\magnification=\magstep1
\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Midterm Exam 2 -- November 17, 2011}
\bigskip
\noindent
I.  Birth weights of babies born in the US are normally distributed
with mean $\mu = 3315$ grams and standard deviation $\sigma = 575$ grams.
\medskip
\item{A)}  (5)  Find the (approximate) probability that a baby will have
birth weight more than $3700$ grams.
\medskip
\item{} {\it Solution:} Let $Y = $ birth weight of a randomly selected
baby, so by the given information, $Y \sim Normal(3315,575^2)$.  Standardizing, we have
$$P(Y > 3700) = P\left({Y - 3315\over 575} > {3700 - 3315 \over 575}\right)
\doteq P(Z > .67) \doteq .2514 $$
(since ${Y - 3315\over 575} = Z \sim Normal(0,1)$).
\medskip
\item{B)}  (10)  If 20 newborn babies are selected randomly, how
many of them would be expected to have weights between $2740$ and $3890$ grams?
\medskip
\item{} {\it Solution:}
We have $2740 = 3315 - 575 = \mu - \sigma$ and $3890 = \mu +\sigma$.
The probability that a single baby has a weight in this range is 
$$P(2740 < Y < 3890) = P(-1 < Z < 1) = 1 - 2P(Z > 1) = 1 - 2(.1587) = .6826$$
(Note that this interval is where the $68\%$ in the Empirical Rule comes from.)
If $X$ is the number of babies in the 20 that have weights in this
range, then technically $X$ would have a {\it hypergeometric} distribution,
since the selection {\it would not be done with replacement}.  However,
we don't know the size of the entire population of babies in question.   
In any case that would be much larger than
$20$.  So the {\it binomial} formulas can be used to estimate this.  
Say $X \sim Binomial(20,.6826)$.  Then
$$E(X) = n\cdot p = 20 \cdot 0.6826  = 13.652.$$ 
So we expect this many babies to have weights in this range.  
\item{C)}  (10)  If we knew the mean and standard deviation, but we did not 
know the precise distribution of the birth weights, what could 
we say in general about the probability that a randomly selected baby 
has weight differing from $3315$ grams by at least $862.5$ grams?
\medskip
\item{} {\it Solution:}  By Tchebysheff's Theorem, since $862.5 = 1.5 \cdot 575$
$$P(|Y - \mu| \ge 1.5 \sigma) \le {1\over (1.5)^2} = .4444.$$   
\bigskip
\noindent
II.  (15) The number of defects $Y$ in a 1-foot segment
of a magnetic tape is a Poisson random variable with 
mean $3$.  What is the probability that a 1-foot segment 
contains more than 4 defects?    
\medskip
\noindent
{\it Solution:}  This is 
$$P(Y > 4) = \sum_{y=5}^\infty {3^y e^{-3}\over y!} = 1 - \sum_{y=0}^4 {3^y e^{-3}\over y!} \doteq .1847.$$
\bigskip
\noindent
III. (10) (From an actuarial ``P Exam'' problem) The warranty on a machine 
specifies that it will be replaced at failure or at age 4, whichever 
occurs first.  The machine's age at failure, $X$, is a random variable 
with probability density function
$$f(x) = \cases{1/5 & if $0 < x < 5$\cr
                 0  & otherwise\cr}$$
Let $Y$ be the age of the machine {\it at replacement}.  What is $E(Y)$?
\medskip
\noindent
{\it Solution:}  Note that $X$ has a uniform distribution on
the interval $[0,5]$.  The age of the machine at replacement is given
by this function of $X$:
$$Y = \cases{X & if $0 \le X < 4$ \cr
             4 & if $4 \le X \le 5$\cr
             0 & otherwise\cr}$$
So 
$$E(Y) = \int_0^4 x \cdot {1\over 5}\ dx + \int_4^5 4 \cdot {1\over 5}\ dx = {16\over 10} + {4\over 5} = {12\over 5}.$$
\bigskip
\noindent
IV. 
\item{A)} (10) Let 
$$g(x) = \cases{ {1\over 16} x^2 e^{x/2} & if $x < 0$\cr
                         0 & otherwise (that is, if $x \ge 0$).\cr}$$
Show that $g$ is a legal pdf for a random variable $X$. 
\medskip
\item{} {\it Solution:}  $g(x) \ge 0$ for all $x$ by inspection.  Moreover
$$\int_{-\infty}^{\infty} g(x)\ dx = \int_{-\infty}^0 {1\over 16} x^2 e^{x/2}\ dx$$
Let $y = -x/2$.  Then this integral becomes
$${1\over 2}\int_0^\infty y^2 e^{-y}\ dy = {1\over 2} \Gamma(3) = 1.$$
Therefore all the requirements for a pdf are satisfied.     
