Solution
Mean μ = ∑ X i N \mu = \dfrac{\sum X_i}{N} μ = N ∑ X i
Standard deviation σ = ∑ ( X i − μ ) 2 N \sigma=\sqrt{\dfrac{\sum (X_i-\mu)^2}{N}} σ = N ∑ ( X i − μ ) 2
(a) Mean
μ = 265 + 270 + 290 + 300 + 315 + 320 6 \mu =\frac{265+270+290+300+315+320}{6} μ = 6 265 + 270 + 290 + 300 + 315 + 320
μ = 1760 6 = 293.33 \mu =\dfrac{1760}{6}=293.33 μ = 6 1760 = 293.33
(b) Standard deviation
X X − μ ( X − μ ) 2 265 − 28.33 802.77 270 − 23.33 544.44 290 − 3.33 11.11 300 6.67 44.44 315 21.67 469.44 320 26.67 711.11 ∑ 2583.33 \def\arraystretch{1.5}\begin{array}{c:c:c}X&X-\mu&(X-\mu)^2\\\hline265&-28.33&802.77\\\hdashline270&-23.33&544.44\\\hdashline290&-3.33&11.11\\\hdashline300&6.67&44.44\\\hdashline315&21.67&469.44\\\hdashline320&26.67&711.11\\\hline\sum &&2583.33\\\hline\end{array} X 265 270 290 300 315 320 ∑ X − μ − 28.33 − 23.33 − 3.33 6.67 21.67 26.67 ( X − μ ) 2 802.77 544.44 11.11 44.44 469.44 711.11 2583.33
σ = 2583.33 6 \sigma=\sqrt{\dfrac{2583.33}{6}} σ = 6 2583.33
σ = 20.75 \sigma=20.75 σ = 20.75