2. The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:

\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline & High & Low & Q1 & Q3 & IQR & Median & Mean & [tex]$\sigma$[/tex] \\
\hline College & 50 & 6 & 8.5 & 17 & 8.5 & 12 & 15.4 & 11.7 \\
\hline High School & 28 & 3 & 4.5 & 15 & 10.5 & 11 & 10.5 & 5.8 \\
\hline
\end{tabular}

Which of the choices below best describes how to measure the spread of this data?
(Hint: Use the minimum and maximum values to check for outliers.)

A. Both spreads are best described with the IQR.
B. Both spreads are best described with the standard deviation.
C. The college spread is best described by the IQR. The high school spread is best described by the standard deviation.
D. The college spread is best described by the standard deviation. The high school spread is best described by the IQR.



Answer :

To determine the best measure of spread for the given data, we need to evaluate whether each dataset has outliers. We typically use two measures of spread: the Interquartile Range (IQR) and the Standard Deviation (σ). The presence of outliers can influence which measure is more appropriate. Generally:

- IQR is more robust to outliers.
- Standard Deviation is sensitive to outliers.

Let's check for outliers using the IQR method. A data point is considered an outlier if it is below [tex]\(Q1 - 1.5 \times \text{IQR}\)[/tex] or above [tex]\(Q3 + 1.5 \times \text{IQR}\)[/tex].

### College data analysis:
- Minimum (Low): 6
- Maximum (High): 50
- Mean: 15.4
- Standard Deviation (σ): 11.7
- Q1: 8.5
- Q3: 17
- IQR (Q3 - Q1): 8.5

Lower bound to detect outliers:\
[tex]\[ Q1 - 1.5 \times \text{IQR} = 8.5 - 1.5 \times 8.5 = 8.5 - 12.75 = -4.25 \][/tex]

Upper bound to detect outliers:\
[tex]\[ Q3 + 1.5 \times \text{IQR} = 17 + 1.5 \times 8.5 = 17 + 12.75 = 29.75 \][/tex]

The minimum value (6) lies above -4.25, which is fine.\
The maximum value (50) lies above 29.75, indicating an outlier.

### High School data analysis:
- Minimum (Low): 3
- Maximum (High): 28
- Mean: 10.5
- Standard Deviation (σ): 5.8
- Q1: 4.5
- Q3: 15
- IQR (Q3 - Q1): 10.5

Lower bound to detect outliers:\
[tex]\[ Q1 - 1.5 \times \text{IQR} = 4.5 - 1.5 \times 10.5 = 4.5 - 15.75 = -11.25 \][/tex]

Upper bound to detect outliers:\
[tex]\[ Q3 + 1.5 \times \text{IQR} = 15 + 1.5 \times 10.5 = 15 + 15.75 = 30.75 \][/tex]

The minimum value (3) lies above -11.25, which is fine.\
The maximum value (28) lies below 30.75, which is also fine.

### Conclusions:

For the college data, since there is an outlier (maximum value of 50 is more than 29.75), the best measure of spread is the standard deviation.

For the high school data, there are no outliers detected, so the IQR is a suitable measure of spread.

Therefore, the correct choice is:
Both spreads are best described with the standard deviation.