\documentclass[11pt]{article}
\def\Z{\mathbb{Z}}
\def\Q{\mathbb{Q}}
\def\R{\mathbb{R}}
\def\C{\mathbb{C}}
\usepackage{latexsym,amsmath,amssymb}
\usepackage[all]{xy}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.0in}
\begin{document}
\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 7}
\centerline{November 10, 2009}
\vskip 10pt
4.58.  Using the standard normal table,  
(a) $P(0 \le Z \le 1.2) = .5 - P(Z > 1.2) = .5 - .1151 - .3849$.

(b) By symmetry, $P(-.9 \le Z \le 0) = P(0 \le Z \le .9) = .5 - P(Z > .9) = 
.5 - .1841 = .3159$.

(c) $P(.3 \le Z \le 1.56) = P(Z \ge .3) - P(Z > 1.56) = .3821 - .0594 = .3227$.

(d) By symmetry, $P(-.2 \le Z \le .2) = 2P(0 \le Z \le .2) = 2(.5 - P(Z > .2))
= 1 - .8414 = .1586$.  

(e) By symmetry, $P(-1.56 \le Z \le -.2) = P(.2 \le Z \le 1.56) = P(Z \ge .2)
- P(Z > 1.56) = .4207 - .0594 = .3613$.
\vskip 10pt
\noindent
4.59.  (a)  $P(Z > z_0) = .5$ for $z_0 = 0$.

(b) $P(Z < z_0) = .8643 \Leftrightarrow P(Z \ge z_0) = 1 - .8643 = .1357$,
so $z_0 = 1.1$.

(c)  $P(-z_0 < Z < z_0) = .9 \Leftrightarrow P(Z \ge z_0) = \frac{1}{2}(1 - .9) =
.05$.  Hence, $z_0 \doteq 1.645$.  (We need to interpolate between
two entries in the table here.)

(d)  $P(-z_0 < Z < z_0) = .99 \Leftrightarrow P(Z \ge z_0) = \frac{1}{2}(1 - .99) =
.005$.  Hence, $z_0 \doteq 2.575$.  (We also need to interpolate between
two entries in the table here.)
\vskip 10pt
\noindent
4.61.  The median is the $y_0$ such that $P(Y \ge y_0) = \frac{1}{2}$.
By the symmetry of the normal density function, this happens
when $y_0 = \mu$.  
\vskip 10pt
\noindent
4.68.  This is given by the normal probability
$P(Y > 3) = P(Z > \frac{3 - 2.4}{.8}) = P(Z > .75) = .2266$.  So 
about $23\%$ of the students will have GPA's above 3.  (Note:  This
problem was written either before the current state of 1 ``grade inflation''
or else at a large university where the student body is not as strong
as at a college like HC!) 
\vskip 10pt
\noindent
4.69.  $P(Y < 1.9) = P(Z < \frac{1.9 - 2.4}{.8}) = P(Z < -.625) = P(Z > .625)$
(by symmetry).  This is $\doteq .2660$ from the table.  So about $27\%$
of the students would be dropped(!)
\vskip 10pt
\noindent
4.70.  This is a binomial probability question with the 
``success probability'' obtained from the normal distribution.
We have $P(Y > 3) = .2266$ from question 4.68.  So if we
pick three students at random, the probability that all 
three have GPA's greater than 3.0 is $\binom{3}{3}(.2266)^3(.7731)^0 \doteq .012$.  
\vskip 10pt
\noindent
4.75.  Let $Y$ be the amount of soda dispensed, which is normally
distributed with mean $\mu$ and standard deviation $\sigma = .3$.
We want to set $\mu$ so $P(Y > 8) = .01$.  So after standardizing,
$P(Z > \frac{8 - \mu}{.3}) = .01$.  From the standard normal
table, $\frac{8 - \mu}{.3} \doteq 2.33$ so $\mu \doteq 7.3$ ounces.  
\vskip 10pt
\noindent
4.76.  The question is asking what $\sigma$ will make
$P(\mu - 1 < Y < \mu + 1) \ge .95$.  Standardizing, this is
equivalent to $P(-\frac{1}{\sigma} < Z < \frac{1}{\sigma}) \ge .95$.
Hence, by symmetry, $P(Z \ge \frac{1}{\sigma}) \le .025$, so $\frac{1}{\sigma} 
\ge 1.96$ from the standard normal table.  
So $\sigma \le \frac{1}{1.96} \doteq .51$.  
\vskip 10pt
\noindent
4.89. (a) We have 
\begin{align*}
.0821 &=  P(Y > 2)\\
 &= \int_2^\infty \frac{1}{\beta} e^{\frac{-y}{\beta}}\ dy\\
 &= e^{\frac{-2}{\beta}}.
\end{align*}
So we solve by taking logarithms:
$\beta = \frac{-2}{\ln(.0821)} \doteq .8$.  

