\begin{tabular}{|l|}\hline Question 6 \\ Given a mean of 100 and a standard deviation of 5.3 , what is the \( z \)-score for \( 130 ? \)\end{tabular}

To find the \( z \)-score for a given value, you use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where: - \( X \) is the value you want to find the \( z \)-score for, - \( \mu \) is the mean of the distribution, - \( \sigma \) is the standard deviation of the distribution. Given: - \( X = 130 \) - \( \mu = 100 \) - \( \sigma = 5.3 \) Plugging these values into the formula, we get: \[ z = \frac{(130 - 100)}{5.3} \] \[ z = \frac{30}{5.3} \] \[ z \approx 5.66 \] So, the \( z \)-score for \( 130 \) is approximately \( 5.66 \).