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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Midterm Exam 2a}
\centerline{November 19, 2005}
\bigskip
\noindent
I. (10) In a shipment of 14 television sets, 5 have defective ``on''
switches and will not power up.  Out of the shipment, 7 sets are
selected at random.   What is the probability that at most 3 will 
not power up?
\bigskip
\item{} {\it Solution:}  The number $Y$ that will not power up
is hypergeometric with $N = 14$, $n = 7$, $r = 5$.  So
$$P(Y \le 3) = {{5\choose 0}{9\choose 7} + {5\choose 1}{9\choose 6} + 
{5\choose 2}{9\choose 5} + {5\choose 3}{9\choose 4}\over {14 \choose 7}}$$


\noindent
II.  
\item{A)} (10) The number of defects $Y$ in any 1-foot segment
of a magnetic tape is a Poisson random variable with 
mean $5$.  What is the probability that a 1-foot segment 
contains exactly 4 defects?  
\bigskip
\item{} {\it Solution:}  $\lambda = 5$ in the 
Poisson formula.  The probability is ${5^4 e^{-5}\over 4!} \doteq .175$. 
\bigskip
\item{B)} (15) Show, by any applicable method, that
if number of defects in one 1-foot segment is $Y_1$ and 
the number in a second 1-foot segment (not overlapping the first)
is $Y_2$, then the total number of defects, $Y = Y_1 + Y_2$, has a Poisson
distribution with mean 10.  (Assume defects
in the two segments are independent.)
\bigskip
\item{} {\it Solution:} We show in general that if $Y_1$
is Poisson with mean $\lambda_1$, $Y_2$ is Poisson with
mean $\lambda_2$, and $Y_i$ are independent, then $Y = Y_1 + Y_2$
also has a Poisson distribution, with mean $\lambda = \lambda_1 + \lambda_2$.
One method for this is to actually work out the form of the 
probability mass function for $Y$.  Note that we need to add
up all the ways to get a sum of $y$ for $Y$:
$$\eqalign{
P(Y = y) &= \sum_{y_1=0}^y P(Y_1 = y_1)P(Y_2 = y-y_1)\cr
&=\sum_{y_1=0}^y {\lambda_1^{y_1}e^{-y_1}\over y_1!} 
{\lambda_2^{y-y_1}e^{-y_2}\over (y - y_1)!}\cr
&= {e^{-(\lambda_1+\lambda_2)}\over y!}\sum_{y_1=0}^y {y \choose y_1} \lambda_1^{y_1} \lambda_2^{y-y_1}\cr
&= {e^{-(\lambda_1+\lambda_2)}\over y!}(\lambda_1+\lambda_2)^y\cr}$$
(where we used the binomial theorem at the last step).  This 
last line is exactly the form of the Poisson probability 
mass function with $\lambda = \lambda_1+\lambda_2$.  So we are done.
(Alternately, we could apply the method of moment generating functions
as we discussed in class(!))
\bigskip
\noindent
III.  In a biomedical research experiment, the survival time $Y$
(in weeks) of an animal exposed to a certain fatal disease is a random 
variable with a gamma distribution with $\alpha = 3$ and $\beta = 2$.
\item{A)} (5) What is the mean survival time of a randomly selected 
exposed animal?
\bigskip
\item{} {\it Solution:} $E(Y) = \alpha\beta = 6$.  For future
reference, note that $V(Y) = \alpha\beta^2 = 12$.
\bigskip
\item{B)} (10) Use Tchebysheff's Theorem to estimate
the probability that a single exposed animal will 
survive longer than 29 weeks.
\bigskip
\item{} {\it Solution:} We want to use the statement 
$$P(|Y - \mu| > k\sigma) = P(Y > \mu + k\sigma) + P(Y < \mu - k\sigma) \le {1\over k^2}$$
to estimate
$P(Y > 29)$.  This means that we take $\mu + k\sigma = 29$, so 
$6 + k \sqrt{12} = 29$, or $k \doteq 6.639$.  This gives
$P(Y > 29) < {1\over (6.639)^2} = .023$.  
\bigskip
\item{C)} (15) If 6 exposed animals are chosen at random, use
your answer from part B to estimate the probability
at least 2 of them survive longer than 29 weeks.
\bigskip
\item{} {\it Solution:}  The number $N$ that survive is a
binomial random variable, so 
$$P(N \ge 2) = \sum_{n=2}^6 {6\choose n} (.023)^n (.977)^{6-n}$$
\bigskip
\item{D)} (15) The cost of treating an exposed animal is 
$C = 5 + .2 Y^2$.  Find the mean and variance of $C$.
\bigskip
\item{} {\it Solution:} First 
$$E(C) = 5 + (.2)E(Y^2) = 5 + (.2)(V(Y) + (E(Y))^2) = 5 + (.2)(12+36) = 14.6.$$
Then to compute the variance of the cost:
$$\eqalign{
E(C^2) &= \int_0^\infty (5+.2y^2)^2 {y^2 e^{-y/2}\over \Gamma(3)\cdot 8}\ dy \cr
&={1\over 16} \left( 25\int_0^\infty y^2 e^{-y/2}\ dy + 2\int_0^\infty y^4 e^{-y/2}\ dy + 
(.04) \int_0^\infty y^6 e^{-y/2}\ dy\right)\cr
&= {1\over 16}\left(25\cdot 2^3\Gamma(3) + 2\cdot 2^5\Gamma(5) + (.04)\cdot 2^7 \Gamma(7)\right)\cr
&=351.4\cr}$$
Then $V(C) = E(C^2) - (E(C))^2 = 351.4 - (14.6)^2 \doteq 138.24$.
\bigskip
\noindent
IV. 
\item{A)} (10) The lifetime of a certain type of small motor is 
{\it uniformly} distributed on the interval $[0,30]$.  What
is the probability that the motor will last more than 10 years?
\bigskip
\item{} {\it Solution:}  $P(Y > 10) = \int_{10}^{30} {dy\over 30 - 0} = {2\over 3}$.
\bigskip
\item{B)}  (10) Now, assume the lifetime of the motors is 
{\it normally} distributed with mean 15 years and standard deviation 
4 years.  The manufacturer replaces free all motors that fail while 
under warrantee.  But to avoid going out of business because of the 
cost of doing the free replacements, the manufacturer must set the 
warrantee so that they will have to replace 
no more than 3\% of the motors that fail.  What should the length 
of the warrantee be?
\bigskip
\item{} {\it Solution:}  Let $w$ be the length of the warrantee. 
Then we want $P(Y < w) = .03$.  After standardizing, this says
$P\left(Z < {w - 15\over 4}\right) = .03$.  From the standard
normal table, this says ${w - 15\over 4} = -1.88$, so 
$w = 15 - (4)(1.88) = 7.48$ years.
\bigskip
\noindent
{\it Extra Credit} (10) 
Let $Z$ be a standard normal random variable.  What is the 
moment-generating function for $Y = 8Z - 4.3$?  What is the 
distribution of $Y$?
\bigskip
\item{} {\it Solution:}  
$$m_Y(t) = e^{-4.3t} m_Z(8t) = e^{-4.3t + {64 t^2\over 2}}.$$
By the Uniqueness Theorem, $Y$ has a normal distribution too, with 
$\mu = -4.3$ and $\sigma = 8$.

\bye
