Resultados de la Guía - Probabilidad y Estadística (61.06 y 61.09) [Foros-FIUBA::Wiki]

Traza: • Parcial - 63.01. Química X - xx/xx/20xx • Resultados de la Guía - Probabilidad y Estadística (61.06 y 61.09)

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Resultados de la Guía - Probabilidad y Estadística (61.06 y 61.09)

Resultados de la Guía - Probabilidad y Estadística (61.06 y 61.09)

Guía 1: Probabilidad

1. 0.56
2. 0.14
3. 0.04
4. 0.26
$<tex>\mbox{Mujeres solteras no universitarias}=-57</tex>$
1. 0.36
2. 0.64
3. 0.53
4. 0.17
5. 0.75
1. Roja
2. Roja
3. Blanca
1. 0.45
2. $<tex>B=\{1,2,5\}</tex>$
3. $<tex>C=\{\{3\}\}</tex>$
4. $<tex>D=\{2\}</tex>$ o $<tex>\{3\}</tex>$ o $<tex>\{2,3\}</tex>$ o $<tex>\{3,4\}</tex>$ o $<tex>\{2,4\}</tex>$ o $<tex>\{2,3,4\}</tex>$
1. 2/7
2. 16/35
3. 19/35
4. 2/5
El esposo.
1. 1/6
2. 13/18
3. 1/12
4. 1/4
5. 5/18
1. 0.5
2. $<tex>\frac{\pi}{4}</tex>$
3. $<tex>P(C_{k})=\frac{\left(0.5+\frac{1}{k}\right)^{2}}{2}</tex>$ con $<tex>k\geq2</tex>$ ; $<tex>P(\bigcap_{k=1}^{\infty}C_{k})=\frac{1}{8}</tex>$
4. $<tex>P(D_{k})=\pi\left(\frac{1}{k}\right)^{2}</tex>$ ; $<tex>P(D)=0</tex>$
5. $<tex>D_{k}</tex>$ representa el evento de que el dardo caiga dentro del círculo con centro en $<tex>\left(\frac{1}{2},\frac{1}{2}\right)</tex>$ y radio $<tex>\frac{1}{k}</tex>$ .
  $<tex>D</tex>$ representa el evento de que el dardo caiga en el centro del círculo.
2. $<tex>P(\mbox{gane Juan})=P(\mbox{gane Maria})=\frac{5}{14}</tex>$
  $<tex>P(\mbox{gane Pedro})=\frac{2}{7}</tex>$
  $<tex>P(\mbox{nadie gana})=0</tex>$
1. 1/1000
2. 1/59049
3. 0
1. Con reposición:
  1. 1/64
  2. 1/16
  3. 1/1600
  4. 3/8
2. Sin reposición:
  1. 3/247
  2. 12/247
  3. 0
  4. 100/247
1. 0.7015
2. 0.0489
1. 0.818
2. $<tex>P(k)=\frac{{100-k \choose 10}{k \choose 0}}{{100 \choose 10}}+\frac{{100-k \choose 9}{k \choose 1}}{{100 \choose 10}}</tex>$ si $<tex>0<k\leq90</tex>$
3. 0.3329
0.916
Conviene apostar al resultado del evento “A”
1. 1/6
2. 1/6
3. 0
4. 1/2
5. 1/4
1. 0.56
2. 5/28
3. 8/55
4. 17/28
5. 7/32
6. 1/5
1. 60/143
2. 5/12
1. 12/248
2. 12/248
3. 0.04858
4. 158/63973
5. 0.0508
1. 1/3
2. 1/2
1. 3/4
2. 1/2
3. 3/8
4. 1/5
1. 0.44
2. 19/22
9899/9900
1. 1/3
2. 40/63
3. 9/23
4. 17/21
5. Van a dar los mismos resultados porque las proporciones son las mismas.
24/49
1. 0.049
2. 0.00006734
1. 0.049
2. 0.000067
1. 0.0411
2. 0.086
4. 0.0611
1. Al menos 5 respuestas correctas.
2. 0.3671
1. 1/2
2. No son independientes.
$<tex>\begin{array}{|c|c|c|c|}\hline k & N=k & N\leq k & k\leq N\tabularnewline\hline \hline 0 & 400 & 400 & 900\tabularnewline\hline 1 & 400 & 800 & 500\tabularnewline\hline 2 & 100 & 900 & 100\tabularnewline\hline \end{array}</tex>$
1. 17/100
2. 1/150
3. 1/25
4. 23/25
5. $<tex>\frac{\pi}{100}</tex>$
1. $<tex>\Omega=\{1,2,3,\ldots,\}</tex>$
2. $<tex>P\left(A_{1}\right)=\frac{1}{6}</tex>$
  $<tex>P\left(A_{7}\right)=\left(\frac{5}{6}\right)^{6}\times\left(\frac{1}{6}\right)^{1}</tex>$
  $<tex>P\left(A_{1017}\right)=\left(\frac{5}{6}\right)^{1016}\times\left(\frac{1}{6}\right)^{1}</tex>$
3. $<tex>P(B_{1})=1</tex>$
  $<tex>P(B_{5})=0.4822</tex>$
  $<tex>P(B_{1000})=0</tex>$
5. $<tex>P(B)=0</tex>$
1. $<tex>P(\mbox{par})=0.422569</tex>$
2. $<tex>P(\mbox{dos pares})=0.47539</tex>$
3. $<tex>P(\mbox{terna})=0.021128</tex>$
4. $<tex>P(\mbox{escalera})=0.003940</tex>$
5. $<tex>P(\mbox{color})=0.001981</tex>$
6. $<tex>P(\mbox{full})=0.001441</tex>$
7. $<tex>P(\mbox{poker})=0.000240</tex>$
8. $<tex>P(\mbox{escalera real})=0.000015</tex>$
1. 1/2
2. 1/3
- $<tex>p_{0}=\frac{1}{4}</tex>$
- $<tex>p_{8}=\frac{1}{4}</tex>$
- $<tex>p_{5}=\frac{3}{8}</tex>$
1. 1/2
2. 1/2
3. $<tex>\left(\frac{1}{2}\right)^{11}</tex>$
4. 4/5
1. $<tex>\frac{p}{p+q}</tex>$
2. $<tex>\frac{p}{1-r^{3}}</tex>$ siendo $<tex>r=1-p-q</tex>$
3. $<tex>\frac{p}{p+q}</tex>$