(b)  Then 
$$P(Y \le 1.7) = \int_0^{1.7} \frac{1}{.8} e^{\frac{-y}{.8}}\ dy \doteq .8806.$$   
\vskip 10pt
\noindent
4.91.  The mean is $100$ cfs, so $\beta = 100$ in the exponential density.

(a)  $P(Y > 200) = \int_{200}^\infty \frac{1}{100} e^{\frac{-y}{100}}\ dy = 
e^{-2} \doteq .1353$.

(b)  Let $c$ be the desired capacity.  The question is asking: 
What $c$ will make $P(Y > c) = .01$.  As in part (a), 
$P(Y > c) = \int_c^\infty \frac{1}{100} e^{\frac{-y}{100}}\ dy = 
e^{\frac{-c}{100}}$. So we want $.01 = e^{\frac{-c}{100}}$.
Solve by taking logarithms:  $c = -100\ln(.01) \doteq 460.5$.  
\vskip 10pt
\noindent
4.95.  (a) We have by the definition of the discrete random variable $X$:
\begin{align*}
P(X=k) &= \int_{k-1}^k \frac{1}{\beta} e^{\frac{-y}{\beta}}\ dy\\
&= \left.-e^{\frac{-y}{\beta}}\right|_{k-1}^k\\
&= -e^{\frac{-k}{\beta}} + e^{\frac{-(k-1)}{\beta}}\\
&= \left(e^{\frac{-(k-1)}{\beta}}\right)\left(1 - e^{\frac{1}{\beta}}\right)\\
&= \left(e^{\frac{-1}{\beta}}\right)^{k-1} \left(1 - e^{\frac{-1}{\beta}}\right).
\end{align*}

