Answer to Question #95527 in Statistics and Probability for Nandhini

Question

Suppose two rats A and B have been trained to navigate a large maze.
X=Time taken by rat A to navigate the maze X~N(80,100)
Y=Time taken by rat B to navigate the maze, Y~N(78,169)
On any given day, what is the probability that rat A runs the maze faster than rat B?

Answer 1

We will suppose that $X$ and $Y$ are independent variables. So we need to find$P(X<Y)=\iint\limits_{x<y}\rho_X(x)\rho_Y(y)dxdy=$ $\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^y\rho_X(x)\rho_Y(y)dxdy=$ $=\frac{1}{100\sqrt{2\pi}}\frac{1}{169\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^ye^{-\frac{(x-80)^2}{2*100^2}}e^{-\frac{(y-78)^2}{2*169^2}}dxdy$.Let $u=\frac{x-80}{100}$, then $\frac{1}{100\sqrt{2\pi}}\frac{1}{169\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^ye^{-\frac{(x-80)^2}{2*100^2}}e^{-\frac{(y-78)^2}{2*169^2}}dxdy=$ $=\frac{1}{100\sqrt{2\pi}}\frac{1}{169\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{\frac{y-80}{100}}e^{-\frac{u^2}{2}}e^{-\frac{(y-78)^2}{2*169^2}}d(100u+80)dy=$ $=\frac{1}{\sqrt{2\pi}}\frac{1}{169\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{\frac{y-80}{100}}e^{-\frac{u^2}{2}}e^{-\frac{(y-78)^2}{2*169^2}}dudy$.Let $v=\frac{y-78}{169}$ , then $\frac{y-80}{100}=\frac{169v-2}{100}$ and $\frac{1}{\sqrt{2\pi}}\frac{1}{169\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{\frac{y-80}{100}}e^{-\frac{u^2}{2}}e^{-\frac{(y-78)^2}{2*169^2}}dudy=$ $=\frac{1}{\sqrt{2\pi}}\frac{1}{169\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{\frac{169v-2}{100}}e^{-\frac{u^2}{2}}e^{-\frac{v^2}{2}}dud(169v+78)=$ $=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{\frac{169v-2}{100}}e^{-\frac{u^2}{2}}e^{-\frac{v^2}{2}}dudv$

We have the following region of integrate in image. Make orthogonal change of variables, where the line $u=\frac{169v}{100}$ will be the axis $u'$ and orthogonal line $u=-\frac{100}{169}v$ will be axis $v'$ .

So $\begin{pmatrix} v \\ u \end{pmatrix}= \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} \begin{pmatrix} v' \\ u' \end{pmatrix}$ , where $\tan\alpha=-\frac{100}{169}$ , that is $\cos\alpha=\frac{169}{\sqrt{38561}}$ and $\sin\alpha=-\frac{100}{\sqrt{38561}}$ .

We obtain that line $u=\frac{169v-2}{100}$ will be transformed into $-\frac{100}{\sqrt{38561}}v'+\frac{169}{\sqrt{38561}}u'=\frac{169}{100}\bigl(\frac{169}{\sqrt{38561}}v'+\frac{100}{\sqrt{38561}}u'\bigr)-\frac{1}{50}$ or $v'=\frac{2}{\sqrt{38561}}$

Line $u=-\frac{100}{169}v$ will be transformed into $u'=0$.

We obtain the new region of integration $\bigl\{(u',v'): v'\ge\frac{2}{\sqrt{38561}}\bigr\}$

Jacobian of our transformation is $J=\bigl|\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}\bigr|=1$

Since our transformation is orthogonal, we have $u^2+v^2=(u')^2+(v')^2$ , so $\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{\frac{169v-2}{100}}e^{-\frac{u^2}{2}}e^{-\frac{v^2}{2}}dudv=$

$=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}\int\limits_{\frac{2}{\sqrt{38561}}}^{+\infty}Je^{-\frac{(u')^2}{2}}e^{-\frac{(v')^2}{2}}du'dv'=$ $=\frac{1}{\sqrt{2\pi}}\int\limits_{\frac{2}{\sqrt{38561}}}^{+\infty}e^{-\frac{(u')^2}{2}}du'\cdot\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}e^{-\frac{(v')^2}{2}}dv'=$ $=\Bigl(1-\Phi\bigl(\frac{2}{\sqrt{38561}}\bigr)\Bigr)\Phi(+\infty)=1-\Phi\bigl(\frac{2}{\sqrt{38561}}\bigr)$ , where $\Phi(x)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^x e^{-\frac{y^2}{2}}dy$

https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf - Normal Disribution Table

Since $\frac{2}{\sqrt{38561}}\approx 0.01$ , from table we see that $\Phi\bigl(\frac{2}{\sqrt{38561}}\bigr)\approx\Phi(0.01)\approx 0.50399$ , so $1-\Phi\bigl(\frac{2}{\sqrt{38561}}\bigr)\approx 0.49601$

Answer: $0.49601$