\magnification=\magstep1
\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions -- Midterm Exam 2 Practice Problems}
\centerline{November 9, 2011}
\bigskip
\noindent
I.  If $Y$ denotes the number of underage students who get carded,
then $Y$ is hypergeometric with $N = 10$, $r = 4$, $n = 5$.  So
$$P(Y = 2) = {{4 \choose 2} {6\choose 3}\over {10 \choose 5}} = {10\over 21} \doteq .48 $$
\bigskip
\noindent
II.  By examining the form of the density function, we see
that $Y$ has a {\it beta distribution} with $\alpha = \beta = 3$.
Hence:
\item{A)}  The constant $c$ must be 
$$c = {1\over B(3,3)} = {\Gamma(6)\over \Gamma(3)\Gamma(3)} = {5!\over 2!2!} = 30$$
\item{B)}  The variance is 
$$V(Y) = {\alpha\beta\over (\alpha+\beta)^2(\alpha+\beta+1)} = {3\cdot 3\over 36\cdot 7}
 = {1\over 28}$$
\item{C)}  The cumulative distribution function is the antiderivative
$F(y)$ of the density with $F(-\infty) = 0$ and $F(+\infty) = 1$.  This
is
$$F(y) = \cases{0 & if $y < 0$\cr
                10y^3 - 15y^4 + 6y^5 & if $0 \le y < 1$\cr
                1 & if $y \ge 1$\cr}$$
\item{D)} This is 
$$\int_{.1}^{.25} 30y^2(1-y)^2\ dy = F(.25) - F(.1) \doteq .09496.$$
\bigskip
\noindent
III.  The lifetime $Y$ of a single switch has the exponential density
$$f(y) = \cases{{1\over 2}e^{-y/2} & if $y > 0$\cr
                    0 & otherwise\cr}$$
\item{A)}  The probability that a single switch fails during the 
first year is the same as the probability that its life is less
than 1:
$$P(Y < 1) = \int_0^1 {1\over 2}e^{-y/2}\ dy = -e^{-y/2}|_0^1 = 1 - e^{-1/2}
\doteq .3935$$
\item{B)}  The probability that at most 30 out of 100 of these switches
will fail in the first year is computed by the {\it binomial} probability 
formula (since we assume the switches operate independently):
$$\sum_{y=0}^{30} {100\choose y} (.3935)^y (.6065)^{100-y}.$$
\bigskip
\noindent
IV.  
\item{A)}  Let $Y$ be the weight of one poodle
$$P(7.3 < Y < 9.1) = P\left({7.3-8\over .9} < {Y-8\over .9} < {9.1-8\over .9}\right)$$
This is the same as 
$$P\left(-.78 < {Y-8\over .9} < 1.22\right)$$
We know $Z = {Y-8\over .9}$ is a standard normal, so we can use
the {\it standard normal table} to find this probability.  
By the symmetry of the normal density,
$$P(-.78 < Z < 0) = P(0 < Z < .78) = .5 - .2177 = .2823$$
We also have 
$$P(0 < Z < 1.22) = .5 - .1112 = .3888$$
Hence
$$P(-.78 < Z < 1.22) = .2823 + .3888 = .6711$$
\bigskip
\item{B)}  Partial credit would be given for an answer like
$$E(Y^3) = \int_{-\infty}^\infty y^3 {1\over \sqrt{2\pi(.9)^2}} e^{-(y-8)^2/2(.9)^2}\ dy.$$
But the only reasonable way to get a numerical value for this is to think of 
using the {\it moment-generating function} for $Y$.  $E(Y^3) = m'''(0)$ where
$m(t)$ is the mgf for a normal random variable.  The  general form
of this is
$$m(t) = e^{{\sigma^2 t\over 2} + \mu t}$$
Hence differentiating with the chain and product rules:
$$\eqalign{
m'(t) &= e^{{\sigma^2 t\over 2} + \mu t} ( \sigma^2 t + \mu)\cr
m''(t) &= e^{{\sigma^2 t\over 2} + \mu t} (( \sigma^2 t + \mu)^2 + \sigma^2)\cr
m'''(t) &= e^{{\sigma^2 t\over 2} + \mu t} (( \sigma^2 t + \mu)^3 + 3\sigma^2(\sigma^2t + \mu))\cr
\Rightarrow m'''(0) &= \mu^3 + 3\sigma^2 \mu\cr}$$
With $\mu = 8$ and $\sigma = .9$, this yields $E(Y^3) = 531.44$.
