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\begin{center}

\textbf{College of the Holy Cross, Spring 2018} \\
\textbf{Math 134 Midterm Exam 4 Solutions} \\
\textbf{Friday, May 4}
\end{center}

\vskip 10pt
\noindent
I. The time $t$ (in seconds) between two successive incoming calls at a 
very busy consumer service call center is a random variable with pdf
$$f(t) = \begin{cases}
         5 e^{-0.2 t} & {\rm if}\ t \ge 0\\
              0 & {\rm if}\ t < 0
\end{cases}
$$
{\it Notes:}  
\begin{enumerate}
\item[1.] This was a \emph{defective problem}.  The given function is \emph{not} a 
valid pdf because the factor of $5$ should be $\frac{1}{5}$ instead.  This was 
entirely my fault and I gave people full credit on B for the correct value of the 
integral with the 5 rather than the $\frac{1}{5}$.  However that was $20.47$, which of course
is not a possible value for a probability -- probabilities have to be between
0 and 1.  
\item[2.] In addition, only a few people (two or three) made any real progress on part
C, so I threw that part out and gave people Extra Credit if they did that correctly
or close to correctly given the information you had.
\end{enumerate}
The solutions below use the correct form of the pdf I wanted:
$$f(t) = \begin{cases}
         \frac{1}{5} e^{-0.2 t} & {\rm if}\ t \ge 0\\
              0 & {\rm if}\ t < 0
\end{cases}
$$

