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George Weigt 2006-11-14 09:30:56 -07:00
parent db28fe9fce
commit c6ac2704cc
1 changed files with 8 additions and 8 deletions

16
150.tex
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@ -16,15 +16,15 @@ tube will last more than 1950 hours?
$$y=0.098$$
\medskip\noindent
Solution: See the book, page 306, except flip $X\leftrightarrow Y$.
$$\rho={Cov(X,Y)\over\sigma_X\sigma_Y}={4.35\over50\times0.1}=0.87$$
$$\eqalign{
E(X|y)
&=\mu_X+\rho(\sigma_X/\sigma_Y)(y-\mu_Y)\cr
&=2000+(4.35)(2500/0.01)^{1/2}(0.098-0.1)\cr
&=1995.65
&=2000+(0.87)(2500/0.01)^{1/2}(0.098-0.1)\cr
&=1999.13
}$$
$$\rho={Cov(X,Y)\over\sigma_X\sigma_Y}={4.35\over500\times0.1}=0.0870$$
$$\sigma_{X|y}^2=\sigma_X^2(1-\rho^2)=2500(1-0.0870^2)=2481.08$$
$$P(X>1950|y)=\hbox{\tt1-NORMDIST(1950,1995.65,SQRT(2481.08),TRUE)}=0.8203$$
$$\sigma_{X|y}^2=\sigma_X^2(1-\rho^2)=2500(1-0.87^2)=607.75$$
$$P(X>1950|y)=\hbox{\tt1-NORMDIST(1950,1999.13,SQRT(607.75),TRUE)}=0.9769$$
\beginsection 7. (b)
@ -35,11 +35,11 @@ Solution:
$$\eqalign{
E(X|y)
&=\mu_X+\rho(\sigma_X/\sigma_Y)(y-\mu_Y)\cr
&=2000+(4.35)(2500/0.01)^{1/2}(0.104-0.1)\cr
&=2008.7
&=2000+(0.87)(2500/0.01)^{1/2}(0.104-0.1)\cr
&=2001.74
}$$
Note that the variance $\sigma_{X|y}^2$ is the same as above.
$$P(X>2000|y)=\hbox{\tt1-NORMDIST(2000,2008.7,SQRT(2481.08),TRUE)}=0.5693$$
$$P(X>2000|y)=\hbox{\tt1-NORMDIST(2000,2001.74,SQRT(607.75),TRUE)}=0.5281$$
\end