\medskip
\item{B)} (10)  Determine the moment-generating function of $X$.  On
which interval of $t$ values is your formula valid?
\medskip
\item{} {\it Solution:}  The mgf is 
$$\eqalign{
E(e^{tX}) &= \int_{-\infty}^0 e^{tx} \cdot {1\over 16} x^2 e^{x/2}\ dx\cr
&= \int_{-\infty}^0 {1\over 16} x^2 e^{(1+2t)x/2}\ dx\cr}$$
Let $y = -(1 + 2t)x/2$ or $x = {- 2y\over 1+2t}$.  Making the substitution,
we get (provided $1 + 2t > 0$), 
$$\eqalign{ 
&= \int_0^\infty {1\over 2 (1 + 2t)^3} y^2 e^{-y}\ dy\cr
&= {1\over 2(1 + 2t)^3} \Gamma(3)\cr
&= {1\over (1 + 2t)^3}\cr}$$
In order for this to be valid, the improper integral must converge,
and for that $1 + 2t > 0$, or $t > -1/2$.
\bigskip
\noindent
V.  Let $Y_1,Y_2$ be jointly continuous random variables with joint density
$$f(y_1,y_2) = \cases{c y_1 & if $0 \le y_1 \le 1$ and $y_1^2 \le y_2 \le 1$\cr
                   0 & otherwise\cr}$$
\item{A)}  (15)  What is $P( Y_1 \le Y_2 )$? 
\medskip
\item{} {\it Solution:}
The region where the density is nonzero is the region with $0 \le y_1 \le 1$, 
above the parabola $y_2 = y_1^2$, and below the line $y_2 = 1$. 
The value of $c$ is determined as follows.  We must have 
$$\eqalign{1 &= \int_0^1 \int_{y_1^2}^1 cy_1\ dy_2\ dy_1\cr
&= c\int_0^1 y_1 - y_1^3\ dy_1\cr
&= c\left({1\over 2} - {1\over 4}\right)\cr}$$
So $c = 4$.  

The condition
$Y_1 \le Y_2$ then defines the triangle with vertices at $(0,0), (1,1), (0,1)$.
$$\eqalign{
P(Y_1 \le Y_2) &= \int_0^1\int_{y_1}^1 4y_1\ dy_2\ dy_1\cr
&= \int_0^1 4y_1 - 4y_1^2\ dy_1\cr
&= \left. 2y_1^2 - {4\over 3} y_1^3 \right|_0^1\cr
&= {2\over 3}.\cr}$$  
\item{B)}  (15)  Find the marginal density $f_1(y_1)$ and use it to 
compute $V(Y_1)$.
\medskip
\item{} {\it Solution:}  The marginal density is
$$f_1(y_1) = \cases{\int_{y_1^2}^1 4y_1\ dy_2 = 4y_1(1 - y_1^2) & if $0 \le y_1 \le 1$\cr
                                            0 & otherwise\cr}$$
Then 
$$E(Y_1) = \int_0^1 4y_1^2 - 4y_1^4\ dy_1 = {4\over 3} - {4\over 5} =  {8\over 15}.$$
Similarly
$$E(Y_1^2) = \int_0^1 4y_1^3 - 4y_1^5\ dy_1 = 1 - {2\over 3} =  {1\over 3}.$$
So 
$$V(Y_1) = {1\over 3} - \left({8\over 15}\right)^2 = {11\over 225}.$$
\bigskip
\noindent
{\it Extra Credit.} (10)   
What is $E(X)$ for the random variable $X$ from question IV? 
\medskip
\noindent
{\it Solution:}  The answer is $E(X) = -6$.  This can be computed either directly 
by evaluating $E(X) = \int_{-\infty}^0 x\cdot g(x)\ dx$, or more cleverly by 
using the mgf from the answer to IV B):
$$E(X) = {d\over dt}\left. {1\over (1 + 2t)^3}\right|_{t = 0} = {(-3)(2)\over (1 + 2\cdot 0)^4} = -6.$$
(Note: the density $g(x)$ has the form of a gamma density with $\alpha = 3$ and $\beta = 2$, but reflected about $x = 0$.  So the expected value should be the 
negative of the expected value of the corresponding standard gamma density: $\alpha\beta = 6$.) 
\bye