(b) This is the form ($q^{k-1}p$) for the probability  function 
for a {\it geometric} random variable with $p = 1 - e^{\frac{-1}{\beta}}$.
\vskip 10pt
\noindent
4.109.  First, let's derive the general formula for $E(Y^k)$ when 
$Y$ has a Gamma-distribution with parameters $\alpha, \beta$.  We
apply the substitution $u = \frac{y}{\beta}$, so $dy = \beta\ du$,
then apply the functional equation $\Gamma(x+1) = x\Gamma(x)$ repeatedly:
\begin{align*}
E(Y^k) &= \frac{1}{\Gamma(\alpha) \beta^\alpha}
\int_0^\infty y^k \cdot y^{\alpha-1} e^{\frac{-y}{\beta}}\ dy\\
&=\frac{1}{\Gamma(\alpha) \beta^\alpha}
\int_0^\infty y^{\alpha+k-1} e^{\frac{-y}{\beta}}\ dy\\
&=\frac{1}{\Gamma(\alpha) \beta^\alpha}\cdot \beta^{\alpha+k} 
\int_0^\infty u^{\alpha+k-1} e^{-u}\ du\\
&=\frac{1}{\Gamma(\alpha) \beta^\alpha}\cdot \beta^{\alpha+k} \cdot \Gamma(\alpha+k)\\
&=\beta^k (\alpha+k-1)(\alpha+k-2)\cdots (\alpha+1)(\alpha).
\end{align*}
In our problem, with $\alpha=3, \beta =2$, we get
\begin{align*}
           E(Y) &= \alpha\beta = 6\\
         E(Y^2) &= (\alpha+1)(\alpha)\beta^2 = 48\\
         E(Y^3) &= (\alpha+2)(\alpha+1)\alpha\beta^3 = 480\\
         E(Y^4) &= (\alpha+3)(\alpha+2)(\alpha+1)\alpha\beta^4=5760.
\end{align*}
Then
$$E(L) = 30 E(Y) + 2E(Y^2) = 30\cdot 6 + 2\cdot 48 = 276$$
and
$$E(L^2) = 900 E(Y^2) + 120E(Y^3) + 4E(Y^4) = 900\cdot 48 + 120\cdot 480 + 4\cdot
5760 = 123840.$$
Hence
$$V(C) = E(L^2) - (E(L))^2 = 123840 - 276^2 = 47664.$$
\vskip 10pt
\noindent
4.129.  Instead of computing the expected values directly 
from the integral formulas, we can use the general results
for $\beta$-distributed random variables from Theorem 4.11.
We have 
$$E(C) = E(10 + 20Y + 4Y^2) = 10 + 20E(Y) + 4E(Y^2).$$
From Theorem 4.11, we have $E(Y) = \frac{\alpha}{\alpha+\beta} = \frac{1}{3}$.
Then 
$$
E(Y^2) = V(Y) + E(Y)^2 = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} +
\frac{1}{9} = \frac{1}{18} + \frac{1}{9} = \frac{1}{6}.
$$
So $E(C) = 10 + \frac{20}{3} + \frac{4}{6} = \frac{52}{3}$.
To get $V(C) = E(C^2) - E(C)^2$, we can proceed as follows:
$E(C^2) = 100 + 400E(Y) + 480 E(Y^2) + 160 E(Y^3) + 16 E(Y^4)$.
We already know $E(Y),E(Y^2)$ from the above.  To get $E(Y^3)$
and $E(Y^4)$ proceed like this:
$$
E(Y^3) = 2\int_0^1 y^3 (1 - y)\ dy.
$$
Apart from the constant multiple, this is the integral of a 
beta density function with $\alpha = 4$ and $\beta = 2$.  
So 
$$
E(Y^3) = 2B(4,2) = 2\frac{\Gamma(4)\Gamma(2)}{\Gamma(6)} = 
2\frac{3!1!}{5!} = \frac{1}{10}.
$$
Similarly,
$$
E(Y^4) = 2B(5,2) = 2\frac{\Gamma(5)\Gamma(2)}{\Gamma(7)} = 
2\frac{4!1!}{6!} = \frac{1}{15}.
$$
So then
$$
V(C) = 10 + \frac{400}{3} + \frac{480}{6} + \frac{160}{10} + \frac{16}{15} - 
\left(\frac{52}{3}\right)^2 = \frac{1348}{45}.
$$
\vskip 10pt
\noindent
4.130. We use the same ``trick'' as in the last problem. 
We know $E(Y) = \frac{\alpha}{\alpha+\beta}$ from Theorem 4.11
and work in class. 
\begin{align*}
E(Y^2) &= \frac{1}{B(\alpha,\beta)} \int y^2\cdot y^{\alpha-1} (1 - y)^{\beta - 1}\ dy\\
&= \frac{1}{B(\alpha,\beta)} \int y^{\alpha+1} (1 - y)^{\beta - 1}\ dy\\
&= \frac{B(\alpha+2,\beta)}{B(\alpha,\beta)}\\
&= \frac{\Gamma(\alpha+2)\Gamma(\beta) \Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta) \Gamma(\alpha+\beta+2)}.
\end{align*}
 Then using the functional equation for
the $\Gamma$-function
\begin{align*}
&=\frac{(\alpha+1)\alpha\Gamma(\alpha)\Gamma(\beta) \Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta) (\alpha+\beta+1)(\alpha+\beta)\Gamma(\alpha+\beta)}\\
&=\frac{(\alpha+1)\alpha}{(\alpha+\beta+1)(\alpha+\beta)}.
\end{align*}
So then
\begin{align*}
V(Y) &= \frac{(\alpha+1)\alpha}{(\alpha+\beta+1)(\alpha+\beta)} - 
\left(\frac{\alpha}{\alpha+\beta}\right)^2\\
&= \frac{(\alpha + \beta)(\alpha + 1)(\alpha) - (\alpha + \beta + 1)(\alpha^2)}{(\alpha+\beta)^2(\alpha + \beta + 1)}\\
&= \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha + \beta + 1)},
\end{align*}
which is what we wanted to show.  
\vskip 10pt
\noindent
4.133.  (a)  This is the form for a beta density function with
$\alpha = 3$ and $\beta = 5$, so $c = \frac{1}{B(3,5)} = \frac{7!}{2!4!} = 105$.

