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Statistics · real student question

Find the standard deviation of 2.9, 2.5, 2.9.

Question

Find the standard deviation of

2.9,2.5,2.92.9,\quad 2.5,\quad 2.9

Step-by-step solution

  1. Compute the mean, keeping it as a fraction. Two of the three values are identical:

    xˉ=2.9+2.5+2.93=8.33=2.76‾≈2.76667\bar x=\frac{2.9+2.5+2.9}{3}=\frac{8.3}{3}=2.7\overline{6}\approx 2.76667

    Because the mean repeats, rounding it to 2.772.77 before squaring would visibly distort the answer — keep at least five digits, or work with 8330\tfrac{83}{30}.

  2. Find the deviations.

    2.9−8330=87−8330=430=0.133‾(twice),2.5−8330=75−8330=−830=−0.266‾2.9-\frac{83}{30}=\frac{87-83}{30}=\frac{4}{30}=0.13\overline{3}\quad(\text{twice}),\qquad 2.5-\frac{83}{30}=\frac{75-83}{30}=-\frac{8}{30}=-0.26\overline{6}

    They sum to 4+4−830=0  ✓\tfrac{4+4-8}{30}=0\;\checkmark.

  3. Square and add, using the fractions.

    2(430)2+(830)2=32+64900=96900=875≈0.1066672\left(\frac{4}{30}\right)^2+\left(\frac{8}{30}\right)^2=\frac{32+64}{900}=\frac{96}{900}=\frac{8}{75}\approx 0.106667

  4. Divide by n−1n-1 or by nn.

    s2=8/752=475≈0.053333,σ2=8/753=8225≈0.035556s^2=\frac{8/75}{2}=\frac{4}{75}\approx 0.053333,\qquad \sigma^2=\frac{8/75}{3}=\frac{8}{225}\approx 0.035556

  5. Take the square roots.

    s=475=275≈0.23094,σ=8225=2215≈0.18856s=\sqrt{\frac{4}{75}}=\frac{2}{\sqrt{75}}\approx 0.23094,\qquad \sigma=\sqrt{\frac{8}{225}}=\frac{2\sqrt2}{15}\approx 0.18856

    Report s≈0.231s\approx 0.231 for a sample and σ≈0.189\sigma\approx 0.189 for a complete population. As a check, the range is 2.9−2.5=0.42.9-2.5=0.4, and both values are comfortably below it ✓\checkmark.

Answer

s=275≈0.2309 (sample),σ=2215≈0.1886 (population)s=\frac{2}{\sqrt{75}}\approx 0.2309\ \text{(sample)},\qquad \sigma=\frac{2\sqrt2}{15}\approx 0.1886\ \text{(population)}

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