) is randomly selected with a mean between 236.7 and 236.8 . \( P(236.7

Ask by Munoz Moran. in the United States

Mar 16,2025

To find the probability that a sample mean \( M \) falls between 236.7 and 236.8, we can use the Central Limit Theorem. According to the theorem, the distribution of the sample mean will be normally distributed with mean \( \mu \) equal to the population mean and a standard deviation \( \sigma_M \) equal to the population standard deviation divided by the square root of the sample size \( n \). First, we calculate the standard deviation of the sample mean: \[ \sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{6.9}{\sqrt{206}} \approx \frac{6.9}{14.3578} \approx 0.4802 \] Next, we calculate the \( z \)-scores for the values 236.7 and 236.8: \[ z_1 = \frac{236.7 - \mu}{\sigma_M} = \frac{236.7 - 237.1}{0.4802} \approx \frac{-0.4}{0.4802} \approx -0.8330 \] \[ z_2 = \frac{236.8 - \mu}{\sigma_M} = \frac{236.8 - 237.1}{0.4802} \approx \frac{-0.3}{0.4802} \approx -0.6244 \] Now, we can use the standard normal distribution table (or a calculator) to find the probabilities corresponding to these \( z \)-scores. Using the standard normal distribution table or a calculator: 1. For \( z_1 = -0.8330 \): \( P(Z < -0.8330) \approx 0.2023 \) 2. For \( z_2 = -0.6244 \): \( P(Z < -0.6244) \approx 0.2659 \) Now we subtract the two probabilities to get the probability that the sample mean falls between 236.7 and 236.8: \[ P(236.7 < M < 236.8) = P(Z < -0.6244) - P(Z < -0.8330) \approx 0.2659 - 0.2023 \approx 0.0636 \] Thus, the probability that the sample mean \( M \) is between 236.7 and 236.8 is: \[ P(236.7 < M < 236.8) \approx 0.0636 \] Final answer: \( P(236.7