Glass Garbage Generation A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 55 families will be between 17 and 18 pounds.

To find the probability that the mean glass garbage generated by a sample of 55 American families is between **17** and **18** pounds, we'll use the **Central Limit Theorem (CLT)**. Here's a step-by-step breakdown: ### **Given:** - **Population Mean (μ):** 17.2 pounds - **Population Standard Deviation (σ):** 2.5 pounds - **Sample Size (n):** 55 families - **Desired Range for Sample Mean (x̄):** 17 ≤ x̄ ≤ 18 pounds ### **Step 1: Calculate the Standard Error (SE)** The standard error of the mean is given by: \[ SE = \frac{\sigma}{\sqrt{n}} \] Plugging in the values: \[ SE = \frac{2.5}{\sqrt{55}} \approx \frac{2.5}{7.4162} \approx 0.3372 \text{ pounds} \] ### **Step 2: Convert to Z-Scores** We'll convert the desired range into Z-scores using the formula: \[ Z = \frac{(X - \mu)}{SE} \] - **For X = 17 pounds:** \[ Z_1 = \frac{17 - 17.2}{0.3372} \approx \frac{-0.2}{0.3372} \approx -0.593 \] - **For X = 18 pounds:** \[ Z_2 = \frac{18 - 17.2}{0.3372} \approx \frac{0.8}{0.3372} \approx 2.372 \] ### **Step 3: Find the Probabilities** Using the standard normal distribution table: - **P(Z ≤ -0.593):** Approximately **0.2766** - **P(Z ≤ 2.372):** Approximately **0.9912** ### **Step 4: Calculate the Probability Between the Two Z-Scores** \[ P(17 \leq x̄ \leq 18) = P(Z \leq 2.372) - P(Z \leq -0.593) \] \[ P \approx 0.9912 - 0.2766 = 0.7146 \] ### **Conclusion:** There is approximately a **71.5%** probability that the mean glass garbage generated by a sample of 55 American families will be between **17** and **18** pounds annually.