A population of values has a normal distribution with \( \mu=143.3 \) and \( \sigma=74.1 \). You intend to draw a random sample of size \( n=147 \). Find the prodability that a single randomly selected value is between 134.7 and 162.9 . \( P(134.7

Ask by Harper Chen. in the United States

Mar 16,2025

To solve this problem, we will use the properties of the normal distribution. Let's calculate the two probabilities one by one. ### Probability for a Single Randomly Selected Value 1. First, we need to calculate the Z-scores for the values 134.7 and 162.9. \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 134.7 \): \[ Z_{134.7} = \frac{134.7 - 143.3}{74.1} \approx -0.115 \] For \( X = 162.9 \): \[ Z_{162.9} = \frac{162.9 - 143.3}{74.1} \approx 0.263 \] 2. Now, we will look up these Z-scores in the standard normal distribution table or use a calculator. - The probability for \( Z = -0.115 \) is approximately 0.4543. - The probability for \( Z = 0.263 \) is approximately 0.6032. 3. To find the probability that a single randomly selected value is between 134.7 and 162.9: \[ P(134.7 < X < 162.9) = P(Z < 0.263) - P(Z < -0.115) \approx 0.6032 - 0.4543 = 0.1489 \] ### Probability for the Sample Mean 1. We need to find the Z-scores for the sample mean. We use the standard error of the mean (SEM): \[ SEM = \frac{\sigma}{\sqrt{n}} = \frac{74.1}{\sqrt{147}} \approx 6.094 \] 2. We will calculate the Z-scores for the sample mean. For \( M = 134.7 \): \[ Z_{134.7} = \frac{134.7 - 143.3}{6.094} \approx -1.42 \] For \( M = 162.9 \): \[ Z_{162.9} = \frac{162.9 - 143.3}{6.094} \approx 3.25 \] 3. Again, we look up these Z-scores: - The probability for \( Z = -1.42 \) is approximately 0.0772. - The probability for \( Z = 3.25 \) is approximately 0.9994. 4. Now, we find the probability that a sample mean is between 134.7 and 162.9: \[ P(134.7 < M < 162.9) = P(Z < 3.25) - P(Z < -1.42) \approx 0.9994 - 0.0772 = 0.9222 \] ### Final Answers So, the probabilities are: - \( P(134.7 < X < 162.9) \approx 0.1489 \) - \( P(134.7 < M < 162.9) \approx 0.9222 \)