The stem-and-leaf plot displays the amount of time, in minutes, that a student spent practicing their musical instrument over 10 days.

\begin{tabular}{|l|l|}
\hline 1 & 5 \\
\hline 2 & [tex]$0, 2, 5$[/tex] \\
\hline 3 & 2, 4 \\
\hline 4 & 5 \\
\hline 5 & 3, 6 \\
\hline 6 & 0 \\
\hline & Key: [tex]$2 \mid 0$[/tex] means 20 \\
\hline
\end{tabular}

Part A: Calculate the mean and median for the data given. (2 points)

Part B: A student would like to show their teacher that they have practiced long enough for the day. Which measure of center should the student give to their teacher? Explain your answer. (2 points)



Answer :

Certainly! Let's go through both parts of the question step-by-step.

### Part A: Calculate the Mean and Median

#### Data Extraction
From the stem-and-leaf plot, we can extract the following data points:
- 15 (1 | 5)
- 20, 22, 25 (2 | 0, 2, 5)
- 32, 34 (3 | 2, 4)
- 45 (4 | 5)
- 53, 56 (5 | 3, 6)
- 60 (6 | 0)

So the dataset is: [15, 20, 22, 25, 32, 34, 45, 53, 56, 60]

#### Mean Calculation
The mean is calculated by summing all the data points and dividing by the number of data points.

[tex]\[ \text{Mean} = \frac{15 + 20 + 22 + 25 + 32 + 34 + 45 + 53 + 56 + 60}{10} = \frac{362}{10} = 36.2 \][/tex]

#### Median Calculation
To find the median, we need to arrange the data in ascending order (which it already is):
[15, 20, 22, 25, 32, 34, 45, 53, 56, 60]

Since there are 10 data points (an even number), the median is the average of the 5th and 6th values:

[tex]\[ \text{Median} = \frac{32 + 34}{2} = \frac{66}{2} = 33.0 \][/tex]

### Part B: Appropriate Measure of Center

To determine which measure of center (mean or median) the student should present to the teacher, we need to consider the implications of each measure. Typically:

- The mean is affected by extremely high or low values (outliers). If the data is symmetric and there are no outliers, the mean is a good measure of central tendency.
- The median is more robust to outliers and skewed data. It represents the middle value of a dataset.

In this case:
- If the dataset is reasonably symmetric (which appears to be the case here), either the mean or the median could be a reasonable measure. However, since we are not given detailed information about potential outliers or data skewness, the median might still be considered a safer choice to ensure robustness.

Because we do not have clear indicators of skewness or outliers, presenting a measure that is less sensitive to potential extreme values would be prudent.

#### Recommendation:
If the student wants to argue that their practice time is adequately represented, they should present the median (33.0 minutes). This measure is less likely to be influenced by any potential anomalies in their practice times, thus providing a more reliable indicator of their typical daily practice duration.

### Conclusion

The student should give the median (33.0 minutes) as the measure of center to demonstrate their practice time to their teacher because it gives a more robust and reliable representation of their practice duration, minimizing the effect of any potential extreme values.