To find the range and standard deviation of the data set, we can follow these steps:

1. Finding the Range:

• Range is the difference between the maximum and minimum values in the data set.

• First, identify the minimum and maximum values from the data set.

• Minimum value = 11

• Maximum value = 92

• Range = Maximum value - Minimum value

• Range = $92 - 11 = 81$

2. Finding the Standard Deviation:

The formula for the standard deviation for a sample is:

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

where:

• $s$ = sample standard deviation
• $n$ = number of data points
• $x_i$ = each individual data point
• $\bar{x}$ = mean of the data set

Let's calculate it step by step:

1. Calculate the mean ( $\bar{x}$ ): $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$

2. Calculate each deviation from the mean, square it, and then find the sum of those squared deviations.

3. Divide by $n-1$ (degrees of freedom).

4. Take the square root to find the standard deviation.

Let's compute the mean and the standard deviation.

Calculations:

1. Mean ( $\bar{x}$ ): $\bar{x} = \frac{71 + 64 + 47 + 58 + 82 + 86 + 76 + 30 + 74 + 65 + 33 + 27 + 44 + 49 + 23 + 56 + 21 + 43 + 47 + 61 + 41 + 21 + 92 + 11}{24}$ $\bar{x} = \frac{1223}{24} \approx 50.96$

2. Standard Deviation ( $s$ ): $s = \sqrt{\frac{(71-50.96)^2 + (64-50.96)^2 + \dots + (11-50.96)^2}{24-1}}$

Let's compute the standard deviation.### Results:

• Range = 81
• Standard Deviation = 22.59

Would you like more details or have any other questions?

Here are five related questions that might interest you:

1. How would the range change if the maximum value increased by 10?
2. What is the variance of this data set, and how is it related to the standard deviation?
3. How does the standard deviation differ when calculated for the entire population versus a sample?
4. How would removing the outlier (the minimum value) affect the standard deviation?
5. Can you determine the interquartile range (IQR) of this data set?

Tip: The range is a simple measure of spread, while the standard deviation gives a more comprehensive picture of the data's variability.