Problem 5 ( 10 points). The height of fifth-grade students in elementary school are normally distributed with mean ( \( \mu \) ) equal to 48 inches and a standard deviation equal to 2.8 inches. How tall in inches would a fifth-grade student need to be to have a height at the 91 st percentile? \( 0.90 \rightarrow 2 \) scole table 0.90

To find the height at the 91st percentile for fifth-grade students in elementary school, we need to use the standard normal distribution table. Given: - Mean (\( \mu \)) = 48 inches - Standard deviation (\( \sigma \)) = 2.8 inches - We want to find the height at the 91st percentile The z-score corresponding to the 91st percentile can be found using the standard normal distribution table. The z-score for the 91st percentile is approximately 1.34. The formula to convert a z-score back to the original height is: \[ x = \mu + z \times \sigma \] Substitute the values: \[ x = 48 + 1.34 \times 2.8 \] Now, we can calculate the height at the 91st percentile. Calculate the value by following steps: - step0: Calculate: \(48+1.34\times 2.8\) - step1: Multiply the numbers: \(48+3.752\) - step2: Add the numbers: \(51.752\) The height at the 91st percentile for fifth-grade students in elementary school is approximately 51.75 inches.