To analyze the given dot plot data for the ages of students in a school play, let's summarize the information:
- Ages and counts based on the dot plot:
- Age 12: 1 student
- Age 13: 2 students
- Age 14: 6 students
- Age 15: 2 students
- Age 16: 1 student
Now, let's evaluate each statement:
A. The distribution is symmetric.
- To determine if the distribution is symmetric, we can consider the shape of the distribution:
- The data shows a peak at age 14 and tapers off on either side (1 student at 12 and 16, and 2 students at 13 and 15). This suggests that the distribution is not symmetric; it is skewed to the right (or positively skewed) because there are more students at 14 compared to ages 15 and 16.
- This statement is false.
B. The maximum age would best describe the ages of the students in the play.
- The maximum age is 16, but this age only has 1 student, so it does not represent the majority of the data.
- Describing a dataset with the maximum age can be misleading if it is not also a common age among the students.
- This statement is false.
C. The mean and the median are the same.
- To find the mean, we sum the ages multiplied by their frequency:
- \( (12 \times 1) + (13 \times 2) + (14 \times 6) + (15 \times 2) + (16 \times 1) = 12 + 26 + 84 + 30 + 16 = 168 \)
- Number of students = \( 1 + 2 + 6 + 2 + 1 = 12 \)
- Mean = \( \frac{168}{12} = 14 \)
- To find the median, we list the ages in order:
- Ages: 12, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 16 (total of 12 ages)
- The median is the average of the 6th and 7th values (both are 14): \( \frac{14 + 14}{2} = 14 \).
- Since the mean is 14 and the median is also 14, they are equal.
- This statement is true.
D. The mean and the median are different.
- Based on the calculations from statement C, this is not true as both the mean and the median are the same (14).
- This statement is false.
Therefore, the only true statement regarding the data is: C. The mean and the median are the same.