Question

Find the mean and standard deviation of the data set. $$ 25,30,32,32,41,45,57,62 $$

Short answer

Answer: The mean of the data set is 40.5, and the standard deviation is approximately 12.5.

Step-by-step solution

Calculate the mean

First, we'll add all the data points and divide by the total number of data points to find the mean: $$ \text{mean} = \frac{25+30+32+32+41+45+57+62}{8} $$

Calculate the deviations from the mean

Now, calculate the difference between each data point and the mean we found in step 1: $$ \text{deviations} = x_i - \text{mean} $$

Square the deviations

Square each of the deviations found in step 2: $$ \text{squared deviations} = (\text{deviations})^2 $$

Calculate the mean of the squared deviations

Next, find the mean of the squared deviations by adding them all together and then dividing by the total number of data points: $$ \text{mean of squared deviations} = \frac{\sum(\text{squared deviations})}{8} $$

Calculate the standard deviation

Finally, take the square root of the mean of squared deviations we found in step 4 to get the standard deviation: $$ \text{standard deviation} = \sqrt{\text{mean of squared deviations}} $$ Now we can apply this process to our given data set: 1. Calculate the mean: $$ \text{mean} = \frac{25+30+32+32+41+45+57+62}{8} = \frac{324}{8} = 40.5 $$ 2. Calculate the deviations: $$ \text{deviations} = [-15.5, -10.5, -8.5, -8.5, 0.5, 4.5, 16.5, 21.5] $$ 3. Square the deviations: $$ \text{squared deviations} = [240.25, 110.25, 72.25, 72.25, 0.25, 20.25, 272.25, 462.25] $$ 4. Calculate the mean of squared deviations: $$ \text{mean of squared deviations} = \frac{240.25+110.25+72.25+72.25+0.25+20.25+272.25+462.25}{8} = \frac{1250}{8} = 156.25 $$ 5. Calculate the standard deviation: $$ \text{standard deviation} = \sqrt{156.25} \approx 12.5 $$ So, the mean of the data set is 40.5, and the standard deviation is approximately 12.5.

Key concepts

Data Set

A data set is simply a collection of numbers or values that you want to analyze. In this exercise, our data set consists of eight numbers: 25, 30, 32, 32, 41, 45, 57, and 62.
Analyzing a data set helps us understand various properties of the data, such as the average, variability, and patterns.
Here are a few important points to remember about data sets:

• Each data point is an individual value within the set.
• The entire group of data points makes up the data set.
• Data sets are the foundation for calculating statistical measures like mean and standard deviation.

Understanding and organizing your data set is the first step toward data analysis.

Mean Calculation

The mean, also known as the average, provides us with a single value that represents the center of a data set. Calculating the mean gives us an overall idea of what a typical data point might look like in the set. To find the mean:

• Add up all the data points in the set.
• Then divide by the total number of data points.

For example, in our exercise, we calculate the mean of the data set using the formula:\[\text{mean} = \frac{25+30+32+32+41+45+57+62}{8} = \frac{324}{8} = 40.5\]This tells us that, on average, a data point in the set is 40.5. The mean gives us one kind of central location for the data and helps compare additional sets to this one.

Standard Deviation Calculation

The standard deviation is a measure of how spread out the numbers in a data set are. It tells us how much the data points typically deviate from the mean. A larger standard deviation means the data is more spread out, whereas a smaller standard deviation means the data is closely packed around the mean.
To compute the standard deviation, follow these steps:

• Calculate the mean of the data set.
• Find the deviation of each data point from the mean (difference between the data point and the mean).
• Square each deviation to make them positive, emphasizing larger differences.
• Calculate the mean of these squared deviations.
• Finally, take the square root of the mean of the squared deviations to get the standard deviation.

In our example, we calculated the standard deviation as follows:\[\text{standard deviation} = \sqrt{156.25} \approx 12.5\]This value indicates that the typical distance of a data point from the mean is about 12.5. Understanding the standard deviation helps us assess the variability within the data set, making it possible to predict or compare different data sets more effectively.