Guía 2: Funciones de densidad, distribución, conjuntas y marginales

$<tex>F_{X}</tex>$ es monótona no decreciente
$<tex>F_{X}</tex>$ es continua a derecha
$<tex>\lim_{x\to-\infty}=0</tex>$
$<tex>\lim_{x\to\infty}=1</tex>$
1. En $<tex>X=\pm2</tex>$ , donde $<tex>P(X=\pm2)=\frac{1}{3}</tex>$
2. $<tex>\frac{2}{3},1,\frac{2}{3},\frac{1}{3}</tex>$
3. $<tex>0,\frac{1}{3},0</tex>$
4. $<tex>1,\frac{2}{3}</tex>$
5. $<tex>\frac{1}{2}</tex>$
1. $<tex>F_{X}(0)=0</tex>$
2. $<tex>F_{X}(0.8)=0.1</tex>$
3. $<tex>F_{X}(2)=0.25</tex>$
4. $<tex>F_{X}\left(2\sqrt{2}\right)=0.5</tex>$
5. $<tex>F_{X}(4)=1</tex>$
1. $<tex>F_{X}(x)=\begin{cases}0 & \mbox{si }0<x\\\frac{2}{3}x^{2}+\frac{1}{3}x & \mbox{si }0\leq x\leq1\\1 & \mbox{si }x>1\end{cases}</tex>$
2. $<tex>\frac{1}{3}</tex>$ , $<tex>\frac{4}{9}</tex>$
3. $<tex>\frac{1}{10}</tex>$
1. $<tex>f_{T*}=\begin{cases}0 & \forall\mbox{ otro }t*\\e^{-t*} & \mbox{si }0\leq t*<1\end{cases}</tex>$
2. $<tex>e^{-1}</tex>$
1. $<tex>\begin{array}{|c|c|}\hline k & P(X=k)\tabularnewline\hline \hline 0 & 0.0609\tabularnewline\hline 1 & 0.2284\tabularnewline\hline 2 & 0.3427\tabularnewline\hline 3 & 0.2570\tabularnewline\hline 4 & 0.0963\tabularnewline\hline 5 & 0.0144\tabularnewline\hline \end{array}</tex>$
2. $<tex>\begin{array}{|c|c|}\hline k & P(X=k)\tabularnewline\hline \hline 1 & 0.1428\tabularnewline\hline 2 & 0.5714\tabularnewline\hline 3 & 0.2857\tabularnewline\hline \end{array}</tex>$
1. $<tex>P(N=n)=\left(\frac{1}{4}\right)^{n-1}\left(\frac{3}{4}\right)</tex>$ para $<tex>n\in N</tex>$
2. 4/5
1. $<tex>\theta\in\left[-\frac{1}{2},\frac{1}{2}\right]</tex>$
2. $<tex>\mbox{Mediana}=\frac{-1+\sqrt{1+4\theta^{2}}}{2\theta}</tex>$
3. $<tex>P(X<1)=1</tex>$
  $<tex>P(0<X\leq1)=\frac{1+\theta}{2}</tex>$
  $<tex>P\left(X<0|-\frac{1}{2}<X<\frac{1}{2}\right)=\frac{1}{2}-\frac{\theta}{4}</tex>$
3/4
2/3
1. $<tex>f_{X}(x)=\frac{3x(20-x)}{3514}</tex>$ para $<tex>3\leq x\leq17</tex>$
2. $<tex>f_{X}(x)=\frac{x(20-x)}{162}</tex>$ para $<tex>0<x<3</tex>$ o $<tex>17<x<20</tex>$
1. $<tex>f_{X|X<3}(x)=\begin{cases}0.5 & \mbox{si }0\leq x<2\\2-\frac{2}{5}x & \mbox{si }2\leq x<3\end{cases}</tex>$
2. $<tex>f_{X|X>3}(x)=\frac{5}{2}-\frac{x}{2}</tex>$ si $<tex>3<x<5</tex>$
1. $<tex>f_{T}(t)=\beta\cdot\left(\frac{t}{\alpha}\right)^{\beta-1}\cdot e^{-\left(\frac{t}{\alpha}\right)^{\beta}}</tex>$
2. 1. Fallas casuales.
  2. Fallas por desgaste.
  3. Fallas tempranas.
  4. Fallas tempranas.
  5. Fallas por desgaste.
  6. Fallas casuales.
4. $<tex>P(T>4|T>3)=1-e^{-\left(\frac{t}{\alpha}\right)^{\beta}}</tex>$
1. Sin reposición
  $<tex>\begin{array}{|c||c|c|c|c}\cline{1-4} X\backslash Y & 0 & 1 & 2 & \mbox{Marginal de }X\tabularnewline\hline \hline 0 & 0 & 0.0285 & 0.0428 & \mathit{0.0714}\tabularnewline\hline 1 & 0.0428 & 0.2571 & 0.1285 & \mathit{0.4285}\tabularnewline\hline 2 & 0.1285 & 0.2571 & 0.0428 & \mathit{0.4285}\tabularnewline\hline 3 & 0.0428 & 0.0285 & 0 & \mathit{0.0714}\tabularnewline\hline \multicolumn{1}{c||}{\mbox{Marginal de }Y} & \mathit{0.2142} & \mathit{0.5714} & \mathit{0.2142} & \tabularnewline\end{array}</tex>$
  $<tex>P(X+Y\leq2)=\frac{1}{2}</tex>$
2. Con reposición
  $<tex>\begin{array}{|c||c|c|c|c|c|c}\cline{1-6} X\backslash Y & 0 & 1 & 2 & 3 & 4 & \mbox{Marginal de }X\tabularnewline\hline \hline 0 & 0.0197 & 0.0527 & 0.0527 & 0.0234 & 0.039 & 0.3164\tabularnewline\hline 1 & 0.0791 & 0.1582 & 0.1054 & 0.0234 & 0 & 0.4218\tabularnewline\hline 2 & 0.1186 & 0.1582 & 0.0527 & 0 & 0 & 0.2109\tabularnewline\hline 3 & 0.0791 & 0.0527 & 0 & 0 & 0 & 0.0468\tabularnewline\hline 4 & 0.0197 & 0 & 0 & 0 & 0 & 0.0039\tabularnewline\hline \multicolumn{1}{c|}{\mbox{Marginal de }Y} & 0.3164 & 0.4218 & 0.2142 & 0.0468 & 0.0039 & \tabularnewline\end{array}</tex>$
  $<tex>P(X+Y\leq2)=0.4812</tex>$
1. 1/2
2. $<tex>f_{X}(x)=\left(\frac{2}{\pi}\right)\sqrt{1-x^{2}}</tex>$ para $<tex>x\in[-1,1]</tex>$
  $<tex>f_{Y}(y)=\left(\frac{4}{\pi}\right)\sqrt{1-y^{2}}</tex>$ para $<tex>y\in[0,1]</tex>$
3. No son independientes.
1. 5/12
2. $<tex>f_{X}(x)=4x(1-x^{2})</tex>$ para $<tex>0\leq x\leq1</tex>$
  $<tex>f_{Y}(y)=4y^{3}</tex>$ para $<tex>0\leq y\le1</tex>$
3. No son independientes.
8/9
1. 0.1035
2. 4/9
3. 0.127
1. 1/50
2. 47/100
3. 17/18
4. $<tex>f_{T}(t)=\begin{cases}\frac{t-22}{136} & \mbox{ si }22<t<34\\\frac{12}{136} & \mbox{ si }34<t<36\\\frac{48-t}{136} & \mbox{ si }36<t<40\end{cases}</tex>$

