How is the corrected sample variance calculated for the six pulse-rate data points?

Conditions

• The sample data consists of the values 71, 83, 67, 74, 70, and 79.
• The sample mean $\bar{x}$ has been calculated as 74.
• The sample size is $n=6$, so the denominator is $n-1=5$.

Reasoning, step by step

1. Subtract the sample mean (74) from each data point to find the deviations: -3, 9, -7, 0, -4, 5.
2. Square each deviation: 9, 81, 49, 0, 16, 25.
3. Sum the squared deviations: $9 + 81 + 49 + 0 + 16 + 25 = 180$.
4. Divide the sum by $n-1 = 5$: $S^2 = \frac{180}{5} = 36$.

Example

The screen displays "S² = $1/(6-1)$[(71-74)^2 + ... + (79-74)^2]", and finally calculates 36.

Common misconceptions

• Dividing the sum of squared deviations by $n=6$ instead of $n-1=5$.
• Forgetting to square the deviations before summing them.

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Connected concepts

Explore next

• How is the sample standard deviation derived from the sample variance?
• Why is the denominator of the corrected sample variance n-1 instead of n?
• What is the sample arithmetic mean of this dataset?