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Worked solution

Estimate mean and variance from seven raw values

Given the data {52,50,65,59,70,63,67}\{52,50,65,59,70,63,67\}, find unbiased estimates of the mean and variance of the population from which the data are drawn.

CJC · Lecture · Example 2

Method 1: raw-data summaries

These are seven raw observations, not a frequency table. ∑x=426\sum x=426 and ∑x2=26268\sum x^2=26268. Therefore xˉ=426/7=60.85714286≈60.9\bar x=426/7=60.85714286\approx60.9 and s2=16(26268−42627)=4007≈57.1.\begin{aligned}&s^2\\ & =\frac1{6}\left(26268-\frac{426^2}{7}\right)\\ & =\frac{400}{7}\approx57.1.\end{aligned} The denominator is n−1=6n-1=6.

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Method 2: calculator and notation

Press STAT → 1:Edit. Highlight L1 and press CLEAR to remove old entries, then enter 52, 50, 65, 59, 70, 63, 67 in L1. Choose STAT → CALC → 1:1-Var Stats, set List L1 with no frequency list, and Calculate → ENTER. The screen gives n=7n=7, xˉ=60.85714286\bar x=60.85714286, ∑x=426\sum x=426, ∑x2=26268\sum x^2=26268, Sx=7.55928946S_x=7.55928946, σx=6.998542122\sigma_x=6.998542122, minimum 50 and first quartile 52. Retrieve VARS → 5:Statistics → 3:Sx and square it with x² → ENTER: Sx2=57.14285714S_x^2=57.14285714. Use unrounded stored values.

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Common mistakes

The teacher’s larger-value rule distinguishes the n−1 and n denominators for the same nonconstant sample. Sx² is unbiased for population variance; Sx is not generally unbiased for population standard deviation, despite the source’s wording. If all observations are equal both displayed SDs are zero.

Solution notes

Teacher page 10 flags raw data; page 11 explicitly contrasts Sx and σx. The source’s “unbiased standard deviation” wording is qualified rather than silently repeated. No printed marks.

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