## Given Information - Estimate the 85th percentile speed on a road - Margin of error (E): 1 mile per hour - Confidence level: 95% - Standard deviation (σ): 3.7 miles per hour ## Solution The formula for sample size \( n \) when estimating a mean with a given margin of error at a certain confidence level is: \[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \] For a 95% confidence level, the z-value is approximately 1.96. Substitute the values: - \( z = 1.96 \) - \( \sigma = 3.7 \) - \( E = 1 \) \[ n = \left( \frac{1.96 \times 3.7}{1} \right)^2 \] \[ n = (7.252)^2 \] \[ n = 52.6 \] Round to the nearest whole number: \[ n = 53 \] ## Final Answer **53 vehicles**

## Given Information - Estimate of the 85th percentile speed on a road. - Margin of error (E): 1 mile per hour. - Confidence level: 95%. - Standard deviation (σ): 3.7 miles per hour. - Percentile (p): 0.85 (for the 85th percentile). ## Sample Size Calculation To estimate the sample size required to determine the 85th percentile speed with a specified margin of error and confidence level, we use the formula: \[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \] ### Determine z-value For a 95% confidence level: - \( z \approx 1.96 \) ### Substitute Values Substituting the known values into the formula: - \( z = 1.96 \) - \( \sigma = 3.7 \) - \( E = 1 \) \[ n = \left( \frac{1.96 \times 3.7}{1} \right)^2 \] ### Calculate 1. Calculate \( 1.96 \times 3.7 \): \[ 1.96 \times 3.7 = 7.252 \] 2. Square the result: \[ n = (7.252)^2 \approx 52.6 \] ### Round to Nearest Whole Number Round 52.6 to the nearest whole number: \[ n = 53 \] ## Final Answer **53 vehicles** ## Variance Calculation for Sample Quantile To compute the variance of a sample quantile \( \hat{q}_p \): \[ \text{Var}(\hat{q}_p) \approx p(1 - p) \frac{n}{f(q_p)^2} \] Where: - \( p \) is the percentile (0.85 for the 85th percentile). - \( f(q_p) \) is the probability density function evaluated at the quantile \( q_p \). - \( n \) is the sample size. This formula provides an estimate of the variance of the sample quantile based on the sample size and the distribution properties.

## Given Information - Percentile (p): 0.85 (for the 85th percentile) - Sample size (n): To be calculated - Standard deviation (σ): 3.7 miles per hour - Margin of error (E): 1 mile per hour - Probability density function \( f(q_p) \): Needs to be determined ## Variance of Sample Quantile The variance of a sample quantile \( \hat{q}_p \) can be calculated using: \[ \text{Var}(\hat{q}_p) \approx p(1 - p) \frac{n}{f(q_p)^2} \] ### Estimate \( f(q_p) \) Assuming the speeds follow a normal distribution, we can use the standard normal distribution to estimate \( f(q_p) \) at the 85th percentile: 1. The 85th percentile corresponds to a z-score of approximately 1.036. 2. The probability density function \( f(q_p) \) for a normal distribution can be calculated using the formula: \[ f(q_p) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(z^2)}{2}} \] Substituting \(\sigma = 3.7\) and \(z = 1.036\): \[ f(q_p) = \frac{1}{3.7 \sqrt{2\pi}} e^{-\frac{(1.036^2)}{2}} \] ### Calculate \( f(q_p) \) 1. Calculate \( e^{-\frac{(1.036^2)}{2}} \): \[ e^{-\frac{(1.036^2)}{2}} \approx e^{-0.537} \approx 0.584 \] 2. Calculate \( \sqrt{2\pi} \approx 2.506 \): \[ f(q_p) = \frac{1}{3.7 \times 2.506} \times 0.584 \approx \frac{0.584}{9.284} \approx 0.0629 \] ### Substitute into Variance Formula Now, substituting \( f(q_p) \) back into the variance formula: \[ \text{Var}(\hat{q}_p) \approx 0.85(1 - 0.85) \frac{n}{(0.0629)^2} \] \[ \text{Var}(\hat{q}_p) \approx 0.85(0.15) \frac{n}{0.00395} \] \[ \text{Var}(\hat{q}_p) \approx 0.1275 \frac{n}{0.00395} \] ### Calculate Sample Size \( n \) To ensure the variance is less than the square of the margin of error, set \( \text{Var}(\hat{q}_p) \) equal to \( E^2 \): \[ 0.1275 \frac{n}{0.00395} \leq 1^2 \] Solving for \( n \): \[ 0.1275n \leq 0.00395 \] \[ n \leq \frac{0.00395}{0.1275} \approx 30.98 \] ### Round to Nearest Whole Number The required sample size is: \[ n = 31 \] ## Final Answer **31 vehicles**