Calculate The Variance and Standard Deviation for The Following Sample

Variance and standard deviation are fundamental measures of statistical dispersion that help quantify how spread out numbers in a data set are. This guide explains how to calculate them for a sample, provides an interactive calculator, and offers practical interpretation tips.

What is variance?

Variance measures how far each number in a data set is from the mean (average) of the set. A higher variance indicates that the numbers are more spread out, while a lower variance indicates they are closer to the mean.

For a sample (a subset of a larger population), we use the sample variance formula, which divides by n-1 (degrees of freedom) to provide an unbiased estimate of the population variance.

What is standard deviation?

Standard deviation is the square root of the variance. It provides a measure of dispersion in the same units as the original data, making it more interpretable than variance alone.

Like variance, we calculate sample standard deviation using the sample mean and dividing by n-1.

How to calculate variance and standard deviation

To calculate variance and standard deviation for a sample:

Find the mean (average) of your data set
For each number, subtract the mean and square the result (these are the squared differences)
Sum all the squared differences
Divide the sum by n-1 (where n is the number of data points) to get the sample variance
Take the square root of the variance to get the standard deviation

Sample Variance Formula

s² = Σ(xᵢ - x̄)² / (n - 1)

Where:

s² = sample variance
xᵢ = each individual data point
x̄ = sample mean
n = number of data points

Sample Standard Deviation Formula

s = √[Σ(xᵢ - x̄)² / (n - 1)]

Where:

s = sample standard deviation
All other variables are as defined above

Note: For population variance and standard deviation, we divide by n instead of n-1. This calculator uses the sample formulas with n-1 in the denominator.

Example calculation

Let's calculate variance and standard deviation for the following sample of test scores: 85, 90, 78, 92, 88.

Step 1: Calculate the mean

Mean = (85 + 90 + 78 + 92 + 88) / 5 = 433 / 5 = 86.6

Step 2: Calculate squared differences from the mean

(85 - 86.6)² = (-1.6)² = 2.56
(90 - 86.6)² = (3.4)² = 11.56
(78 - 86.6)² = (-8.6)² = 73.96
(92 - 86.6)² = (5.4)² = 29.16
(88 - 86.6)² = (1.4)² = 1.96

Step 3: Sum the squared differences

Sum = 2.56 + 11.56 + 73.96 + 29.16 + 1.96 = 119.24

Step 4: Calculate variance

Variance = 119.24 / (5 - 1) = 119.24 / 4 = 29.81

Step 5: Calculate standard deviation

Standard Deviation = √29.81 ≈ 5.46

Results

For the sample [85, 90, 78, 92, 88]:

Variance: 29.81

Standard Deviation: 5.46

Interpreting the results

A variance of 29.81 means that, on average, each score in the sample differs from the mean by about 5.46 points. This indicates moderate variability in the test scores.

The standard deviation of 5.46 provides a more intuitive measure of dispersion. It means that most scores fall within about 5.46 points of the mean (approximately 81.14 to 92.06 in this case).

These measures help assess the consistency of the test scores. A higher standard deviation would suggest more variability in the data, while a lower value would indicate more consistency.

FAQ

What's the difference between variance and standard deviation?

Variance measures the average squared deviation from the mean, while standard deviation is the square root of variance. Standard deviation is in the same units as the original data, making it more interpretable.

When should I use sample vs. population formulas?

Use sample formulas (with n-1 in the denominator) when your data represents a subset of a larger population. Use population formulas when you have data for the entire population.

What does a high standard deviation mean?

A high standard deviation indicates that the data points are spread out over a wider range of values. This suggests more variability or inconsistency in the data.

Can I calculate variance and standard deviation with negative numbers?

Yes, the formulas work with negative numbers. The squared differences ensure all values are positive, and the final standard deviation will be in the same units as your original data.