\bigskip\noindent
V. This is the density for a Gamma-distributed random 
variable with $\alpha = 2$ and $\beta = 1$.  Hence
$$m(t) = (1 - t)^{-2}$$
(This can also be computed directly as
$$E(e^{tY}) = \int_0^\infty e^{ty}\cdot y e^{-y}\ dy.)$$
\bigskip
\noindent
VI.  From the situation, it is clear that the selection of
missiles is being done {\it without replacement}.  Hence
the required probabilities will be computed using the 
hypergeometric formulas with $N = 10$, $r = 3$, $n = 4$.
Let $Y$ be the number that {\it will not} fire out of the four selected
\item{A)} The probability that all four of the chosen missiles will fire is
$$P(Y = 0) = {{3\choose 0} {7\choose 4} \over {10\choose 4}} = {1\over 6}$$
\item{B)}  The probability that at most 2 will not fire is 
$$P(Y \le 2) = {{3\choose 2} {7\choose 2} + {3\choose 1}{7\choose 3} + {3 \choose 0}{7\choose 4} \over {10\choose 4}} = {29\over 30}.$$
\bigskip
\noindent
VII.  
\item{A)}  The mean survival time is $E(Y) = \alpha\beta = 12$ weeks.
\item{B)}  This is not possible since $12\sqrt{2} = 2 \sigma$.  By
Tchebysheff's Theorem, $P(|Y - \mu| < 2\sigma) \ge .75$ for every
random variable (independent of its distribution).  
\item{C)}  The probability that a single animal survives 15 or 
more weeks is computed like this (integrating by parts):
$$\eqalign{
p &= \int_{15}^\infty  {y e^{-y/6}\over 36}\ dy\cr 
  &= \left. {-1\over 6} y e^{-y/6}\right|_{15}^\infty + {1\over 6} \int_{15}^\infty e^{-y/6}\ dy\cr
  &= {5\over 2} e^{-5/2} + e^{-5/2}\cr
  &= {7\over 2} e^{-5/2} \doteq .2873\cr}$$
Then if $Y$ is the number out of the 6 that survive longer than $15$ weeks,
$$P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1) = 1 - {6\choose 0}p^0(1-p)^6 - {6\choose 1}p^1 (1 - p)^5 \doteq .5520.$$
(Note:  The original posting of this solution contained an error because
the Gamma density used did not incorporate the proper value of $\beta = 6$.) 
\bigskip
\noindent
VIII.  Let $w$ be the warrantee period and $Y$ be the lifetime of a randomly
chosen motor.  We want to choose $w$ so that the probability that a motor
lasts less than the warrantee period is only $.03$, or in other words:
$P(Y < w) = .03$.  Since $Y \sim Normal(10,4)$, this 
is the same as 
$$.03 = P(Y < w) = P(Z < {w - 10\over 2})$$
From the standard normal table, we see ${w - 10\over 2} \doteq -1.88$ so
$w = 6.24$ years.  
\bigskip
\noindent
IX.   The joint density is nonzero on the triangle in the $y_1,y_2$
plane with vertices at $(0,-1), (1,0)$, and $(0,1)$.
\item{A)}  Then 
$$P(Y_2 > 0) = \int_0^1 \int_0^{1 - y_1} 30 y_1 y_2^2\ dy_2\ dy_1 = {1\over 2}$$
(This can be seen without calculation if you notice the symmetry of the region
and the density function under reflection across the $y_1$-axis; otherwise
just compute!)
\item{B)}  The marginal density of $Y_1$ is $0$ for $y_1 \notin [0,1]$ and  
$$\int_{y_1-1}^{1-y_1} 30 y_1 y_2^2\ dy_2 = 20 y_1(1 - y_1)^3$$
if $0 \le y_1 \le 1$.  Since 
$$B(2,5) = {\Gamma(2)\Gamma(4)\over \Gamma(6)} = {3!\over 5\!} = {1\over 20}$$
This is the beta density with $\alpha = 2$ and $\beta = 4$.
\item{C)}  The conditional density is computed as $f(y_1,y_2)/f_1(y_1)$.  
So by part B), this gives
$$\eqalign{
f(Y_2|Y_1 = y_1) &= {30 y_1 y_2^2 \over 20 y_1 (1 - y_1)^3}\cr
                 &= {3y_2^2\over 2(1 - y_1)^3}\cr}$$
for $-1 \le y_2 \le 1$ and zero otherwise.  (Note: we would use this with
a particular value of $y_1$ substituted in, and think of it as 
a function of $y_2$.)
\bye