Guía 3: Esperanza, varianza y covarianza

1. 1.1
2. $<tex>p_{3}=0.6</tex>$
  $<tex>x_{3}=0</tex>$
3. 1.625
4. 2.2
1. 2/3
2. 1
3. 7/9
4. 7.2
$<tex>\frac{2}{3}\theta</tex>$
1. 19/18
2. $<tex>E[X|X<1]=\frac{2}{3}</tex>$
  $<tex>E[X|X\leq1]=\frac{5}{6}</tex>$
3. $<tex>f_{X|X>1}(x)=1</tex>$ para $<tex>1<x<2</tex>$
1. $<tex>p_{1}=p_{3}=0.5</tex>$
  $<tex>p_{2}=0</tex>$
2. $<tex>p_{1}=p_{3}=0</tex>$
  $<tex>p_{2}=1</tex>$
1. $<tex>\frac{1800}{\pi}</tex>$
2. $<tex>\sqrt{15}\pi</tex>$
1. $<tex>\mu_{y}=2</tex>$
  $<tex>\sigma_{y}^{2}=36</tex>$
2. $<tex>\mu_{y}=27</tex>$
3. $<tex>\mu_{y}=16</tex>$
4. $<tex>c=2</tex>$
5. $<tex>a=\pm\frac{1}{3}</tex>$ , $<tex>b=\mp\frac{2}{3}</tex>$
6. $<tex>-11.4<a<11.4</tex>$
$<tex>-\frac{1}{2}</tex>$
1. $<tex>E[N]=\frac{19}{9}</tex>$
  $<tex>var(N)=0.3209</tex>$
  $<tex>cov(N,X_{1})=0</tex>$
3. $<tex>cov(X_{i},X_{i})=var(X_{i})=\frac{2}{3}</tex>$ , $<tex>cov(X_{i},X_{j})=-\frac{1}{3}</tex>$
$<tex>\rho=-0.123</tex>$
1. 1/9
2. 2/3
3. 2/3
1. $<tex>E[Z]=\frac{2}{3}</tex>$ ,
  $<tex>var(Z)=\frac{1}{18}</tex>$
  $<tex>E[W]=\frac{1}{3}</tex>$
  $<tex>var(W)=\frac{1}{18}</tex>$
2. 7/18
3. 1/36
1. $<tex>E[S_{n}]=2n</tex>$
  $<tex>var(S_{n})=9n</tex>$
3. $<tex>n>9000</tex>$

