Problem 36 - Algebra 2

Question

Consider the two sets of data. A = {1, 2, 2, 2, 2, 3, 3, 3, 3, 4}, B = {1, 1, 2, 2, 2, 3, 3, 3, 4, 4} Find the mean, median, variance, and standard deviation of each set of data.

Step-by-Step Solution

Answer

Both sets have a mean and median of 2.5. Set A has variance 0.6 and std deviation 0.775, Set B has variance 1 and std deviation 1.

Step 1: Calculate the Mean of Set A

To find the mean, sum all the elements in set A and divide by the number of elements. Set A = \{1, 2, 2, 2, 2, 3, 3, 3, 3, 4\}. Sum = 1 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 4 = 25. Number of elements = 10. Thus, the mean is \(\frac{25}{10} = 2.5\).

Step 2: Calculate the Median of Set A

Order set A (it's already ordered). With 10 elements, the median is the average of the 5th and 6th elements. The 5th element is 2 and the 6th is 3, so the median is \(\frac{2 + 3}{2} = 2.5\).

Step 3: Calculate the Variance of Set A

First, calculate the deviations from the mean, square them, and take the average. Deviations: \((-1.5)^2, (-0.5)^2, (-0.5)^2, (-0.5)^2, (-0.5)^2, (0.5)^2, (0.5)^2, (0.5)^2, (0.5)^2, (1.5)^2\) = \{2.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 2.25\}. Sum of squared deviations = 6. Multiply by \(\frac{1}{n}\) (since population variance is requested), \(\frac{6}{10} = 0.6\).

Step 4: Calculate the Standard Deviation of Set A

The standard deviation is the square root of the variance: \(\sqrt{0.6} \approx 0.775\).

Step 5: Calculate the Mean of Set B

Sum all the elements in set B: 1 + 1 + 2 + 2 + 2 + 3 + 3 + 3 + 4 + 4 = 25. There are 10 elements, so the mean is \(\frac{25}{10} = 2.5\).

Step 6: Calculate the Median of Set B

Set B is \{1, 1, 2, 2, 2, 3, 3, 3, 4, 4\}. The median is the average of the 5th and 6th elements: \(\frac{2 + 3}{2} = 2.5\).

Step 7: Calculate the Variance of Set B

Deviations from the mean: \{(-1.5)^2, (-1.5)^2, (-0.5)^2, (-0.5)^2, (-0.5)^2, (0.5)^2, (0.5)^2, (0.5)^2, (1.5)^2, (1.5)^2\} = \{2.25, 2.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 2.25, 2.25\}. Sum of squared deviations = 10. Thus, variance is \(\frac{10}{10} = 1\).

Step 8: Calculate the Standard Deviation of Set B

The standard deviation is the square root of the variance: \(\sqrt{1} = 1\).

Key Concepts

Mean

The mean, often referred to as the average, is one of the fundamental measures in statistics that provides us with a general sense of the central tendency of a data set. To calculate the mean, we simply add up all the numbers in our data set and then divide by the number of elements.
For example, in data set A = \{1, 2, 2, 2, 2, 3, 3, 3, 3, 4\}, the process involves summing all numbers: 1 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 4 = 25.
With a total of 10 numbers in the set, the mean is calculated as \( \frac{25}{10} = 2.5 \). The same calculation applies to data set B = \{1, 1, 2, 2, 2, 3, 3, 3, 4, 4\}, which also results in a mean of 2.5 as its total sum equals 25, divided by 10.
The mean provides a useful insight into the "center" of the data but may not always reflect disparities due to outliers.

Median

The median is a measure of central tendency that indicates the middle value of a sorted data set, providing insight into its center. Unlike the mean, the median is less sensitive to extreme values.
To find the median, first, arrange the data in ascending order, which is often already done in exercises. Then, if the number of data points is odd, the median is the middle number. If the number of data points is even, the median is the average of the two central numbers.
For set A = \{1, 2, 2, 2, 2, 3, 3, 3, 3, 4\}, an even total of 10 elements means the median is the mean of the 5th and 6th numbers. Thus, \( \frac{2+3}{2} = 2.5 \).
Similarly, for set B = \{1, 1, 2, 2, 2, 3, 3, 3, 4, 4\}, the median is also 2.5 for the same reason.
The median is particularly useful in skewed distributions, as it provides an accurate reflection of the "typical" value.

Variance

Variance measures how spread out a data set is. It's calculated based on the average of the squared deviations from the mean.
The formula involves taking each data point, subtracting the mean, squaring the result, and then averaging these squared differences.
For set A, with a mean of 2.5, the deviations are squared and summed to get 6, giving a variance of \( \frac{6}{10} = 0.6 \).
In set B, the variance is determined similarly, resulting in a higher variance of \( \frac{10}{10} = 1 \).
A larger variance indicates that the data points are more spread out from the mean, while a smaller variance suggests they are closer to the mean. Variance is vital in statistics to understand data dispersion quantitatively.

Standard Deviation

The standard deviation is a widely used measure to quantify the amount of variation or dispersion in a set of data values. It is literally the square root of the variance.
This measure gives us a sense of how much each data point deviates from the mean, on average.
For set A, where the variance is 0.6, the standard deviation is calculated as \( \sqrt{0.6} \approx 0.775 \).
Meanwhile, for set B with a variance of 1, the standard deviation is \( \sqrt{1} = 1 \).
The standard deviation is highly useful because it gives an understandable scale to measure spread or dispersion in the same units as the data.
When data points are close to the mean, the standard deviation is smaller, indicating lower variability. Conversely, a larger standard deviation reflects higher variability and a wider spread of data points.