To calculate the range and standard deviation of the given sample of entertainment expenses, follow these steps:

1. Range Calculation:

The range is the difference between the maximum and minimum values in the data set.

• Maximum value = 773
• Minimum value = 682

$\text{Range} = 773 - 682 = 91$

2. Standard Deviation Calculation:

To compute the sample standard deviation, use the formula:

$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \overline{x})^2}$ Where:

• $n$ is the number of data points (25 in this case),
• $x_i$ are the individual data points,
• $\overline{x}$ is the mean of the data points.

Step-by-step:

• Step 1: Calculate the mean ($\overline{x}$) of the data. $\overline{x} = \frac{692 + 726 + 757 + \cdots + 731}{25} = \frac{17868}{25} = 714.72$

• Step 2: Calculate each deviation from the mean, square them, and sum them up. $\sum (x_i - \overline{x})^2 = (692 - 714.72)^2 + (726 - 714.72)^2 + \cdots + (731 - 714.72)^2$

The sum of the squared deviations is approximately $6513.28$.

• Step 3: Divide the sum of squared deviations by $n - 1 = 24$, then take the square root to get the standard deviation: $s = \sqrt{\frac{6513.28}{24}} = \sqrt{271.39} \approx 16.47$

Final Results:

• Range = 91
• Standard Deviation = 16.47 (rounded to 2 decimal places)

Would you like further explanation on any part of this?

Here are 5 follow-up questions:

1. What is the importance of calculating the range in a data set?
2. How does the standard deviation help in understanding data variability?
3. What are the differences between population and sample standard deviation?
4. How would the standard deviation change if we had more data points?
5. Why is it important to round the standard deviation in practical applications?

Tip: When analyzing data, the standard deviation tells you how spread out the data points are around the mean. The larger the standard deviation, the greater the variability in the data set.