Q. 11

\[ u=1000 \] \[ n=100 \] \[ E[x]=u/n = 10 \]

Q. 14

For \[ X_2 \geqslant X_1 \]

\[ fz(z)=\int_{-\infty}^{\infty}fx_1(z+x_2)fx_2(x_2)dx_2 \] \[ fz(z)=\int_{-\infty}^{0}\lambda e^{-\lambda (z+x_2)}\lambda e^{-\lambda x_2}dx_2\] \[fz(z)=\int_{-\infty}^{0}\lambda e^{-\lambda (z)}\lambda e^{-2\lambda x_2}dx_2\] \[fz(z)=\frac{-\lambda}2(e^{-\lambda z})\]

For \[ X_1 \geqslant X_2 \]

\[ fz(z)=\int_{-\infty}^{\infty}fx1(z+x_2)fx_2(x_2)dx_2\] \[fz(z)=\int_{0}^{\infty} \lambda e^{-\lambda (z+x_2)}\lambda e^{-\lambda x_2}dx_2\] \[fz(z)=\int_{0}^{\infty}\lambda e^{-\lambda (z)}\lambda e^{-2\lambda x_2}dx_2\] \[fz(z)=\frac{\lambda}2(e^{-\lambda z})\]

We know case 1 has

\[ z < 0 \]

and case 2 has

\[ z \geqslant 0\]

Therefore, we can rewrite this as

\[ f(z)= \frac{\lambda}{2} e^{-\lambda \lvert z \rvert} \]

Pg. 320 Q. 1

  1. \[ P(|X-10| \geqslant 2)=P(|X-10|\geqslant.2\sqrt3\frac{10}{\sqrt3})\leqslant\frac{1}{(.2\sqrt3)^2}\leqslant8.33 \]
  2. \[ P(|X-10|\geqslant5)=P(|X-10|\geqslant.5\sqrt3\frac{10}{\sqrt3})\leqslant\frac{1}{(.5\sqrt3)^2}\leqslant1.33 \]
  3. \[ P(|X-10|\geqslant9)=P(|X-10|\geqslant.9\sqrt3\frac{10}{\sqrt3})\leqslant\frac{1}{(.9\sqrt3)^2}\leqslant0.412 \]
  4. \[ P(|X-10|\geqslant20)=P(|X-10|\geqslant2\sqrt3\frac{10}{\sqrt3})\leqslant\frac{1}{(2\sqrt3)^2}\leqslant0.083 \]