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\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Problem Set 6}
\centerline{October 28, 2011}
\vskip 10pt
\noindent\vskip 10pt
\noindent
4.21.  The mean is 
\begin{align*}
E(Y) &= \int_{-\infty}^\infty yf(y)\ dy\\
&= \int_0^1 \frac{3 y^3}{2} + y^2\ dy\\
&= \left.\frac{3y^4}{8} + \frac{y^3}{3}\right|_0^1\\
&= \frac{3}{8} + \frac{1}{3} = \frac{17}{24}.
\end{align*}
Then the variance is $V(Y) = E(Y^2) - E(Y)^2$.
\begin{align*}
E(Y^2) &= \int_{-\infty}^\infty y^2 f(y)\ dy\\
&= \int_0^1 \frac{3 y^4}{2} + y^3\ dy\\
&= \left.\frac{3y^5}{10} + \frac{y^4}{4}\right|_0^1\\
&= \frac{3}{10} + \frac{1}{4}\\ 
&= \frac{11}{20}.
\end{align*}
So $V(Y) = \frac{11}{20} - \left(\frac{17}{24}\right)^2 = \frac{139}{2880}$.
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\noindent
4.22.  We have
$$E(Y) = \int_{-1}^0 y(.2)\ dy + \int_0^1 y(.2 + 1.2y)\ dy =  .4.$$
Then 
$$E(Y^2) = \int_{-1}^0 y^2(.2)\ dy + \int_0^1 y^2(.2 + 1.2y)\ dy \doteq  .433.$$
So $V(Y) \doteq .433 - (.4)^2 = .273$.  
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\noindent
4.32.  (a)  $E(Y) = \frac{3}{64}\int_0^4 y^3(4 - y)\ dy = 
\frac{3}{64}\int_0^4 4y^3  - y^4\ dy = 2.4$.
Then $E(Y^2) = \frac{3}{64}\int_0^4 4y^4  - y^5\ dy = 6.4$
and $V(Y) = E(Y^2) - E(Y)^2 = .64$.

(b)  Easiest method here is to use general facts
about expected values of linear combinations as in 4.26 above:
$E(C) = 200(2.4) = \$480$.  $V(C) = 200^2(.64) = 25600$
(units of variance are dollars${}^2$), so $\sigma = 160$.

(c) $P(C > 600) = P(Y > 3) = \frac{3}{64}\int_3^4 y^2(4-y)\ dy \doteq .26$, 
so this can certainly happen.
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\noindent
4.34.  We have by properties of iterated and double integrals,
\begin{align*}
E(Y) &= \int_0^\infty y f(y)\ dy\\
     &= \int_0^\infty \left(\int_0^y\ dt\right) f(y)\ dy\\
     &= \int_0^\infty \int_0^y f(y)\ dt\ dy\\
     &= \int_0^\infty \int_t^\infty f(y)\ dy\ dt\\
     &= \int_0^\infty 1 - F(t)\ dt.
\end{align*}
(The inner integral represents the probability $P(Y > t) = 1 - F(t)$.)  The
integration variable may be renamed to $y$ to match the formula to be proved. 
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\noindent
4.39.  Closer to $A$ means that $Y = $ location she lands satisfies
$Y < \frac{A + B}{2}$, the midpoint.  
$$P\left(Y < \frac{A + B}{2}\right) = \int_A^{\frac{A+B}{2}} \frac{1}{B - A}\ dy = \frac{1}{2}.$$
Similarly, distance to $A$ is more than three times the distance to $B$
means $\frac{A + 3B}{4} \le Y \le B$.  The probability is $\frac{1}{4}$, since
this is one quarter of the interval from $A$ to $B$ (this can be done
with an integral too, of course!) 
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\noindent
4.40.  This is a binomial probability, $Y = $ number that land closer to $B$
is binomial with $n = 3$, $p = 1/2$.  So $P(Y = 1) = \binom{3}{1} (1/2)^1(1/2)^2 =
\frac{3}{8} = .375$.   
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\noindent
4.42.  We have 
$$
.5 = P(Y \le \phi_{.5}) = \int_{\theta_1}^{\phi_{.5}} \frac{1}{\theta_2 - \theta_1}\ dy
= \frac{\phi_{.5} - \theta_1}{\theta_2 - \theta_1}.
$$
Solving this for $\phi_{.5}$, we get $\displaystyle{\phi_{.5} = \frac{\theta_1 + \theta_2}{2}}$.  This is the 
median, by definition.
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\noindent
4.43.  The mean of the area is 
$\overline{A} = E(A) =  \int_0^1 \pi r^2\ dr = \frac{\pi}{3}$, 
and the variance is 
\begin{align*}
V(A) &= E(A^2) - E(A)^2\\
&= \int_0^1 (\pi r^2)^2\ dr - \left(\frac{\pi}{3}\right)^2\\
&= \frac{\pi^2}{5} - \frac{\pi^2}{9}\\
&= \frac{4\pi^2}{45}.
\end{align*}
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\noindent
4.47.  The density function for $Y$ is $f(y) = \frac{1}{4}$ for 
$1 \le y \le 5$, and zero otherwise.  Hence

