Let's solve for both the standard deviation and the interquartile range (IQR) of the given data set. Here’s the list of values representing monthly phone use in minutes:

Data:
323, 480, 548, 450, 492, 475, 345, 513, 378, 419, 514, 342, 512, 459, 484, 405, 394, 527, 468, 326

Step 1: Calculating the Standard Deviation

1. Find the mean (average): $\text{Mean} = \frac{\sum \text{Data}}{N} = \frac{323 + 480 + 548 + \ldots + 468 + 326}{20}$

2. Calculate each data point's deviation from the mean, square each deviation, and then find the average of those squared deviations: $\text{Variance} = \frac{\sum (\text{Data} - \text{Mean})^2}{N}$

3. Take the square root of the variance to find the standard deviation.

Step 2: Calculating the Interquartile Range (IQR)

1. Sort the data in ascending order.

2. Find the first quartile (Q1) and third quartile (Q3). These are the 25th and 75th percentiles of the data, respectively.

3. Subtract Q1 from Q3 to get the IQR: $\text{IQR} = Q3 - Q1$

Let me perform these calculations for you.The calculations yield the following results:

• Standard Deviation: $69.54$ minutes (rounded to two decimal places)
• Interquartile Range (IQR): $107$ minutes

These measures help in understanding the spread and variability of the monthly phone usage data.

Would you like more details or have any questions?

Here are five related questions to further your understanding:

1. How does the standard deviation help in assessing the consistency of monthly phone usage?
2. What can be inferred if the IQR is large in relation to the range of the data?
3. How would outliers in the data affect the calculation of standard deviation and IQR?
4. What is the significance of using population versus sample standard deviation?
5. How can quartiles help in understanding the distribution of data?

Tip: To reduce the impact of outliers on measures of variability, consider using the interquartile range instead of the standard deviation.