Ch 7 - 23, 25

7.23

Question

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$X$ is syntax errors, $Y$ is logic errors:

Proof
a.

\begin{aligned} P (X > Y) & = \sum_{x > y} p (x, y) \\ = .03 + .02 + .01 \\ + .03 + .01 \\ + .01 \\ = .11 \end{aligned}

$x$	0	1	2	3
$p_{X} (x)$	.78	.12	.07	.03

$y$	0	1	2
$p_{Y} (y)$	.77	.14	.09
c.
Notice that $p (0, 0) = .71 \neq .6006 = p_{X} (0) \cdot p_{Y} (0)$ so then they are independent.

μ_{X} = \sum_{x} x p_{X} (x) = 0 \cdot .78 + 1 \cdot .12 + \dots = .35 syntax err.s

μ_{Y} = \sum_{y} y p_{Y} (y) = 0 \cdot .77 + 1 \cdot .14 + 2 \cdot .09 = .32 logic err.s

E (P) = E (100 - 4 X - 9 Y) = 100 - 4 μ_{X} - 9 μ_{Y} = 95.72 pts

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Question

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Proof

$x_{1}$	0	1	2	3	4
$p_{X_{1}} (x_{1})$	.19	.30	.25	.14	.12

$x_{2}$	0	1	2	3
$p_{X_{2}} (x_{2})$	.19	.30	.28	.23

μ_{X_{1}} = 0 \cdot .19 + 1 \cdot .30 + \dots = 1.7 customers

μ_{X_{2}} = 0 \cdot .19 + 1 \cdot .30 + \dots = 1.55 customers

μ_{X_{1} - X_{2}} = μ_{X_{1}} - μ_{X_{2}} = 0.15 customers

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