Guía 4: Transformación de variables aleatorias

1. $<tex>f_{Y}(y)=\frac{1}{9}\left(\frac{y-b}{a}+1\right)^{2}</tex>$ si $<tex>b-a<y<2a+b</tex>$
2. $<tex>f_{Y}(y)=\frac{\left(\sqrt[3]{-y}+1\right)^{2}}{27(-y)^{2/3}}</tex>$ si $<tex>-8<y<0</tex>$ y $<tex>0<y<1</tex>$
3. $<tex>f_{Y}(y)=\frac{\left(\frac{3}{2}-\sqrt{9+4y}\right)^{2}+\left(\frac{3}{2}+\sqrt{9+4y}\right)^{2}}{18\sqrt{9+4y}}\mbox{ si }0<y<1,25</tex>$
4. $<tex>f_{Y}(y)=\begin{cases}\frac{\left(1+\sqrt{y}\right)^{2}+\left(1-\sqrt{y}\right)^{2}}{18\sqrt{y}} & \mbox{si }0<y<1\\\frac{\left(1+\sqrt{y}\right)^{2}}{18\sqrt{y}} & \mbox{si }1<y<4\end{cases}</tex>$
5. $<tex>F_{Y}(y)=\begin{cases}\frac{1}{27}\left(y+1\right)^{2} & \mbox{si }-1<y<1\\1 & \mbox{si }y\geq1\end{cases}</tex>$
$<tex>f_{X}(x)=\frac{1}{\pi(1+x^{2})}</tex>$ para todo $<tex>x</tex>$
$<tex>f_{L}(l)=\frac{l}{30\pi}e^{-\frac{l^{2}}{60\pi}}</tex>$ si $<tex>l>0</tex>$
$<tex>y=\begin{cases}\frac{1-e^{-x}}{3} & \mbox{si }0<x<\ln(4)\\1-3e^{-x} & \mbox{si }x>\ln(4)\end{cases}</tex>$
$<tex>f_{T}(t)=\begin{cases}\frac{1}{10} & \mbox{si }0<t<5\\\frac{1}{20} & \mbox{si }5\leq t<15\end{cases}</tex>$
$<tex>F_{V_{2}}(v_{2})=\begin{cases}\frac{1}{2}v_{2}+\frac{1}{4} & \mbox{si }0\leq v_{2}<1\\1 & \mbox{si }v_{2}\geq1\end{cases}</tex>$
$<tex>P(N=k)=e^{-\lambda k}(e^{\lambda}-1)</tex>$ para $<tex>k\in N</tex>$
$<tex>f_{Z}(z)=\begin{cases}z+1 & \mbox{si }-1\leq z<0\\1-z & \mbox{si }0\leq z<1\end{cases}</tex>$
1. $<tex>f_{X}(x)=1</tex>$ si $<tex>0<x<1</tex>$
  $<tex>f_{Y}(y)=1</tex>$ si $<tex>0<y<1</tex>$
2. $<tex>Z=X+Y</tex>$
  $<tex>f_{Z}(z)=\begin{cases}z & \mbox{si }0<z<1\\2-z & \mbox{si }1<z<2\end{cases}</tex>$
$<tex>f_{Z}(z)=\begin{cases}\frac{45}{4}z^{2} & \mbox{si }0<z<0.5\\\frac{3}{4z^{2}}-\frac{3}{4}z^{2} & \mbox{si }0.5<z<1\end{cases}</tex>$
$<tex>f_{U}(u)=\begin{cases}1.6(1-u) & \mbox{si }0<u<0.5\\3.2(1-u) & \mbox{si }0.5<u<1\end{cases}</tex>$
1. $<tex>f_{U}(u)=\left(\lambda_{1}+\lambda_{2}\right)e^{-u\left(\lambda_{1}+\lambda_{2}\right)}</tex>$ para $<tex>u>0</tex>$
2. $<tex>P(j=1)=\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}</tex>$
  $<tex>P(j=2)=\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}</tex>$
0.2314
$<tex>f_{Z|Z<\frac{1}{4}}(z)=4</tex>$ para $<tex>0<z<\frac{1}{4}</tex>$
1. $<tex>f_{Z}(z)=\begin{cases}2 & \mbox{si }0<z<\frac{1}{4}\\\frac{2\left(1-\sqrt{z}\right)}{\sqrt{z}} & \mbox{si }\frac{1}{4}<z<1\end{cases}</tex>$
2. $<tex>f_{V}(v)=\begin{cases}-\frac{148}{9}v+\frac{14}{3} & \mbox{si }0<v<0.25\\-\frac{4}{9}v+\frac{2}{3} & \mbox{si }0.25<v<1.5\end{cases}</tex>$
$<tex>f(u,v)=u\lambda^{2}e^{-\lambda u}</tex>$ para $<tex>u>0</tex>$ y $<tex>0<v<1</tex>$
$<tex>f(u)=u\lambda^{2}e^{-\lambda u}</tex>$ para $<tex>u>0</tex>$
$<tex>f(v)=1</tex>$ para $<tex>0<v<1</tex>$
Son independientes.
$<tex>f_{X}(x)=\frac{1}{\pi(x^{2}+1)}</tex>$ si $<tex>x\in(-\infty,\infty)</tex>$
1. 0.0128
  0.08
2. $<tex>F_{Y}(y)=\begin{cases}0 & \mbox{ si }y<0\\0.0128 & \mbox{ si }y=0\\\frac{2}{625}y^{2}+\frac{8}{625}y+0.0128 & \mbox{ si }0<y<10.5\\-\frac{2}{625}y^{2}+\frac{92}{625}y-0.6928 & \mbox{ si }10.5<y<18\\1 & \mbox{ si }y\geq18\end{cases}</tex>$

Guía 5: Esperanza y varianza condicional, suma de variables aleatorias, mezcla, truncamiento

