Final answer:
The question is about interpreting Julie's final exam score in her statistics class using the normal distribution. Julie's score of 84 is compared to the mean of 75 and the standard deviation of 6 to calculate her Z-score, which indicates her performance relative to her classmates.
Step-by-step explanation:
The question centers on the concept of normal distribution, a foundational concept within the field of statistics. In this scenario, Julie's score in her statistics class is given, along with the mean and standard deviation of the class scores.
To interpret Julie's performance, we use her score and compare it to the class distribution characterized by the given mean (75) and standard deviation (6).
When a score follows a normal distribution, you can determine how well a student performed relative to peers by calculating the Z-score. This involves subtracting the mean from the individual score and then dividing by the standard deviation: Z = (X - μ) / σ. For Julie, her Z-score would be (84 - 75) / 6, which equals 1.5. This Z-score indicates how many standard deviations above or below the mean her score lies.
In Julie's case, her score is 1.5 standard deviations above the mean, typically seen as a good performance within the context of a normal distribution.