hw5 110590049

tags probability

2023PB_HW5.pdf

1

$$\begin{gathered} P(X=0)=\frac{6}{36},P(X=1)=\frac{10}{36},P(X=2)=\frac{8}{36}, \\ P(X=3)=\frac{6}{36},P(X=4)=\frac{4}{36},P(X=5)=\frac{2}{36} \end{gathered}$$

2

$$\begin{gathered} P(X<1)=F(1^{-})=\frac{1}{2} \\ P(X=1)=F(1)-F(1^{-})=\frac{1}{6} \\ P(0\le X<1)=F(1^{-})-F(0^{-})=\frac{1}{4} \\ P(X>\frac{1}{2})=1-F(\frac{1}{2})=\frac{1}{2} \\ P(X=\frac{3}{2})=F(\frac{3}{2})-F(\frac{3^{-}}{2})=0 \\ P(1<X\le 6)=F(6)-F(1)=\frac{1}{3} \end{gathered}$$

3

$$\begin{gathered} q=1-p=0.88 \\ 1-q^x\ge 0.6 \\ {q}^x\le 0.4 \\ 0.88^x\le 0.4 \\ x\times ln(0.88)\ge ln(0.4) \\ x\ge 7.1,x=8 \end{gathered}$$

4

$$\begin{gathered} P(1)=\frac{11}{36},P(2)=\frac{9}{36},P(3)=\frac{7}{36} \\ P(5)=\frac{5}{36},P(5)=\frac{3}{36},P(6)=\frac{1}{36} \\ f(x)=\{\begin{array}{ll}0& x<1\\ \frac{11}{36}& 1\le x<2\\ \frac{5}{9}& 2\le x<3\\ \frac{3}{4}& 3\le x<4\\ \frac{8}{9}& 5\le x<6\\ 1& 6\le x\end{array} \end{gathered}$$

5

$$\frac{2000000}{2000000}\ast -1+\frac{4000}{2000000}\ast 30+\frac{500}{2000000}\ast 800+\frac{1}{2000000}\ast 1200000=-0.14$$

6

$$\begin{gathered} E[x(11-x)]=E[-x^2]+11E[x] \\ E[-x^2]=\sum _{i=1}^{10}\frac{-i^2}{10}=-38.5 \\ 11E[x]=11\sum _{i=1}^{10}\frac{i}{10}=60.5 \\ E[x(11-x)]=22 \end{gathered}$$

7

$$\begin{gathered} \text{First one.} \\ \text{Because the standard deviation is smaller} \end{gathered}$$

8

$$\begin{gathered} E[x]=\sum _{i=0}^{\infty }i\ast p(x=i),E[x^2]=\sum _{i=0}^{\infty }{i}^2\ast p(x=i) \\ E[x]=-3\ast \frac{3}{8}+0\ast \frac{3}{8}+6\ast \frac{1}{4}=\frac{3}{8} \\ E[x^2]=9\ast \frac{3}{8}+0\ast \frac{3}{8}+36\ast \frac{1}{4}=\frac{99}{8} \\ var(x)=E[x^2]-(E[x]{)}^2=12.23 \\ \sigma =\sqrt{var(x)}=3.498 \end{gathered}$$

9

$$\begin{gathered} E[x^2-2x]=E[x^2]-2E[x]=3 \\ E[x^2]=5 \\ var(x)=E[x^2]-(E[x]{)}^2=5-1=4 \\ var(-3x+4)=(-3{)}^{2}var(x)=9\ast 4=36 \end{gathered}$$