(a)  $P(Y \ge 2) = \int_2^5 \frac{1}{4}\ dy = \frac{3}{4}$.

(b)  The expected cost is $E(C) = E(c_0 + c_1 y^2) = c_0 + c_1 E(Y^2)$.
We can compute $E(Y^2)$ directly:
$$E(Y^2) = \int_1^5 y^2\cdot \frac{1}{5-1}\ dy = \left. \frac{y^3}{12}\right|1^5 = 
\frac{31}{3}.$$  
(Alternative method:  
$$E(Y^2) = V(Y) + E(Y)^2 = \frac{(5-1)^2}{12} + 9 = \frac{31}{3}.)$$
Therefore,
$$E(C) = c_0 + \frac{31 c_1}{3}.$$  

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\noindent
4.51.  This is a conditional probability:  
$$
P(Y > 65 \mid Y > 55) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}.
$$
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\noindent
4.59. Using the standard normal table, (a)  $z_0 = 0$.  (b)  $z_0 = 1.1$.  (c) $z_0 \doteq 1.645$.  (d)  $z_0 \doteq 2.575$.
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\noindent
4.62.  (a)  We have $P(Z^2 < 1) = P(-1 < Z < 1) = 1 - 2(.1587) = .6826$.
(b) Similarly, $P(Z^2 < 3.84146) = P(-1.96 < Z < 1.96) = 1 - 2(0.025) = .95$.
(Note that these correspond with the $68\%$ and $95\%$ from the ``empirical
rule'' we saw in Chapter 1.)
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\noindent
4.71. (a)  Let $Y$ be resistance of a wire produced by company $A$.  We compute
the desired probability by ``standardizing'' and consulting the standard
normal table:
$$P(.12 < Y < .13) = P\left(\frac{.12 - .13}{.005} < Z < \frac{.14 - .13}{.005}\right) = 
P(-2 < Z < 2) \doteq 1 - 2(.0228) = .9544.$$ 
(b)  This can be thought of as a binomial probability:
$\binom{4}{4}(.9544)^4(.0456)^0 = .8297$. 
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\noindent
4.74.  (a)  We have $\sigma = 6$, so $P(Y > 72) = P(Z > -1) = .8413$.
(b)  We want $b$ so $P(Z > \frac{b - 78}{6}) \doteq .1$.  From the normal 
table, $\frac{b - 78}{6} \doteq 1.28$ so  $b \doteq 85.68$.  If
the scores are to be whole numbers, then $b \ge 86$. 
(c)  $P(Z > z_0) = .281$ when $z_0 = .58$.  So we want $\frac{b - 78}{6} = .58$
and the cutoff is $b = 81.48$ (or $82$ if scores must be whole numbers).  
(d)  The cutoff for the lowest $25\%$ occurs at the where $P(Z > z_0) = .75$,
so $z_0 \doteq -0.675$.  This corresponds to a score of 74.  The proportion
of students with scores $> 74 + 5 = 79$ is about 
$100\% \times P(Z > \frac{79 - 78}{6}) = 100\% \times P(Z > .167) \doteq 43\%$.
(f) $P(Y > 84 \mid Y > 72) = P(Z > 1 \mid Z > -1) = .1587/.8413 \doteq .1886$.   
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\noindent
4.80.  We have $E(A) = E(LW) = E(3|Y|^2) = 3E(Y^2) = 3(\sigma^2 + \mu^2)$.  

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