The following are data obtained from rolling a die several times. Compute the mean and variance of the frequency distribution.
Outcome Number of Occurrences
1 4
2 8
3 6
4 12
5 3
6 7
Xˉ=∑xif(xi)∑f(xi)=(1×4)+(2×8)+(3×6)+(4×12)+(5×3)+(6×7)40=4+16+18+48+15+4240=3.575S2=∑(X−Xˉ)2fn=((1−3.575)2×4)+((2−3.575)2×8)+((3−3.575)2×6)+((4−3.575)2×12)+((5−3.575)2×3)+((6−3.575)2×7)40=26.522+4.961+1.983+4.961+19.891+89.46440=147.78240=3.69\bar{X} = \frac{\sum x_if(x_i)}{\sum f(x_i)} = \frac{(1\times 4) + (2 \times 8) + (3 \times 6) + (4 \times 12) + (5 \times 3) + (6 \times 7)}{40} \\ = \frac{4+16+18+48+15+42}{40} \\ = 3.575 \\ S^2 = \frac{\sum (X- \bar{X})^2f}{n}\\ = \frac{((1-3.575)^2 \times 4) + ((2-3.575)^2 \times 8) + ((3-3.575)^2 \times 6) + ((4-3.575)^2 \times 12) + ((5-3.575)^2 \times 3) + ((6-3.575)^2 \times 7)}{40} \\ = \frac{26.522+4.961+1.983+4.961+19.891+89.464}{40} \\ = \frac{147.782}{40} \\ = 3.69Xˉ=∑f(xi)∑xif(xi)=40(1×4)+(2×8)+(3×6)+(4×12)+(5×3)+(6×7)=404+16+18+48+15+42=3.575S2=n∑(X−Xˉ)2f=40((1−3.575)2×4)+((2−3.575)2×8)+((3−3.575)2×6)+((4−3.575)2×12)+((5−3.575)2×3)+((6−3.575)2×7)=4026.522+4.961+1.983+4.961+19.891+89.464=40147.782=3.69
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