3 replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ edited Apr 13 '17 at 12:55 Here is how you can apply the same principles used in this answerthis answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  $$\$$ 21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  $$\$$ 1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All] RandomVariate[dist, 10]  $$\$$ {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} Here is how you can apply the same principles used in this answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  $$\$$ 21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  $$\$$ 1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All] RandomVariate[dist, 10]  $$\$$ {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} Here is how you can apply the same principles used in this answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  $$\$$ 21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  $$\$$ 1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All] RandomVariate[dist, 10]  $$\$$ {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} 2 deleted 32 characters in body edited Aug 5 '16 at 21:11 Karsten 7. 25.5k55 gold badges5555 silver badges113113 bronze badges Here is how you can apply the same principles used in this answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  21.6074 $$\$$ 21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  1.94362 $$\$$ 1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All]  RandomVariate[dist, 10]  {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} $$\$$ {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} Here is how you can apply the same principles used in this answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All] RandomVariate[dist, 10]  {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} Here is how you can apply the same principles used in this answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  $$\$$ 21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  $$\$$ 1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All] RandomVariate[dist, 10]  $$\$$ {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182} 1 answered Aug 5 '16 at 20:48 Karsten 7. 25.5k55 gold badges5555 silver badges113113 bronze badges Here is how you can apply the same principles used in this answer to your problem: distSum = {0.167467 E^(-(1/2) (-21.5 + x)^2), 0.160772 E^(-(1/2) (-19.7 + x)^2), 0.233762 E^(-(1/2) (-21.7 + x)^2), 0.0930353 E^(-(1/2) (-21.9 + x)^2), 0.373293 E^(-(1/2) (-22.4 + x)^2), 0.126876 E^(-(1/2) (-19. + x)^2), 0.348056 E^(-(1/2) (-22.7 + x)^2), 0.393082 E^(-(1/2) (-21.4 + x)^2), 0.35867 E^(-(1/2) (-20.9 + x)^2), 0.383496 E^(-(1/2) (-22.3 + x)^2)}; dist = ProbabilityDistribution[ Plus @@ distSum, {x, -Infinity, Infinity}, Method -> "Normalize"]; mean = NExpectation[x, x \[Distributed] dist]  21.6074 variance = Expectation[(x - mean)^2, x \[Distributed] dist] // Chop  1.94362 Plot[PDF[dist, x], {x, 12, 30}, PlotRange -> All] RandomVariate[dist, 10]  {20.7231, 21.2831, 20.6987, 22.486, 22.7043, 25.5388, 21.5406, 21.9175, 20.5078, 21.9182}