1. $<tex>f_{Y|X=0.75}(y)=\frac{4}{3}</tex>$ si $<tex>0<y<0.75</tex>$
  $<tex>f_{Y|X=0.4}(y)=\frac{5}{2}</tex>$ si $<tex>0<y<0.4</tex>$
2. No son independientes.
3. $<tex>\frac{1}{3}</tex>$ , 0,375
1. $<tex>f_{Y|X=x}(y)=\begin{cases}1 & \mbox{ si }0<y<1;0<x<0.5\\\frac{2}{3} & \mbox{ si }0<y<0.5;0.5<x<1\\\frac{4}{3} & \mbox{ si }0.5<y<1;0.5<x<1\end{cases}</tex>$
2. 0.5
1. $<tex>f(y_{1},y_{2})=\lambda^{2}y_{1}e^{-\lambda y_{1}}\frac{1}{y_{2}^{2}+2y_{2}+1}\mbox{ si }y_{1}>0;y{}_{2}>0</tex>$
2. $<tex>f(y_{1},y_{3})=\frac{\lambda^{2}}{2}e^{-\lambda y_{1}}\begin{array}{ccc} {} & {} & {}\end{array}y_{1}>0;y_{1}>y{}_{3};y_{1}>-y{}_{3}</tex>$
3. $<tex>\begin{array}{l}f_{y_{1}|y_{2}=1}(y_{1})=f(y_{1})=\lambda^{2}y_{1}e^{-\lambda y_{1}}\mbox{ si }y_{1}>0\\f_{y1|y3=0}(y_{1})=\lambda e^{-\lambda y_{1}}\mbox{ si }y_{1}>0\end{array}</tex>$
4. $<tex>A=2e^{-1}</tex>$
5. $<tex>B=e^{-1}</tex>$
$<tex>f(Cc3)=\begin{cases}0.2689x & \mbox{ si }0<x<2\\0.2689(4-x) & \mbox{ si }2<x<3\\0.1176(4-x) & \mbox{ si }3<x<4\end{cases}</tex>$
1. 0,439
2. 0,851
$<tex>\begin{array}{l}f(x)=\frac{1}{2\sqrt{2\pi}\sigma}\left(e^{-\frac{1}{2}\left(\frac{x+1}{\sigma}\right)^{2}}+e^{-\frac{1}{2}\left(\frac{x-1}{\sigma}\right)^{2}}\right)\\\psi(x)=\frac{e^{-x/\sigma^{2}}}{e^{-x/\sigma^{2}}+e^{x/\sigma^{2}}}\end{array}</tex>$
1. $<tex>F_{Y|X=x}(y)=1-e^{xy}</tex>$ si $<tex>x>0</tex>$ , $<tex>y>0</tex>$
2. $<tex>\varphi(x)=\frac{1}{x}</tex>$ si $<tex>x>0</tex>$
3. $<tex>E[Y|X]=\frac{1}{X}</tex>$
4. $<tex>F(z)=e^{-\frac{1}{z}}</tex>$ si $<tex>z>0</tex>$
5. 0,5811
1. $<tex>F{}_{Y|X=x}(y)=\frac{y}{x}</tex>$ si $<tex>0<y<x</tex>$ , $<tex>0<x<1</tex>$
2. $<tex>\varphi(x)=\frac{x}{2}</tex>$ si $<tex>0<x<1</tex>$
  $<tex>\varphi(x)=\frac{x^{2}}{12}</tex>$ si $<tex>0<x<1</tex>$
3. $<tex>E[Y|X]=\frac{X}{2}</tex>$
  $<tex>E[Y|X]=\frac{X^{2}}{12}</tex>$
4. $<tex>\frac{1}{4}</tex>$ , $<tex>\frac{9}{16}</tex>$
$<tex>Z=E[Y|X]</tex>$
$<tex>W=var[Y|X]</tex>$ Z
$<tex>P\left(Z=\frac{3}{8}\right)=\frac{9}{10}</tex>$
$<tex>P\left(Z=\frac{7}{8}\right)=\frac{1}{10}</tex>$
$<tex>P\left(W=\frac{3}{64}\right)=\frac{9}{10}</tex>$
$<tex>P\left(W=\frac{1}{192}\right)=\frac{1}{10}</tex>$
$<tex>P\left(W<\frac{1}{32}\right)=\frac{1}{10}</tex>$
1. $<tex>P_{Y|X=x}(y)=\frac{{4 \choose y}{2 \choose 3-x-y}}{{6 \choose 3-x}}\mbox{ si }x=0,1,2,3;\,0\le y\le3-x</tex>$ .
2. $<tex>E[Y|X=x]=(3-x)\frac{4}{6}</tex>$
  $<tex>var[Y|X=x]=\left(9-x^{2}\right)\frac{2}{45}</tex>$
3. $<tex>E[Y|X]=(3-X)\frac{4}{6}</tex>$
  $<tex>V[Y|X]=\left(9-X^{2}\right)\frac{2}{45}</tex>$
$<tex>\varphi(x)=E[Y|X]=\frac{e^{x/\sigma^{2}}-e^{-x/\sigma^{2}}}{e^{-x/\sigma^{2}}+e^{x/\sigma^{2}}}</tex>$
1. $<tex>E[Y|X]=0</tex>$
  Son independientes.
2. $<tex>E[Y|X]=0</tex>$
  No son independientes.
14
0
$<tex>Y=100-X</tex>$
$<tex>cov(X,Y)=-25</tex>$
2
5 mm a la derecha.
1
$<tex>E[Y]=7.5</tex>$
$<tex>var(Y)=5.75</tex>$
$<tex>E[T]=1.266</tex>$
$<tex>var(T)=0.8434</tex>$
1. $<tex>E[Y|X]=\frac{X^{2}}{2}</tex>$ si $<tex>0<x<2</tex>$
2. $<tex>Y=4X-\frac{14}{5}</tex>$
1. $<tex>E[Y|X]=X^{2}</tex>$
2. $<tex>Y=20X-50</tex>$

