Solutions - 5180-03

(a)

(1) Write $X=\sigma Z+\mu$ and substitute into the probability:

$$\begin{array}{c}X=2Z+3\\ \\ P[2<X<6]\gg \gg P[2<2Z+3<6]\gg \gg P[-\frac{1}{2}<Z<\frac{3}{2}]\end{array}$$

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(2) Express in terms of $\mathrm{\Phi }$ values:

$$\begin{array}{c}\mathrm{\Phi }\left(\frac{3}{2}\right)-\mathrm{\Phi }(-\frac{1}{2})\gg \gg \mathrm{\Phi }\left(\frac{3}{2}\right)+\mathrm{\Phi }\left(\frac{1}{2}\right)-1\end{array}$$

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(3) Evaluate using a $\mathrm{\Phi }$ table:

$$\begin{array}{c}0.9332+0.6915-1\gg \gg 0.6247\end{array}$$

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(b)

(1) Use the $\mathrm{\Phi }$ table to find $z$ given $\mathrm{\Phi }(z)=0.67$:

$$\mathrm{\Phi }(z)=0.67\gg \gg z\approx 0.4399$$

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(2) Solve for $c$ using $X=2Z+3$:

$$c=2(0.4399)+3\gg \gg 3.8798$$

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(c)

(1) Recall formula for $\text{Var}[X]$ and solve for $E\left[X^2\right]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2\\ \\ \gg \gg E\left[X^2\right]=\text{Var}[X]+{\left(E[X]\right)}^2\end{array}$$

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(2) Plug in $\mu =3$ and ${\sigma }^2=4$:

$$E\left[X^2\right]=4+3^2\gg \gg 13$$