Problem 9 - Algebra 2

Question

Find the variance and standard deviation of each set of data to the nearest tenth. {4.3, 6.4, 2.9, 3.1, 8.7, 2.8, 3.6, 1.9, 7.2}

Step-by-Step Solution

Answer

Variance: 4.8; Standard Deviation: 2.2.

Step 1: Calculate the Mean

To find the variance, we first need to calculate the mean (average) of the data set. Add up all the numbers in the set and then divide by the number of data points. Mean, \( \mu = \frac{4.3 + 6.4 + 2.9 + 3.1 + 8.7 + 2.8 + 3.6 + 1.9 + 7.2}{9} = \frac{40.9}{9} = 4.5444 \approx 4.5 \) to the nearest tenth.

Step 2: Calculate Each Data Point's Deviation from the Mean

Subtract the mean from each data point to find the deviation of each point. \( 4.3 - 4.5 = -0.2 \) \( 6.4 - 4.5 = 1.9 \) \( 2.9 - 4.5 = -1.6 \) \( 3.1 - 4.5 = -1.4 \) \( 8.7 - 4.5 = 4.2 \) \( 2.8 - 4.5 = -1.7 \) \( 3.6 - 4.5 = -0.9 \) \( 1.9 - 4.5 = -2.6 \) \( 7.2 - 4.5 = 2.7 \).

Step 3: Square Each Deviation

Square each of the deviations calculated in Step 2.\( (-0.2)^2 = 0.04 \) \( (1.9)^2 = 3.61 \) \( (-1.6)^2 = 2.56 \) \( (-1.4)^2 = 1.96 \) \( (4.2)^2 = 17.64 \) \( (-1.7)^2 = 2.89 \) \( (-0.9)^2 = 0.81 \) \( (-2.6)^2 = 6.76 \) \( (2.7)^2 = 7.29 \).

Step 4: Calculate the Variance

The variance is the average of these squared deviations. Add up all the squared deviations and divide by the number of data points:Variance, \( \sigma^2 = \frac{0.04 + 3.61 + 2.56 + 1.96 + 17.64 + 2.89 + 0.81 + 6.76 + 7.29}{9} = \frac{43.56}{9} \approx 4.840 \approx 4.8 \) to the nearest tenth.

Step 5: Calculate the Standard Deviation

The standard deviation is the square root of the variance. Taking the square root of the variance found in Step 4:Standard Deviation, \( \sigma = \sqrt{4.8} \approx 2.19 \approx 2.2 \) to the nearest tenth.

Key Concepts

Standard Deviation

The standard deviation is a key concept in statistics that helps us understand how spread out the numbers in a data set really are. Essentially, it tells us how much the individual numbers differ from the average of the entire data set.
To compute the standard deviation for a sample, we first need to find the variance, which is a measure of the spread squared.
Then, taking the square root of the variance gives us the standard deviation. This value helps us measure the data set's spread in the same units as the initial data points making it easily interpretable.

• If the standard deviation is small, the data points are close to the mean.
• If it's large, the data points are more spread out.

This information can help us make better decisions based on the pattern of the data.

Mean Calculation

The first step in assessing the variance and standard deviation is to calculate the mean—commonly known as the average. The mean is calculated by adding all data points together and then dividing by the number of data points in the set.
For example, in our exercise, the mean is found by summing all the values and dividing by nine (the total number of data points), which yields approximately 4.5.

• The mean provides a central value around which all observations are compared.
• It's crucial to get an accurate mean as it influences all subsequent calculations.

A correctly calculated mean simplifies understanding data spread and leads to precise results in finding variance and standard deviation.

Data Deviation

Once the mean is calculated, we look into how each data point deviates or differs from this mean.
To find each data point's deviation, subtract the mean from each individual data value. Deviation describes how far a data point is from the average.

• A positive deviation indicates the value is above the mean.
• A negative deviation shows it is below the mean.

Understanding these deviations is important because they form the basis of later calculations for variance and standard deviation.
They tell us how much data points differ from the central value point.

Squared Deviations

With each deviation calculated, the next process in determining variance is to square each deviation. Squaring the deviations ensures all measurements are non-negative and emphasizes larger deviations. This helps in accurately measuring the spread of the data.
In our exercise, each deviation from the mean is squared; these are then summed to form the total of squared deviations.

• Squaring eliminates negative signs allowing for consistent calculations.
• It gives more weight to larger deviations, emphasizing outliers.

Through squaring deviations, we prepare the data set to capture total variance, providing insight into overall data spread.