Solutions - 5180-05

(a)

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

$$\begin{array}{c}X=200Z+1200\\ \\ P[X>1400]\gg \gg P[200Z+1200>1400]\\ \\ \gg \gg P[Z>\frac{1400-1200}{200}]\gg \gg P[Z>1]\end{array}$$

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(2) Evaluate using the $\mathrm{\Phi }$ table:

$$\begin{array}{c}1-\mathrm{\Phi }(1)\gg \gg 1-0.8413\gg \gg 0.1587\end{array}$$

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(b)

(1) Substitute $X=200Z+1200$ into the probability:

$$\begin{array}{c}P[1000<X<1400]\gg \gg P[1000<200Z+1200<1400]\\ \\ \gg \gg P[-1<Z<1]\end{array}$$

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

$$\begin{array}{c}\mathrm{\Phi }(1)-\mathrm{\Phi }(-1)\gg \gg 2\mathrm{\Phi }(1)-1\\ \\ \gg \gg 2(0.8413)-1\gg \gg 0.6826\end{array}$$

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(c)

(1) Set up conditional probability expression:

$$P[X\ge 1300|X\ge 1200]=\frac{P[X\ge 1300\cap X\ge 1200]}{P[X\ge 1200]}=\frac{P[X\ge 1300]}{P[X\ge 1200]}$$

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(2) Substitute $X=200Z+1200$ in numerator and denominator:

$$\begin{array}{c}\frac{P[200Z+1200\ge 1300]}{P[200Z+1200\ge 1200]}\gg \gg \frac{P[Z\ge 0.5]}{P[Z\ge 0]}\\ \\ \gg \gg \frac{1-\mathrm{\Phi }\left(\frac{1}{2}\right)}{1-\mathrm{\Phi }(0)}\end{array}$$

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

$$\begin{array}{c}\frac{1-0.6915}{1-0.5}\gg \gg 0.617\end{array}$$