\begin{enumerate}
\item[A.] (5) Circle the integral that represents the probability that the time
between successive calls is \emph{at least 1} second:
\item[] {\it Solution:}  The correct integral is:
$$\int_1^\infty \frac{1}{5} e^{-0.2 t}\ dt$$ 
\item[B.] (10) Evaluate the integral you circled in part A.  
\item[] {\it Solution:}  The value is:
\begin{align*}
\int_1^\infty \frac{1}{5} e^{-0.2 t}\ dt &= \lim_{b\to\infty} \int_1^b \frac{1}{5} e^{-0.2 t}\ dt\\
&= \lim_{b\to\infty} \left. -e^{-0.2 t}\right|_1^b\\
&= \lim_{b\to\infty} -e^{-0.2 b} + e^{-0.2}\\
&= 0 + .8187\\
\end{align*}
That is, this happens about $82\%$ of the time.  
\item[C.] (10) The median time between calls is the time $m$ for which 
$P(t \le m) = P(t \ge m) = 1/2$.  Determine the median time between calls at this center.
\item[] {\it Solution:}  
We can determine the median time by setting up the equation
$$0.5 = P(t < m) = \int_0^m \frac{1}{5} e^{-0.2 t}\ dt,$$
integrating, then solving for $m$:
\begin{align*}
0.5 &= \int_0^m \frac{1}{5} e^{-0.2 t}\ dt\\
&= \left. -e^{-0.2 t}\right|_0^m\\
&= 1 - e^{-0.2 m}.
\end{align*}
Hence $e^{-0.2 m} = 0.5$, so $m = \frac{\ln(0.5)}{-0.2} \doteq 3.466$
(seconds).  
\end{enumerate}
\vskip 10pt
\noindent
II.  Use the attached table of normal curve areas to answer the following questions. 
\begin{enumerate}
\item[A.]  (10) If $Z$ has a standard normal distribution, what is $P(-1.00 < Z < 1.34)$?
\item[] {\it Solution:}  By the symmetry of the normal pdf about 0, this is equal to 
$$P(0 < Z < 1) + P(0 < Z < 1.34) = .3413 + .4099 = .7512$$
\item[B.]  (10) Bottles of cream soda have a stated volume of 20 fluid ounces, but 
the actual volume of the contents is normally distributed with mean 20.2 fluid oz.  
and SD $= .31$ fluid oz.  What is the probability that a randomly selected bottle 
contains more than 20 fluid oz. of soda?
\item[] {\it Solution:}  We standardize and use the table.  This is 
$$P(Z > \frac{20 - 20.2}{.31}) \doteq P(Z > -.65) = P(0 < Z < +.65) + .5000 = .2422 + .5000 = .7422$$
\end{enumerate}
\vskip 10pt
\noindent
III.  (Differential Equations)
\begin{enumerate}
\item[A.] (20)  Find the general solution of the differential equation 
$$\frac{dy}{dx} = (1 + y^2)x^3\ln(x)$$ 
\item[] {\it Solution:}  We separate variables and integrate like this:
\begin{align*}
\frac{1}{1 + y^2}\ dy &= x^3 \ln(x)\ dx\\
\int \frac{1}{1 + y^2}\ dy &= \int x^3 \ln(x)\ dx\qquad {\rm (use\ parts\ with\ } u = \ln(x)\rm{)}\\
\tan^{-1}(y) &= \frac{x^4}{4}\ln(x) - \frac{x^4}{16} + C\\
y &= \tan\left(\frac{x^4}{4}\ln(x) - \frac{x^4}{16} + C\right)
\end{align*}
\item[B.] (5)  Newton's Law of Heating/Cooling says that if an object is placed into 
a surrounding medium of constant temperature $A$, then the rate of change of the object's
temperature is proportional to the difference between the object's temperature and the 
surrounding temperature.  Circle the differential equation that corresponds to this 
statement:
\item[] {\it Solution:}  The correct differential equation is:
$$\frac{dy}{dt} = k(y - A)$$
\item[C.]  (15) A roasted chicken is taken out of the oven at a temperature of 
$190^\circ C$ and left to cool on a counter in a room maintained at a constant temperature
of $20^\circ$ C. 
After 10 minutes the temperature of the chicken has decreased to $100^\circ$ C. 
How long will it take for the temperature to reach $45^\circ$ C? 
\item[] {\it Solution:}  The general solution is 
$$y = A + c e^{kt} = 20 + c e^{kt}.$$
Taking $t = 0$ to correspond to the time when the chicken comes out of the oven,
$y(0) = 190$, so $190 = 20 + c$ and $c = 170$.  Then 
$y = 20 + 170 e^{kt}$.  At $t = 10$ minutes, the temperature of the chicken is 
$100$, so $100 = 20 + 170 e^{10k}$, so $k = \frac{\ln(8/17)}{10} \doteq -0.0754$.
Then the chicken reaches 45 when
$45 = 20 + 170e^{-0.0754 t}$ so 
$$t = \frac{\ln(25/170)}{-0.0754} \doteq 25.43$$
(minutes).  
\end{enumerate}
IV.  
\begin{enumerate}
\item[A.]  (5)  What is the future value of $\$100,000$ at an interest rate of $5\%$, compounded
monthly for $15$ years?
\item[] {\it Solution:}  This is 
$$\$100,000 \left(1 + \frac{.05}{12}\right)^{15\cdot 12} \doteq \$211,370.41$$
\item[B.]  (5)  Suppose you make monthly payments of $ \$ 700$ for 15 years at $5\%$, what
is the future value of that stream of payments?
\item[] {\it Solution:}  By the geometric series formula, the future value is 
$$\$700\cdot \frac{\left(\left(1 + \frac{.05}{12}\right)^{15\cdot 12} - 1\right)}{\frac{.05}{12}} 
\doteq \$187,122.28$$
\item[C.]  (5)  Suppose you were making those payments to a bank in return for a loan 
of $\$100,000$.  At the end of the 15 years, would you still owe the bank, or would 
you have overpaid? 
\item[] {\it Solution:}  You would still owe the bank the difference of the amounts 
from parts A and B:
$$\$211,370.41 - \$187,122.28 = \$24268.12.$$
(A fully-amortized loan would be set up with a monthly payment greater than $\$700$ to make
the future values from parts A and B \emph{equal}.)

\end{enumerate}
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\MyAnswer{Integral $ = $}{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}%
\end{enumerate}
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