hw5 tags probability 2023PB_HW5.pdf 1 P(X=0)=636,P(X=1)=1036,P(X=2)=836,P(X=3)=636,P(X=4)=436,P(X=5)=236P(X=0)=\frac{6}{36}, P(X=1)=\frac{10}{36}, P(X=2)=\frac{8}{36},\\ P(X=3)=\frac{6}{36}, P(X=4)=\frac{4}{36}, P(X=5)=\frac{2}{36}P(X=0)=366,P(X=1)=3610,P(X=2)=368,P(X=3)=366,P(X=4)=364,P(X=5)=362 2 P(X<1)=F(1−)=12P(X=1)=F(1)−F(1−)=16P(0≤X<1)=F(1−)−F(0−)=14P(X>12)=1−F(12)=12P(X=32)=F(32)−F(3−2)=0P(1<X≤6)=F(6)−F(1)=13P(X<1)=F(1^-)=\frac{1}{2}\\ P(X=1)=F(1)-F(1^-)=\frac{1}{6}\\ P(0 \le X<1)=F(1^-)-F(0^-)=\frac{1}{4}\\ P(X>\frac{1}{2})=1-F(\frac{1}{2})=\frac{1}{2}\\ P(X=\frac{3}{2})=F(\frac{3}{2})-F(\frac{3^-}{2})=0\\ P(1<X\le6)=F(6)-F(1)=\frac{1}{3}\\P(X<1)=F(1−)=21P(X=1)=F(1)−F(1−)=61P(0≤X<1)=F(1−)−F(0−)=41P(X>21)=1−F(21)=21P(X=23)=F(23)−F(23−)=0P(1<X≤6)=F(6)−F(1)=31 3 q=1−p=0.881−qx≥0.6qx≤0.40.88x≤0.4x×ln(0.88)≥ln(0.4)x≥7.1,x=8q=1-p=0.88\\ 1-q^x\ge 0.6\\ q^x\le 0.4\\ 0.88^x\le0.4\\ x\times ln(0.88)\ge ln(0.4)\\ x\ge 7.1,x=8q=1−p=0.881−qx≥0.6qx≤0.40.88x≤0.4x×ln(0.88)≥ln(0.4)x≥7.1,x=8 4 P(1)=1136,P(2)=936,P(3)=736P(5)=536,P(5)=336,P(6)=136f(x)={0x<111361≤x<2592≤x<3343≤x<4895≤x<616≤xP(1)=\frac{11}{36}, P(2)=\frac{9}{36}, P(3)=\frac{7}{36}\\ P(5)=\frac{5}{36}, P(5)=\frac{3}{36}, P(6)=\frac{1}{36}\\ f(x)=\begin{cases} 0 &x<1\\ \frac{11}{36} & 1 \le x<2\\ \frac{5}{9} & 2 \le x<3\\ \frac{3}{4} & 3 \le x<4\\ \frac{8}{9} & 5 \le x<6\\ 1 & 6 \le x\\ \end{cases}P(1)=3611,P(2)=369,P(3)=367P(5)=365,P(5)=363,P(6)=361f(x)=⎩⎨⎧036119543981x<11≤x<22≤x<33≤x<45≤x<66≤x 5 20000002000000∗−1+40002000000∗30+5002000000∗800+12000000∗1200000=−0.14\frac{200 0000}{2000000}*-1 +\frac{4000}{2000000}*30+\frac{500}{2000000}*800+\frac{1}{2000000}*1200000=-0.1420000002000000∗−1+20000004000∗30+2000000500∗800+20000001∗1200000=−0.14 6 E[x(11−x)]=E[−x2]+11E[x]E[−x2]=∑i=110−i210=−38.511E[x]=11∑i=110i10=60.5E[x(11−x)]=22E[x(11-x)]=E[-x^2]+11E[x]\\ E[-x^2]=\sum_{i=1}^{10}\frac{-i^2}{10}=-38.5\\ 11E[x]=11\sum_{i=1}^{10}\frac{i}{10}=60.5\\ E[x(11-x)]=22E[x(11−x)]=E[−x2]+11E[x]E[−x2]=i=1∑1010−i2=−38.511E[x]=11i=1∑1010i=60.5E[x(11−x)]=22 7 First one.Because the standard deviation is smaller\text{First one.}\\ \text{Because the standard deviation is smaller}First one.Because the standard deviation is smaller 8 E[x]=∑i=0∞i∗p(x=i),E[x2]=∑i=0∞i2∗p(x=i)E[x]=−3∗38+0∗38+6∗14=38E[x2]=9∗38+0∗38+36∗14=998var(x)=E[x2]−(E[x])2=12.23σ=var(x)=3.498E[x]=\sum_{i=0}^{\infty}i*p(x=i),E[x^2]=\sum_{i=0}^{\infty}i^2*p(x=i)\\ E[x]=-3*\frac{3}{8}+0*\frac{3}{8}+6*\frac{1}{4}=\frac{3}{8}\\ E[x^2]=9*\frac{3}{8}+0*\frac{3}{8}+36*\frac{1}{4}=\frac{99}{8}\\ var(x)=E[x^2]-(E[x])^2=12.23\\ \sigma=\sqrt{var(x)}=3.498E[x]=i=0∑∞i∗p(x=i),E[x2]=i=0∑∞i2∗p(x=i)E[x]=−3∗83+0∗83+6∗41=83E[x2]=9∗83+0∗83+36∗41=899var(x)=E[x2]−(E[x])2=12.23σ=var(x)=3.498 9 E[x2−2x]=E[x2]−2E[x]=3E[x2]=5var(x)=E[x2]−(E[x])2=5−1=4var(−3x+4)=(−3)2var(x)=9∗4=36E[x^2-2x]=E[x^2]-2E[x]=3\\ E[x^2]=5\\ var(x)=E[x^2]-(E[x])^2=5-1=4\\ var(-3x+4)=(-3)^2var(x)=9*4=36 E[x2−2x]=E[x2]−2E[x]=3E[x2]=5var(x)=E[x2]−(E[x])2=5−1=4var(−3x+4)=(−3)2var(x)=9∗4=36