Standard Deviation: Measuring Spread

April 12, 2026

Problem

Find the standard deviation of {4, 8, 6, 5, 3}. Show each point's distance from the mean.

Explanation

What does standard deviation measure?

Standard deviation (σ\sigma or ss) measures the typical distance of data points from the mean. A small SD means data is tightly clustered; a large SD means data is spread out.

Step-by-step: {4, 8, 6, 5, 3}

Step 1 — Find the mean:

xˉ=4+8+6+5+35=265=5.2\bar{x} = \frac{4 + 8 + 6 + 5 + 3}{5} = \frac{26}{5} = 5.2

Step 2 — Find each deviation (distance from mean):

45.2=1.24 - 5.2 = -1.2, 85.2=2.88 - 5.2 = 2.8, 65.2=0.86 - 5.2 = 0.8, 55.2=0.25 - 5.2 = -0.2, 35.2=2.23 - 5.2 = -2.2

Step 3 — Square each deviation:

1.44,  7.84,  0.64,  0.04,  4.841.44, \; 7.84, \; 0.64, \; 0.04, \; 4.84

Step 4 — Average the squared deviations (for population SD, divide by nn; for sample SD, divide by n1n-1):

Variance=1.44+7.84+0.64+0.04+4.845=14.85=2.96\text{Variance} = \frac{1.44 + 7.84 + 0.64 + 0.04 + 4.84}{5} = \frac{14.8}{5} = 2.96

Step 5 — Take the square root:

σ=2.961.72\sigma = \sqrt{2.96} \approx 1.72

Interpretation

The data values are, on average, about 1.72 units away from the mean of 5.2. About 68% of data in a normal distribution falls within one SD of the mean.

Population vs Sample SD

  • Population (σ\sigma): divide by nn — used when you have the entire population.
  • Sample (ss): divide by n1n - 1 — used when your data is a sample from a larger population (Bessel's correction).

Try it in the visualization

Each data point is plotted. Distance lines show how far each value is from the mean. Squared deviations are shown as literal squares whose total area is minimized.

Interactive Visualization

Parameters

4.00
8.00
6.00
5.00
3.00
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Standard Deviation: Measuring Spread | MathSpin