WEBVTT

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Here is the problem: Find the sample standard deviation of 88, 92, 79, 85, 96.

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Treat the data as a sample from a larger population (Bessel prime s correction, divide by n minus 1).

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Find the mean.

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Square each deviation from the mean.

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Add the squared deviations.

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Divide by n minus 1 to get the sample variance.

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Take the square root.

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So the answer is s equals 6.519202405.

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Standard deviation measures how spread out the values are around the mean. A small value means the data are clustered near the mean.

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Variance equals 42.5, and 6.519202405 squared equals 42.5.

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Population SD for comparison: 5.830951895.