Guía 6: Procesos Bernoulli, geométrico e hipergeométrico

1. 0,6241
2. 0,6992
1. 0,6651
2. 0,4018
3. 0,2009
1. 1 y 2
2. 2
3. $<tex>\frac{n}{5}</tex>$
0,8233
0.05174
$<tex>n\geq22</tex>$
1. 0.0531
2. 0.2373
$<tex>E[N]=11.4</tex>$
$<tex>var(N)=25.17</tex>$
40
0,03348
0,1273
1. 0,00277
2. 0,1527
3. 0,5042
$<tex>P(r=1)=\begin{cases}0 & \mbox{si }k=0\\\frac{1}{3} & \mbox{si }k=1\\\frac{10}{21} & \mbox{si }k=2\\\frac{15}{28} & \mbox{si }k=3\\\frac{5}{9} & \mbox{si }k=4\\\frac{5}{9} & \mbox{si }k=5\\\frac{6}{81} & \mbox{si }k=6\end{cases}</tex>$
Es más probable cuando $<tex>k=4</tex>$ o $<tex>k=5</tex>$ .
1. $<tex>P(Y=y)=\begin{cases}\frac{1}{6} & \mbox{si }y=0\\\frac{2}{3} & \mbox{si y=1}\\\frac{1}{6} & \mbox{si }y=2\end{cases}</tex>$
2. $<tex>P(X=x)=\begin{cases}\frac{217}{270} & \mbox{si }x=0\\\frac{26}{135} & \mbox{si }x=1\\\frac{1}{270} & \mbox{si }x=2\end{cases}</tex>$
3. 19,63%
1. 0,1297
2. 0,3589
3. 0,9896
4. 0,2969
5. 0,9999
6. 0,00065
0,303
$<tex>E[M]=1</tex>$
$<tex>cov(N,M)=4</tex>$
$<tex>E[X]=\frac{1}{9}</tex>$
$<tex>var(X)=0.1133</tex>$
7,28

Guía 7: Procesos Poisson

1. 0,1429
2. 1 y 2
3. 1,7817
4. 0,218
5. Aumento en 1 las instalaciones
1. $<tex>\sim Binom\left(p=\frac{2}{3},n=60\right)</tex>$
2. 0,108
3. $<tex>\sim Poisson(m=40)</tex>$
1. $<tex>e^{-6}</tex>$
2. $<tex>6e^{-6}</tex>$
3. 1/3
0,9901
- Exponencial
  1. $<tex>e{}^{-0,2}</tex>$
  2. $<tex>e{}^{-0,2}</tex>$
  3. No tiene memoria
- Gamma
  1. 0,9999
  2. 0,6993
1. $<tex>f_{X,S}(x,s)=s\lambda^{3}e^{-\lambda(s+x)}</tex>$
2. $<tex>3\left(1-e^{-\lambda}\right)</tex>$
$<tex>E[T]=7,692</tex>$ minutos
1. 3/4
2. $<tex>P(X=x)=\frac{3}{4^{x+1}}</tex>$ si $<tex>x\text{\ensuremath{\in}N}</tex>$
1. 0,0107
2. 0,0286
3. $<tex>3\times10^{-7}</tex>$
4. 1/2
5. 5/4
1. 0,1025
2. 0,2873
$<tex>e{}^{-60}</tex>$
1. 1/9
2. 5/9
0,07985
$<tex>E[T]=2</tex>$ , $<tex>var(T)=4</tex>$
1. 16,4
2. 0,0656
1/4
240
1. 0
2. 0,0148
3. 552 Kg
4. 331.200 Kg

Guía 8: Distribución normal, TCL

Trivial.
Trivial.
0,676
0,0917
243,53
1. 491.95
2. IC: 394,26; 522,056
3. 0.6157
No son independientes. $<tex>cov(W,Z)\neq0</tex>$
1. 0.9375
2. 0.9327
1. 0.245
2. 0.591
0,5039
0,4348
1. 0,0764
2. 0,318
3. 18.124,89
1. 0,2209
2. 0,612
3. 0,2255; 0,6128
0,9545
0.023
0
103
1684
64

