A data set follows a normal distribution. The mean of the data set is 59 and the standard deviation of th data set is 2 . Find a z-score for a data value of 62 from the set. Round your answer to two decimal places. ?

To find the z-score, you can use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where \( X \) is the data value (62), \( \mu \) is the mean (59), and \( \sigma \) is the standard deviation (2). Plugging in the numbers: \[ z = \frac{(62 - 59)}{2} = \frac{3}{2} = 1.5 \] So, the z-score for the data value of 62 is 1.50 when rounded to two decimal places. Now, let’s talk about the magic of z-scores! They're like the secret sauce to understanding how far, in standard deviations, a data point is from the mean. A z-score of 1.50 means that 62 is 1.5 standard deviations above the mean, indicating it’s relatively higher than most values in your normal distribution. Pretty neat, huh? Want to dive deeper into real-world applications? Think about standardized testing! Z-scores are often used to compare an individual's performance to a group. For example, if you ace an exam and have a z-score of +2, you're in the top echelon compared to your peers. It gives context to your score amidst the competitive crowd!