Problem 38 - Algebra 2

Question

Find the mean and the standard deviation for each data set. $$ 0 \mathrm{km}, 1 \mathrm{km}, 1 \mathrm{km}, 1 \mathrm{km}, 2 \mathrm{km}, 2 \mathrm{km}, 2 \mathrm{km}, 3 \mathrm{km}, 3 \mathrm{km}, 4 \mathrm{km}, 5 \mathrm{km} $$

Step-by-Step Solution

Answer

The mean is approximately \(2.18\) km and the standard deviation is approximately \(1.27\) km.

Step 1: Compute the mean

Sum all the distances together and then divide that sum by the total number of values. Doing this gives: \[ \frac{{0+1+1+1+2+2+2+3+3+4+5}}{{11}} = \frac{{24}}{11} \approx 2.18 \, \mathrm{km} \]

Step 2: Calculate differences from the mean

Subtract the mean from each value and square the result. This will give: \[ (0-2.18)^2, (1-2.18)^2, (1-2.18)^2, (1-2.18)^2, (2-2.18)^2, (2-2.18)^2, (2-2.18)^2, (3-2.18)^2, (3-2.18)^2, (4-2.18)^2, (5-2.18)^2 \]

Step 3: Compute variance

Take the average of these squared differences. This is the variance. \[ \frac{{4.7524 +0.0324 +0.0324 +0.0324 +0.0324 +0.0324 +0.0324 +0.6724 +0.6724 +3.3624 +7.9524}}{11} = \frac{{17.616}}{11} \approx 1.6015 \]

Step 4: Calculate standard deviation

Finally, take the square root of that variance. This will give the standard deviation: \[ \sqrt{1.6015} \approx 1.27 \, \mathrm{km} \]

Key Concepts

Mean Calculation

The mean, often referred to as the average, is a central value for a data set. It's computed by summing all individual values and then dividing by the number of values. In the given exercise, the data set consists of distances:

• 0 km
• 1 km
• 1 km
• 1 km
• 2 km
• 2 km
• 2 km
• 3 km
• 3 km
• 4 km
• 5 km

To find the mean, add these values together: \[ 0 + 1 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 + 5 = 24 \] Then divide by the number of values, which is 11: \[ \frac{24}{11} \approx 2.18 \] This tells us that the average distance for this data set is approximately 2.18 km. Calculating the mean provides a quick sense of what is typical among the values.

Standard Deviation

The standard deviation measures how spread out the numbers in a data set are around the mean. In simpler terms, it tells us about the typical distance each data point is from the mean. To find the standard deviation, first calculate the variance as an intermediate step. Once you have the variance, the standard deviation is the square root of the variance. In our example, the variance was calculated as 1.6015. Therefore, the standard deviation is: \[ \sqrt{1.6015} \approx 1.27 \] The standard deviation of approximately 1.27 km indicates that most of the data points fall within 1.27 km of the mean distance. A small standard deviation implies that the values are close to the mean, whereas a larger standard deviation indicates a wider spread.

Variance

Variance is a statistical measure that represents the degree of spread in a data set. It is calculated by averaging the squared differences of each data point from the mean. This gives insight into the data's variability.For our data set, calculate each data point's deviation from the mean (2.18 km), square these deviations, and then find their average. Here are some examples of squared deviations:

• (0 - 2.18)^2 = 4.7524
• (1 - 2.18)^2 = 1.3924
• (2 - 2.18)^2 = 0.0324

Add all of these squared differences and divide by the total number of data points to find the variance: \[ \frac{17.616}{11} \approx 1.6015 \] This value of approximately 1.6015 tells us about the spread of the data points around the mean. The higher the variance, the more spread out the values are.

Data Analysis

Data analysis involves examining data sets to draw meaningful insights. It often starts by calculating basic statistics such as the mean, variance, and standard deviation. These metrics help summarize central tendencies, variability, and patterns within the data. In our exercise, after calculating the mean of 2.18 km, the variance of approximately 1.6015, and the standard deviation of roughly 1.27 km, we gain valuable insights into our data set.

• The mean indicates the central point of our data set.
• The variance and standard deviation measure the spread of data points, helping gauge data consistency and reliability.

By understanding these elements, we can make data-driven decisions or predictions, ensuring an objective approach to interpreting information. Proper data analysis techniques are crucial in fields like finance, science, and business to make informed decisions.