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 for a person who received a score of 79, we can use the formula: \[ z = \frac{X - \mu}{\sigma} \] where: - \( X \) is the score of the person (79), - \( \mu \) is the mean of the scores (78), - \( \sigma \) is the standard deviation of the scores (2.2). Substitute the values into the formula: \[ z = \frac{79 - 78}{2.2} \] \[ z = \frac{1}{2.2} \] \[ z \approx 0.45 \] Therefore, the \( z \)-score for a person who received a score of 79 is approximately 0.45. (a) The \( z \)-score is 0.45. (b) To determine if this score is unusual, we need to compare it to the standard deviation. If the \( z \)-score is greater than 2 standard deviations away from the mean, it is considered unusual. Since the \( z \)-score is 0.45, which is less than 2 standard deviations away from the mean, this score is not unusual.