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Statistics · real student question

Find the standard deviation of 5.58, 5.53, 6.19, 5.66, 5.87, 5.62.

Question

Find the sample standard deviation of

5.58, 5.53, 6.19, 5.66, 5.87, 5.625.58,\ 5.53,\ 6.19,\ 5.66,\ 5.87,\ 5.62

Step-by-step solution

  1. Find the mean and keep extra digits.

    xˉ=34.456=5.7416‾≈5.74167\bar{x} = \frac{34.45}{6} = 5.741\overline{6} \approx 5.74167

    This mean does not terminate, so carrying only two decimals (5.745.74) would visibly shift the squared deviations. Keep four or five digits until the very end.

  2. Square each deviation from the mean.

    (5.58−5.74167)2≈0.02614,(5.53−5.74167)2≈0.04480(5.58 - 5.74167)^2 \approx 0.02614, \quad (5.53 - 5.74167)^2 \approx 0.04480

    (6.19−5.74167)2≈0.20100,(5.66−5.74167)2≈0.00667(6.19 - 5.74167)^2 \approx 0.20100, \quad (5.66 - 5.74167)^2 \approx 0.00667

    (5.87−5.74167)2≈0.01647,(5.62−5.74167)2≈0.01480(5.87 - 5.74167)^2 \approx 0.01647, \quad (5.62 - 5.74167)^2 \approx 0.01480

  3. Add the squared deviations.

    ∑(x−xˉ)2≈0.30988\sum (x - \bar{x})^2 \approx 0.30988

    Notice that 6.196.19 alone contributes 0.2010.201 of that total — about two thirds. A single value far from the mean dominates the variance because the deviation is squared.

  4. Divide by n − 1 for the sample variance. With n=6n = 6 values treated as a sample:

    s2=0.309885≈0.06198s^2 = \frac{0.30988}{5} \approx 0.06198

    If the six values were the whole population you would divide by 66 instead, giving σ2≈0.05165\sigma^2 \approx 0.05165.

  5. Take the square root.

    s=0.06198≈0.2490s = \sqrt{0.06198} \approx 0.2490

    So the sample standard deviation is about 0.2490.249 (the population value would be σ≈0.227\sigma \approx 0.227). Rounded to two decimals the sample answer is 0.250.25.

Answer

s≈0.249(sample),σ≈0.227(population)s \approx 0.249 \quad \text{(sample)}, \qquad \sigma \approx 0.227 \quad \text{(population)}

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