Let's break down the problem step by step. --- ## **Given:** - The **compressive strength** of concrete, \( X \), is a **lognormal random variable**. - **Mean** (\( \mu_X \)): \( 4.7 \) ksi - **Coefficient of Variation** (\( \text{COV}_X \)): \( 0.21 \) Recall for a lognormal variable \( X \), if \( X = e^Y \) and \( Y \sim N(\mu_Y, \sigma_Y^2) \): - \( E[X] = e^{\mu_Y + \frac{1}{2} \sigma_Y^2} \) - \( \text{Var}[X] = (e^{\sigma_Y^2} - 1) e^{2\mu_Y + \sigma_Y^2} \) - \( \text{COV}_X = \sqrt{e^{\sigma_Y^2} - 1} \) The **10th percentile** value is the value \( x_{0.10} \) such that \( P(X < x_{0.10}) = 0.10 \). --- # **Part (a): Calculate the 10th percentile value ("design value") with COV = 0.21** ## **Step 1: Find the parameters of the underlying normal distribution** Let \( \mu_Y \) and \( \sigma_Y \) be the mean and std dev of \( Y = \ln X \): ### From COV: \[ \text{COV}_X = \sqrt{e^{\sigma_Y^2} - 1} \] \[ 0.21 = \sqrt{e^{\sigma_Y^2} - 1} \] \[ 0.0441 = e^{\sigma_Y^2} - 1 \] \[ e^{\sigma_Y^2} = 1.0441 \] \[ \sigma_Y^2 = \ln(1.0441) = 0.0432 \] \[ \sigma_Y = \sqrt{0.0432} \approx 0.2078 \] ### From the mean: \[ E[X] = e^{\mu_Y + \frac{1}{2}\sigma_Y^2} \] \[ 4.7 = e^{\mu_Y + 0.0216} \] \[ \ln(4.7) = \mu_Y + 0.0216 \] \[ \mu_Y = \ln(4.7) - 0.0216 \approx 1.5476 - 0.0216 = 1.5260 \] --- ## **Step 2: The 10th percentile of X** For a lognormal variable: \[ x_{0.10} = \exp(\mu_Y + z_{0.10} \cdot \sigma_Y) \] Where \( z_{0.10} \) is the 10th percentile of the standard normal distribution (\( z_{0.10} \approx -1.2816 \)). \[ x_{0.10} = \exp(1.5260 + (-1.2816) \times 0.2078) \] \[ = \exp(1.5260 - 0.2666) \] \[ = \exp(1.2594) \] \[ = 3.522 \text{ ksi} \] **Rounded to three significant digits: \(\boxed{3.52 \ \text{ksi}}\)** --- # **Part (b): Repeat calculation for COV = 0.10** ## **Step 1: Update lognormal parameters** \[ \text{COV}_X = 0.10 = \sqrt{e^{\sigma_Y^2} - 1} \] \[ 0.01 = e^{\sigma_Y^2} - 1 \] \[ e^{\sigma_Y^2} = 1.01 \] \[ \sigma_Y^2 = \ln(1.01) = 0.00995 \] \[ \sigma_Y = \sqrt{0.00995} = 0.0997 \] \[ \mu_Y = \ln(4.7) - 0.5 \times 0.00995 = 1.5476 - 0.004975 = 1.5426 \] ## **Step 2: 10th percentile value** \[ x_{0.10} = \exp(\mu_Y + z_{0.10} \cdot \sigma_Y) \] \[ = \exp(1.5426 + (-1.2816 \times 0.0997)) \] \[ = \exp(1.5426 - 0.1278) \] \[ = \exp(1.4148) \] \[ = 4.116 \text{ ksi} \] **Rounded to three significant digits: \(\boxed{4.12 \ \text{ksi}}\)** --- # **Part (c): Discussion** - **With no quality control (COV = 0.21):** Design value \( = 3.52 \) ksi - **With better quality control (COV = 0.10):** Design value \( = 4.12 \) ksi **Interpretation:** - **Reducing the variability (COV) of the compressive strength (e.g., via quality control) raises the 10th percentile value for the same mean.** - This means that with better quality control, you can use a higher design value for compressive strength, potentially allowing for more efficient/less conservative designs. - **Quality control is preferable** because it reduces the risk of low-strength concrete and allows for more economic use of materials. --- ## **Summary Table** | Case | COV | 10th Percentile (ksi) | |-----------------------|-------|:---------------------:| | No Quality Control | 0.21 | 3.52 | | With Quality Control | 0.10 | 4.12 | --- ## **Final Answers (boxed):** **(a) Design value (10th percentile) with COV 0.21:** \[ \boxed{3.52 \ \text{ksi}} \] **(b) Design value (10th percentile) with COV 0.10:** \[ \boxed{4.12 \ \text{ksi}} \] **(c) Discussion:** \[ \boxed{ \begin{minipage}{0.9\linewidth} Improved quality control (lower COV) allows a higher design value for compressive strength at the same mean. This results in more efficient and economical designs, demonstrating that quality control measures are indeed preferable. \end{minipage} } \]