# Step-by-Step Solutions Given: Mean \( \mu = 3.8 \) Standard deviation \( \sigma = 1.7 \) \( x \) is normally distributed. --- ## a. Find \( P(x > 6.5) \) **Step 1: Convert to z-score** \[ z = \frac{x - \mu}{\sigma} = \frac{6.5 - 3.8}{1.7} = \frac{2.7}{1.7} \approx 1.5882 \] **Step 2: Find \( P(z > 1.5882) \)** Look up \( z = 1.59 \) in the standard normal table (or use a calculator): - \( P(Z < 1.59) \approx 0.9441 \) - So, \( P(Z > 1.59) = 1 - 0.9441 = 0.0559 \) **Step 3: Round to 4 decimal places** \[ \boxed{P(x > 6.5) = 0.0559} \] --- ## b. Find \( P(5.5 \leq x \leq 7.5) \) **Step 1: Convert 5.5 and 7.5 to z-scores:** \[ z_1 = \frac{5.5 - 3.8}{1.7} = \frac{1.7}{1.7} = 1.0 \] \[ z_2 = \frac{7.5 - 3.8}{1.7} = \frac{3.7}{1.7} \approx 2.1765 \] **Step 2: Find probabilities from standard normal table:** - \( P(Z < 1.0) = 0.8413 \) - \( P(Z < 2.18) \approx 0.9852 \) (using \( z = 2.18 \) for better accuracy) **Step 3: Subtract for interval probability:** \[ P(5.5 \leq x \leq 7.5) = P(Z < 2.18) - P(Z < 1.0) = 0.9852 - 0.8413 = 0.1439 \] **Step 4: Round to 4 decimal places** \[ \boxed{P(5.5 \leq x \leq 7.5) = 0.1439} \] --- ## c. Find \( x \) such that \( P(x > x) = 0.0869 \) **Step 1: Find z-score corresponding to upper tail 0.0869** - \( P(Z > z) = 0.0869 \implies P(Z < z) = 1 - 0.0869 = 0.9131 \) - Standard normal table: \( z \approx 1.36 \) (as \( P(Z < 1.36) = 0.9131 \)) **Step 2: Convert z-score to x** \[ x = z \cdot \sigma + \mu = 1.36 \cdot 1.7 + 3.8 = 2.312 + 3.8 = 6.112 \] **Step 3: Round to 3 decimal places** \[ \boxed{x = 6.112} \] --- ## d. Find \( x \) such that \( P(x \leq x \leq 3.8) = 0.4719 \) **Step 1: Interpret the interval** This means: \( P(x \leq X \leq 3.8) = 0.4719 \) Since \( 3.8 \) is the mean, we are looking for the value \( x \) below the mean so that the probability from \( x \) to \( 3.8 \) is 0.4719. **Step 2: Standardize the interval** Let \( z_0 \) correspond to \( x \): \[ P(z_0 \leq Z \leq 0) = 0.4719 \] \[ P(Z < 0) - P(Z < z_0) = 0.4719 \] \[ 0.5 - P(Z < z_0) = 0.4719 \] \[ P(Z < z_0) = 0.5 - 0.4719 = 0.0281 \] **Step 3: Find \( z_0 \) for \( P(Z < z_0) = 0.0281 \)** - This is a negative z-score. From the table, \( z_0 \approx -1.90 \) (as \( P(Z < -1.90) = 0.0287 \)), but for \( 0.0281 \), it's closer to \( z_0 \approx -1.91 \). **Step 4: Convert z-score to x** \[ x = z_0 \cdot \sigma + \mu = (-1.91) \cdot 1.7 + 3.8 = -3.247 + 3.8 = 0.553 \] **Step 5: Round to 3 decimal places** \[ \boxed{x = 0.553} \] --- # Summary Table | Part | Solution | Final Answer | |------|----------|--------------| | a | \( P(x > 6.5) \) | **0.0559** | | b | \( P(5.5 \leq x \leq 7.5) \) | **0.1439** | | c | \( x : P(x > x) = 0.0869 \) | **6.112** | | d | \( x : P(x \leq X \leq 3.8) = 0.4719 \) | **0.553** | --- **Note:** All calculations use common approximations from the standard normal table. For more precise results, use a calculator or statistical software.