Guía 9: Estadística Bayesiana

$<tex>\pi(t|\mathbf{x})=6.76202\times0.0468644^{t}\times t^{5}\mbox{ si }t=\{1,\ldots,6\}</tex>$
Media: 1,9356
Moda: 2
1. $<tex>\pi(t|\mathbf{x})=\begin{cases}\frac{125}{62}\cdot\frac{1}{2\sqrt{2\pi}}\cdot e^{-0.5\left(\frac{12.1-\mu}{2}\right)^{2}} & \mbox{ si }\mu=10\\\frac{375}{62}\cdot\frac{1}{2\sqrt{2\pi}}\cdot e^{-0.5\left(\frac{12.1-\mu}{2}\right)^{2}} & \mbox{ si }\mu=14\end{cases}</tex>$
2. $<tex>P(Y>13)=0.2525</tex>$
1. $<tex>\pi(k|\mathbf{x})=\frac{6k-k^{2}}{35}</tex>$ si $<tex>k=\{0,\ldots,6\}</tex>$
2. 3
$<tex>\pi(\theta|5)=\frac{1}{2}(5-\theta)\mbox{ si }\theta\in[3,5]</tex>$
1. $<tex>ab^{3}=3</tex>$
2. $<tex>\pi(t|\bar{\mathbf{x}})=26244t^{3}\mbox{ si }0<t<\frac{1}{9}</tex>$
  Media: $<tex>\frac{4}{45}</tex>$
  Moda: $<tex>\frac{1}{9}</tex>$
$<tex>\frac{13}{6}</tex>$
$<tex>\frac{7}{22}</tex>$
1. Si hay 0 personas irritadas de las 10 encuestadas, la primera opinión (“poca gente irritada”) hará una buena estimación de $<tex>p</tex>$ : $<tex>\hat{p}=\frac{1}{21}</tex>$ .
  Si hay 0 personas irritadas de las 10 encuestadas, la segunda opinión (“mucha gente irritada”) hará una mala estimación de $<tex>p</tex>$ : $<tex>\hat{p}=\frac{10}{21}</tex>$ .
  Si hay 10 personas irritadas de las 10 encuestadas, la primera opinión (“poca gente irritada”) hará una mala estimación de $<tex>p</tex>$ : $<tex>\hat{p}=\frac{10}{21}</tex>$ .
  Si hay 10 personas irritadas de las 10 encuestadas, la segunda opinión (“mucha gente irritada”) hará una buena estimación de $<tex>p</tex>$ : $<tex>\hat{p}=\frac{20}{21}</tex>$ .
2. $<tex>n\geq8989</tex>$
1. Media: $<tex>\frac{\nu}{\lambda}</tex>$
  Moda: $<tex>\frac{\nu-1}{\lambda}</tex>$
2. $<tex>\pi(t|\bar{\mathbf{x}})\sim\Gamma\left(\lambda_{\Gamma}=n+\lambda,k_{\Gamma}=\left(\sum_{i=1}^{n}x_{i}\right)+\nu\right)</tex>$
  Media: $<tex>\frac{1}{1+\frac{\lambda}{n}}\bar{x}+\frac{1}{1+\frac{\lambda}{n}}\frac{\nu}{n}</tex>$
3. Moda: $<tex>\frac{n+\nu-1}{\lambda+\sum_{i=1}^{n}x_{i}}</tex>$
1. $<tex>P\left(Y\geq2\right)\approx0.555</tex>$
1. $<tex>\pi(t|\bar{\mathbf{x}})=295245t^{-6}\mbox{ si }t\geq9</tex>$
  Media: $<tex>11.25</tex>$
  Moda: 9
2. $<tex>\pi(t|\bar{\mathbf{x}})=38880t^{-6}\mbox{ si }t\geq6</tex>$
  Media: $<tex>7.5</tex>$
  Moda: 6
1. $<tex>\pi(t|\mathbf{x})=\frac{e^{-\frac{2}{25}\left(t-173\right)^{2}}}{5\sqrt{\frac{\pi}{2}}}\text{ si }\mu>0</tex>$
2. Media: 173

Guía 10: Estimación Puntual

Dado que $<tex>ECM\left[\hat{\theta}_{1}\right]<ECM\left[\hat{\theta}_{2}\right]<ECM\left[\hat{\theta}_{3}\right]</tex>$ , el mejor estimador es $<tex>\hat{\theta}_{1}</tex>$ .
1. $<tex>\hat{\theta}_{1}</tex>$ es insesgado.
  $<tex>\hat{\theta}_{2}</tex>$ es sesgado.
$<tex>ECM\left[\hat{p}_{2}\right]<ECM\left[\hat{p}_{1}\right]</tex>$ , el estimador $<tex>\hat{p}_{2}</tex>$ es preferible.
1. $<tex>f_{\hat{\Theta}}(t)=\frac{n}{\Theta^{n}}\cdot t^{n-1}</tex>$
2. $<tex>B\left[\hat{\Theta}\right]=\Theta\left(\frac{n}{n+1}-1\right)</tex>$
1. $<tex>P\left(\bar{X}<0.25\right)=0.89435</tex>$
2. $<tex>P\left(S^{2}>0.577\right)=0.95</tex>$
3. $<tex>P\left(0.576\leq\hat{p}\leq0.764\right)=0.68244</tex>$
1. $<tex>\hat{\mu}_{mv}=\bar{X}</tex>$
2. $<tex>\hat{p}_{mv}=\frac{\sum_{i=1}^{n}x_{i}}{n}</tex>$
3. $<tex>\hat{\theta}_{mv}=\max\left(x_{1},\ldots,x_{n}\right)</tex>$
$<tex>\hat{p}_ {mv}=\frac{2}{5}</tex>$
$<tex>\hat{M}_{mv}=50</tex>$
1. -
1. $<tex>\hat{\mu}_{mv}=\frac{1}{n}\left[\sum_{i=1}^{n}x_{i}\right]</tex>$
  $<tex>\sigma^{2}=\frac{1}{n}\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}</tex>$
$<tex>P(Y=1)=\frac{6}{11}</tex>$
1. $<tex>\hat{\sigma^{2}} _{mv}=1.417</tex>$
2. No existe $<tex>\hat{\sigma^{2}} _{mv}</tex>$
- El emv de la media es $<tex> 581.8</tex>$
- El emv de la varianza es $<tex> 338491.24</tex>$
- El emv de la mediana es $<tex> 403.2730</tex>$
$<tex> 0.64</tex>$
$<tex>\hat{\lambda} _{mv}=\frac{1}{15}</tex>$
$<tex>L \left(\lambda|\bar{x}\right)\propto\theta^{3}e^{-2\theta}</tex>$
$<tex>\hat{\theta} _{mv}=\frac{3}{2}</tex>$
$<tex> \hat{p}_{mv}=2\frac{r}{n}-\frac{1}{2}</tex>$

