On a recent quiz, the class mean was 78 with a standard deviation of 2.2 . Calculate the \( z \)-score (to 2 decimal places) for a person who received a score of 79 . (a) \( z \)-score: (b) Is this unusual? Not Unusual Unusual

To calculate the \( z \)-score, use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where \( X \) is the score obtained (79), \( \mu \) is the mean (78), and \( \sigma \) is the standard deviation (2.2). Substituting the values, we get: \[ z = \frac{(79 - 78)}{2.2} = \frac{1}{2.2} \approx 0.45 \] So, the \( z \)-score is approximately 0.45. (b) This \( z \)-score indicates that the score of 79 is 0.45 standard deviations above the mean. In the world of statistics, a \( z \)-score between -2 and 2 is generally considered typical or "not unusual," so this score is indeed not unusual!