# Step-by-Step Solutions Given: Mean \( \mu = 3.8 \) Standard deviation \( \sigma = 2.9 \) \( x \) is normally distributed. --- ## a. Find \( P(x > 6.5) \) **Step 1: Convert to z-score** \[ z = \frac{x - \mu}{\sigma} = \frac{6.5 - 3.8}{2.9} = \frac{2.7}{2.9} \approx 0.9310 \] **Step 2: Find \( P(z > 0.9310) \)** Look up \( z = 0.93 \) in the standard normal table (or use a calculator): - \( P(Z < 0.93) \approx 0.8238 \) - So, \( P(Z > 0.93) = 1 - 0.8238 = 0.1762 \) **Step 3: Round to 4 decimal places** \[ \boxed{P(x > 6.5) = 0.1762} \] --- ## b. Find \( P(5.5 \leq x \leq 7.5) \) **Step 1: Convert 5.5 and 7.5 to z-scores:** \[ z_1 = \frac{5.5 - 3.8}{2.9} = \frac{1.7}{2.9} \approx 0.5862 \] \[ z_2 = \frac{7.5 - 3.8}{2.9} = \frac{3.7}{2.9} \approx 1.2759 \] **Step 2: Find probabilities from standard normal table:** - \( P(Z < 0.5862) \approx 0.7200 \) - \( P(Z < 1.2759) \approx 0.9000 \) **Step 3: Subtract for interval probability:** \[ P(5.5 \leq x \leq 7.5) = P(Z < 1.2759) - P(Z < 0.5862) = 0.9000 - 0.7200 = 0.1800 \] **Step 4: Round to 4 decimal places** \[ \boxed{P(5.5 \leq x \leq 7.5) = 0.1800} \] --- ## c. Find \( x \) such that \( P(x > x) = 0.0918 \) **Step 1: Find z-score corresponding to upper tail 0.0918** - \( P(Z > z) = 0.0918 \implies P(Z < z) = 1 - 0.0918 = 0.9082 \) - Standard normal table: \( z \approx 1.34 \) (as \( P(Z < 1.34) \approx 0.9082 \)) **Step 2: Convert z-score to x** \[ x = z \cdot \sigma + \mu = 1.34 \cdot 2.9 + 3.8 \approx 3.886 + 3.8 = 7.686 \] **Step 3: Round to 3 decimal places** \[ \boxed{x = 7.686} \] --- ## d. Find \( x \) such that \( P(x \leq x \leq 3.8) = 0.4719 \) **Step 1: Interpret the interval** This means: \( P(x \leq X \leq 3.8) = 0.4719 \) Since \( 3.8 \) is the mean, we are looking for the value \( x \) below the mean so that the probability from \( x \) to \( 3.8 \) is 0.4719. **Step 2: Standardize the interval** Let \( z_0 \) correspond to \( x \): \[ P(z_0 \leq Z \leq 0) = 0.4719 \] \[ P(Z < 0) - P(Z < z_0) = 0.4719 \] \[ 0.5 - P(Z < z_0) = 0.4719 \] \[ P(Z < z_0) = 0.5 - 0.4719 = 0.0281 \] **Step 3: Find \( z_0 \) for \( P(Z < z_0) = 0.0281 \)** - This is a negative z-score. From the table, \( z_0 \approx -1.88 \) (as \( P(Z < -1.88) \approx 0.0285 \)), but for \( 0.0281 \), it's closer to \( z_0 \approx -1.88 \). **Step 4: Convert z-score to x** \[ x = z_0 \cdot \sigma + \mu = (-1.88) \cdot 2.9 + 3.8 \approx -5.452 + 3.8 = -1.652 \] **Step 5: Round to 3 decimal places** \[ \boxed{x = -1.652} \] --- # Summary Table | Part | Solution | Final Answer | |------|----------|--------------| | a | \( P(x > 6.5) \) | **0.1762** | | b | \( P(5.5 \leq x \leq 7.5) \) | **0.1800** | | c | \( x : P(x > x) = 0.0918 \) | **7.686** | | d | \( x : P(x \leq X \leq 3.8) = 0.4719 \) | **-1.652** | --- **Note:** All calculations are based on the standard normal distribution and may vary slightly based on the specific table or calculator used.