(b)  $E(Y) = \frac{3}{8} = .375 $  from Theorem 4.11.

(c)  The standard deviation is 
$$\sigma = \sqrt{\frac{15}{64\cdot 9}} \doteq .1614.$$

(d)  We have:
$$P(Y > .375 + 2(.1614)) = \int_{.6978}^1 105 y^2 (1-y)^4\ dy \doteq .02972.$$
\vskip 10pt
\noindent
{\it Additional Problem}
\vskip 10pt
\noindent
A)  1) We must show that when $\alpha$ is a positive integer, 
\begin{equation}
\label{eq1}
\frac{1}{\Gamma(\alpha)}\int_\lambda^\infty y^{\alpha-1} e^{-y}\ dy = 
\sum_{n=0}^{\alpha-1} \frac{\lambda^n}{n!}e^{-\lambda}.
\end{equation}
We will use induction on $\alpha$.  The base case is $\alpha = 1$.  Then
the left of (\ref{eq1}) is
$$
1\cdot\int_\lambda^\infty e^{-y}\ dy = e^{-\lambda}.
$$
With $\alpha = 1$, the sum on the right of (\ref{eq1}) is:
$$
\sum_{n=0}^0 \frac{\lambda^n}{n!}e^{-\lambda} = e^{-\lambda}.
$$
So the base case is established.  

Now for the induction step, assume that (1) is true for $\alpha = k$.
In other words, our induction hypothesis is:
\begin{equation}
\label{eq2}
\frac{1}{\Gamma(k)}\int_\lambda^\infty y^{k-1} e^{-y}\ dy = 
\sum_{n=0}^{k-1} \frac{\lambda^n}{n!}e^{-\lambda}.
\end{equation}
Consider the integral on the left of (\ref{eq1}) for $\alpha = k+1$:
$$
\frac{1}{\Gamma(k+1)} \int_\lambda^\infty y^k e^{-y}\ dy.
$$
Apply integration by parts with $u = y^k$, and $dv = e^{-y}\ dy$.  We
get:
\begin{align*}
\frac{1}{\Gamma(k+1)} \int_\lambda^\infty y^k e^{-y}\ dy
&= \frac{1}{\Gamma(k+1)}\left(\left. -y^k e^{-y}\right|_\lambda^\infty 
+ k\int_\lambda^\infty y^{k-1} e^{-y}\ dy\right)\\
&= \frac{\lambda^k}{k!} e^{-\lambda} 
+ \frac{1}{\Gamma(k)} \int_\lambda^\infty y^{k-1} e^{-y}\ dy.
\end{align*}
Now we use the induction hypothesis (\ref{eq2}) on the remaining integral.
We get that this is equal to 
$$
\frac{\lambda^k}{k!} e^{-\lambda} + \sum_{n=0}^{k-1}\frac{\lambda^n}{n!}e^{-\lambda}
=\sum_{n=0}^k \frac{\lambda^n}{n!}e^{-\lambda}.
$$
Hence our formula also works with $\alpha = k+1$, and by induction
we conclude that it works for all positive integers $\alpha$.

2) Using the Poisson table in the text with $\lambda = 5$, $\alpha = 3$
$$\sum_{n=0}^2 \frac{5^n}{n!}e^{-5} = .125$$

\end{document}