Guía 11: Estimación por Intervalos

$<tex> P\left(\frac{\mathbf{x}}{\sqrt{1-\alpha}}<\theta\right)=1-\alpha</tex>$
1. $<tex> Y=\frac{X_{(n)}}{\theta}</tex>$ es un pivote para $<tex>\theta</tex>$ .
2. $<tex> P\left(X_{(n)}<\theta<\frac{X_{(n)}}{\sqrt[n]{\alpha}}\right)=1-\alpha</tex>$
3. $<tex> k=\sqrt{10}</tex>$
1. $<tex> P\left(7.3845<\mu<8.6912\right)=0.95</tex>$
2. $<tex> n\geq38416</tex>$
1. $<tex> P\left(299836.6<\mu<299868.2\right)=0.95</tex>$
  El intervalo de confianza dado sólo dice lo siguiente: “si realizas otra medición de la velocidad de la luz, hay un 95\% de probabilidad de que $<tex>\mu</tex>$ esté dentro de ese intervalo”.
2. $<tex>P\left(69.4855<\sigma<92.0106\right)=0.95</tex>$
1. $<tex>X_{n+1}-\bar{X}_{n}\sim\mathcal{N}\left(0,\sigma^{2}\left[1-\frac{1}{n}\right]\right)</tex>$
2. $<tex>P\left(2.542<X_{6}<5.4758\right)=0.9</tex>$
1. $<tex>P\left[0.07142<p<0.12878\right]\approx0.9</tex>$
2. $<tex>P\left[0.0778<p\right]\approx0.9</tex>$
9604
1. $<tex>P\left[2.71275<\lambda<7.8525\right]=0.9</tex>$
2. $<tex>P\left[2.78<5.4004\right]=0.9</tex>$
$<tex>P\left[0.00021667<\lambda\right]=0.95</tex>$
1. $<tex>P\left[0.15555<\lambda<1.02861\right]=0.95</tex>$
2. $<tex>P\left[0.15555<\lambda<1.02861\right]=0.95</tex>$
1. $<tex>P\left[0.4321<\Delta<56.31\right]=0.99</tex>$
2. $<tex>P\left[6.84<\Delta<49.9\right]=0.99</tex>$
3. $<tex>P\left[26.613<\Delta<30.128\right]=0.99</tex>$
Proveedor 1.

Guía 12: Test de Hipótesis y Test de Bondad de Ajuste

$<tex>P\left(\mbox{rechazar }H_{0}|\theta\right)=\frac{1}{35}\text{ si }\theta=3</tex>$
$<tex>P\left(\mbox{aceptar }H_{0}|\theta\right)=\frac{31}{35}\mbox{ si }\theta=4</tex>$
1. $<tex>P\left(\text{error tipo I}\right)=\frac{1}{3}</tex>$
  $<tex>P\left(\text{error tipo II}\right)=\frac{1}{2}</tex>$
2. $<tex>P\left(\text{error tipo I}\right)=\frac{1}{3}</tex>$
  $<tex>P\left(\text{error tipo II}\right)=\frac{1}{2}</tex>$
1. $<tex>pot(\theta)=\begin{cases}\left(\frac{2.9}{\theta}\right)^{n} & \text{ si }\theta\in[3,4]\\1 & \text{ si }\theta\in(0;2.9)\\\left(\frac{2.9}{\theta}\right)^{n} & \mbox{ si }\theta\in(2.9,3)\\\left(\frac{2.9}{\theta}\right)^{n}+1-\left(\frac{4}{\theta}\right)^{n} & \text{ si }\theta\in(4,\infty)\end{cases}</tex>$
2. $<tex>NS=\left(\frac{2.9}{3}\right)^{n}</tex>$
3. $<tex>n\geq68</tex>$
$<tex>\delta(\mathbf{x})=\begin{cases}0 & \text{ si }2.3025\geq\mathbf{x}\\1 & \text{ si }2.3025<\mathbf{x}\end{cases}</tex>$
$<tex>P\left[\delta(\mathbf{x})=0|\mu=1.1\right]=0.876</tex>$
$<tex>\delta(\mathbf{x})=\begin{cases}1 & \mbox{ si }5<2.655583\\0 & \text{ si }5\geq2.655558\end{cases}</tex>$
No rechazamos $<tex>H_{0}</tex>$ .
1. F
2. V
3. F
4. F
5. V
6. F
7. F
8. V
1. $<tex>\delta(\mathbf{x})=\begin{cases} 1 & \text{ si }\mu_{0}<\bar{X}-1.65\frac{\sigma}{\sqrt{n}}\\0 & \text{ si }\mu_{0}\geq\bar{X}-1.65\frac{\sigma}{\sqrt{n}} \end{cases}</tex>$
2. $<tex>\beta(\mu)=1-\phi_{\bar{X}_{(\mu,\sigma)}}\left[\mu_{0}+1.65\frac{\sigma}{\sqrt{n}}\right]</tex>$
$<tex>\delta(\mathbf{X})=\begin{cases} 1 & \text{ si }4\notin\left[\max\left\{ X_{i}\right\} ,1.3493\max\left\{ X_{i}\right\} \right]\\ 0 & \text{ si }4\in\left[\max\left\{ X_{i}\right\} ,1.3493\max\left\{ X_{i}\right\} \right]\end{cases}</tex>$
No se puede rechazar $<tex>H_{0}</tex>$ .
$<tex>D^{2}=4.74</tex>$
No se puede rechazar $<tex>H_{0}</tex>$ .
$<tex>D^{2}=3.26</tex>$
No se puede rechazar $<tex>H_{0}</tex>$ .
$<tex>D^{2}=0.9</tex>$
1. No se puede rechazar $<tex>H_{0}</tex>$ .
  $<tex>D^{2}=10.408</tex>$
2. No se puede rechazar $<tex>H_{0}</tex>$ .
  $<tex>D^{2}=6.4</tex>$

materias/61/09/resultados_guia.txt · Última modificación: 2012/09/13 20:25 por sebacuervo

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