Problem 7 - Algebra 2

Question

Find the variance and standard deviation of each set of data to the nearest tenth. {5, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 9}

Step-by-Step Solution

Answer

Variance: 1.7; Standard Deviation: 1.3

Step 1: Calculate the Mean

To find the mean, add all the numbers together and divide by the number of values. Calculate: \( \frac{5 + 4 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 7 + 7 + 7 + 7 + 8 + 9}{16} = \frac{98}{16} = 6.125 \)

Step 2: Calculate Each Deviation from the Mean

Subtract the mean from each number in the data set. For example, for the number 5: \( 5 - 6.125 = -1.125 \). Repeat for all values.

Step 3: Square Each Deviation

Square each result from Step 2. For example, \((-1.125)^2 = 1.2656\). Calculate the square for each deviation.

Step 4: Calculate the Variance

Add all squared deviations from Step 3 and then divide by the number of values (16): \( \frac{12.24 + ... + 8.2656}{16} = 1.7343\). Round to the nearest tenth: 1.7.

Step 5: Calculate the Standard Deviation

Take the square root of the variance obtained in Step 4: \( \sqrt{1.7343} \approx 1.316\), rounded to the nearest tenth: 1.3.

Key Concepts

Mean Calculation

Understanding the mean calculation is the first step in finding variance and standard deviation. The mean represents the average of a set of numbers, providing a central value for the data set. To calculate the mean, you add up all the numbers in the data set and divide by the number of values. In the given data set, we sum up all the numbers like this: \[ 5 + 4 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 7 + 7 + 7 + 7 + 8 + 9 = 98 \] Next, we divide the total by the number of items in the set, which is 16:\[ \frac{98}{16} = 6.125 \]This result informs us that 6.125 is the mean. It represents the balance point of the data set and is essential for calculating deviations.

Deviation from Mean

Once the mean is known, each data point's deviation from the mean can be calculated. A deviation signifies how far each number in the set is from the mean. This step allows us to capture the fluctuation or variation of data points, which is vital for variance and standard deviation calculations. For example, take the number 5 from the data set:\[ 5 - 6.125 = -1.125 \]What this tells us is that 5 is 1.125 units below the mean of 6.125. You perform the same calculation for each number in the data set, resulting in deviations that can be both negative and positive. These deviations provide the raw data needed to compute the variance in the next step.

Squared Deviation

Squaring each deviation helps in eliminating any negative signs, allowing us to focus solely on the magnitude of deviations regardless of direction. Squaring also gives more weight to larger deviations, emphasizing their impact.Taking the deviation for 5 as an example:\[ (-1.125)^2 = 1.2656 \]To calculate variance, each deviation is squared in this way, which ensures that all resulting numbers are positive and suitable for further calculations. The squared deviations offer a comprehensive measure of how dispersed or concentrated data points are around the mean.

Square Root

The square root calculation is the final step in finding the standard deviation, which provides a measure as to how much variation there is from the mean. The square root of the variance returns the measure of standard deviation to the same units as the original data.To illustrate, take the variance, which is calculated by averaging all squared deviations:\[ 1.7343 \] (as computed earlier).Taking the square root of the variance gives:\[ \sqrt{1.7343} \approx 1.316 \]Rounding this to the nearest tenth provides us with a standard deviation of 1.3. This number indicates that, on average, each data point deviates by approximately 1.3 units from the mean. Standard deviation is a critical tool in statistics for assessing how widely spread out the values in a data set are.