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If $$X$$ and $Y$ are independent, normal random variables
with $$E(X)=0, \quad V(X)=4, \quad E(Y)=10, \quad$$ and $$\quad V(Y)=9$$
Determine the following:

$$\begin{array}{ll}{\text { (a) } E(2 X+3 Y)} & {\text { (b) } V(2 X+3 Y)} \\ {\text { (c) } P(2 X+3 Y<30)} & {\text { (d) } P(2 X+3 Y<40)}\end{array}$$

$$x, x \rightarrow$$ independent normal

$$\begin{array}{ll}{E(X)=0} & {V(X)=4} \\ {E(Y)=10} & {v(y)=9}\end{array}$$

(a) $$E(2 X+3 Y)$$

$$=2 \times {0}+3 \times {10}=30$$

(b) $$V(2 X+3 Y)$$

$$=2^{2} \times 4+3^{2} \times 9=97$$

(c) $$P(2 X+3 Y<30)$$

$$\mu=30$$

$$\sigma=\sqrt{97}$$

$$P\left(Z<\frac{30-30}{\sqrt{97}}\right)=P(Z<0)$$

$$=0.5$$

(d) $$P(2 X+3 Y<40)$$

$$=P\left(Z<\frac{40-30}{\sqrt{97}}\right)$$

$$=P(Z<1.02)$$

$$=0.8461$$

A plastic casing for a magnetic disk is composed of
two halves. The thickness of each half is normally distributed
with a mean of 2 millimeters and a standard deviation of
0.1 millimeter and the halves are independent.

(a) Determine the mean and standard deviation of the total
thickness of the two halves.
(b) What is the probability that the total thickness exceeds
4.3 millimeters?

$$\mu=2 \mathrm {m m}$$

$$\sigma=0.1 \mathrm{mm}$$

$$x, y$$

$$T \rightarrow$$ total thickness

$$E(T)=\mu_{x}+\mu_{y}=2+2=4 \mathrm{mm}$$

$$\sigma_{T}^{2}=\sigma_{x}^{2}+\sigma_{y}^{2}=0.1^{2} + 0.1^{2}=0.02 \mathrm{mm}^{2}$$

(a) $$\sigma_{T}=\sqrt{0.02} = 0.1414 \mathrm{mm}$$

(b) $$P\left(T>4.3\right)$$

$$=P\left(Z>\frac{4.3-4}{0.1414}\right)$$

$$=P(Z>2.12)$$

$$=1-P(Z<2.12)$$

$$=1-0.983$$

$$=0.0170$$

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