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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Midterm Exam 2b}
\centerline{November 19, 2005}
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\noindent
I. (10) In a group of 12 chickens, 5 are infected with 
the H5N1 ``bird flu'' virus and the others are not.
If 7 chickens are selected randomly from the group,
what is the probability 2 or more have the virus?
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\item{} {\it Solution:}  The number $Y$ that do have the virus
is hypergeometric with $N = 12$, $n = 7$, $r = 5$.  So
$$P(Y \ge 2) = {{5\choose 2}{7\choose 5} + {5\choose 3}{7\choose 4} + 
{5\choose 4}{7\choose 3} + {5\choose 5}{7\choose 3}\over {12 \choose 7}}$$
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\noindent
II.  
\item{A)} (15) Assume that
$Y_1$ has a Poisson distribution with mean $\lambda_1$
and $Y_2$ has Poisson distribution with mean $\lambda_2$.
Also assume that $Y_1 = y_1$ and $Y_2 = y_2$ are independent 
events for all $y_1 \ge 0$ and $y_2\ge 0$.  Show, by any applicable method, 
that $Y = Y_1 + Y_2$ also has a Poisson distribution, with mean 
$\lambda = \lambda_1 + \lambda_2$.  
\bigskip
\item{} {\it Solution:}
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(!))
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\item{B)} (10) The number of defects $Y$ in any 2-foot segment
of a magnetic tape is a Poisson random variable with 
mean $3$.  Use part A to find the probability that a 4-foot segment 
contains exactly 8 defects, assuming the necessary independence.
(You can use part A even if you didn't completely see how to do
that part.)
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\item{} {\it Solution:}  The number of defects in the 4-foot segment
is Poisson with $\lambda = 6$ by part A.  Hence the probability
is $P(Y = 8) = {6^8 e^{-6}\over 8!} \doteq .103$.
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\noindent
III. 
\item{A)} (10) The lifetime of a certain type of battery is 
{\it uniformly} distributed on the interval $[0,25]$ (years).  What
is the probability that the battery will last more than 10 years?
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\item{} {\it Solution:}  $P(Y > 10) = \int_{10}^{25} {dy\over 25 - 0} = {3\over 5}$.
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\item{B)}  (10) Now, assume the lifetime of the batteries is 
{\it normally} distributed with mean 10 years and standard deviation 
4 years.  The manufacturer replaces free all batteries 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 5\% of the batteries that fail.  What should the length 
of the warrantee be?
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\item{} {\it Solution:}  Let $w$ be the length of the warrantee. 
Then we want $P(Y < w) = .05$.  After standardizing, this says
$P\left(Z < {w - 10\over 4}\right) = .05$.  From the standard
normal table, this says ${w - 10\over 4} = -1.645$, so 
$w = 10 - (4)(1.645) = 3.42$ years.
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\noindent
IV.  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 {\it gamma distribution} with $\alpha = 4$ and $\beta = 3$.
\item{A)} (5) What is the mean survival time of a randomly selected 
exposed animal?
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\item{} {\it Solution:} $E(Y) = \alpha\beta = 12$.  For future
reference, note that $V(Y) = \alpha\beta^2 = 36$.
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\item{B)} (10) Use Tchebysheff's Theorem to estimate
the probability that a single exposed animal will 
survive longer than 30 weeks.
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\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
$P(Y > 30)$.  This means that we take $\mu + k\sigma = 30$, so 
$12 + 6 k  = 30$, or $k = 3$.  This gives
$P(Y > 30) < {1\over 3^2} = .111$.  
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\item{C)} (15) If 6 exposed animals are chosen at random, use
your answer from part B to estimate the probability
no more than 2 of them survive longer than 30 weeks.
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\item{} {\it Solution:}  The number $N$ that survive is a
binomial random variable, so 
$$P(N \le 2) = \sum_{n=0}^2 {6\choose n} (.111)^n (.889)^{6-n}$$
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\item{D)} (15) The cost of treating an exposed animal is 
$C = 4 + .25 Y^2$.  Find the mean and variance of $C$.
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\item{} {\it Solution:} First 
$$E(C) = 4 + (.25)E(Y^2) = 4 + (.25)(V(Y) + (E(Y))^2) = 4 + (.25)(36+144) = 49.$$
Then to compute the variance of the cost:
$$\eqalign{
E(C^2) &= E(16 + 2Y^2 + {1\over 16}Y^4)\cr
&= 16 + 2E(Y^2) + {1\over 16} E(Y^4)\cr
&= 16 + 2(180) + {1\over 16 \cdot \Gamma(4)\cdot 81} \int_0^\infty y^7 e^{-y/3}\ dy\right)\cr
&= 376 + {1\over 96 \cdot 81} \cdot 3^8 \Gamma(8)\cr
&=4628.5\cr}$$
Then $V(C) = E(C^2) - (E(C))^2 = 4628.5 - 49^2 \doteq 2227.5$.
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\noindent
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\noindent
{\it Extra Credit} (10) 
Let $Z$ be a standard normal random variable.  What is the 
moment-generating function for $Y = 3Z+5$?  What is the 
distribution of $Y$?
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\item{} {\it Solution:}  
$$m_Y(t) = e^{5t} m_Z(3t) = e^{5t + {9 t^2\over 2}}.$$
By the Uniqueness Theorem, $Y$ has a normal distribution too, with 
$\mu = 5$ and $\sigma = 3$.
\bye
