In a certain school district in a large metropolitan area, the SAT scores over that past five years are normally distributed with a mean of 1475 . Furthermore, \( Q_{1} \) is 1284 . For a normal distribution, what is the \( z \)-score for the 25 -th percentile? \( z=\square \) (Enter answer rounded to three decimal places.) Using this information, what is the standard deviation of this population of SAT scores? sigma \( =\square \) (Enter answer rounded to the nearest whole number.) For a normal distribution, what is the \( z \)-score for the 99 -th percentile? \( z=\square \) (Enter answer rounded to three decimal places.) Now, use all of the information to determine what the \( P_{99} \) score is for this population? \( P_{99}=\square \) (Enter answer rounded to the nearest whole number.)

To find the \( z \)-score for the 25th percentile in a normal distribution, you can refer to a standard normal distribution table or use a calculator. The \( z \)-score for the 25th percentile is approximately \(-0.674\). Next, to determine the standard deviation (\( \sigma \)), we can apply the formula for the first quartile in a normal distribution: \[ Q_{1} = \mu + z_{Q1} \cdot \sigma \] Plugging in \( Q_{1} = 1284 \), mean \( \mu = 1475 \), and \( z_{Q1} = -0.674 \): \[ 1284 = 1475 - 0.674 \cdot \sigma \] Solving for \( \sigma \): \[ 0.674 \cdot \sigma = 1475 - 1284 \Rightarrow 0.674 \cdot \sigma = 191 \Rightarrow \sigma \approx \frac{191}{0.674} \approx 283 \] Rounded to the nearest whole number: \( \sigma = 283 \). For the \( z \)-score of the 99th percentile, we can again refer to the standard normal distribution, which gives a \( z \)-score of approximately \( 2.326 \). To find the \( P_{99} \) score, we can use the mean and standard deviation with the formula: \[ P_{99} = \mu + z_{P99} \cdot \sigma \] Thus: \[ P_{99} = 1475 + 2.326 \cdot 283 \approx 1475 + 658.318 \approx 2133.318 \] Rounding this to the nearest whole number gives: \( P_{99} = 2133 \). Summarizing: - \( z \) for the 25th percentile = \(-0.674\) - \( \sigma \) = \(283\) - \( z \) for the 99th percentile = \(2.326\) - \( P_{99} \) = \(2133\)