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\textbf{College of the Holy Cross, Spring Semester 2019} \\
\textbf{MONT 105N -- Analyzing Environmental Data} \\
\textbf{Solutions for Midterm Exam, March 25}

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\noindent
I. Both parts of this question deal with the data set:
$$\begin{matrix}
6 & 5 & 11 & 7 & 5 & 10 & 16 & 10 & 9
\end{matrix}$$
\begin{enumerate}
\item[A)]  (15) Find the 5-number summary, and draw the ``box and whisker plot.''
(Use the inclusive method for the quartiles.)
\item[] {\it Solution}  In order, the numbers are
$$5 \le 5 < 6 < 7 < 9 < 10 \le 10 < 11 < 16$$
We have ${\rm min} = 5$, ${\rm max} = 16$, and $Q_1 = 6$, ${\rm median} = 9$, $Q_3 = 10$.
The box extends from $6$ to $10$ with a vertical bar at $9$.  The whiskers extend 
to $5$ on the left and $16$ on the right 
\item[B)]  (15)  Given that the SD of this data set is $S \doteq 3.52$, how many data values
are within two SD's of the mean?
\item[] {\it Solution:}  We need to compute 
$$\overline{x} = \frac{5 + 5 + 6 + 7 + 9 + 10 + 10 + 11 + 16}{9} = \frac{79}{9} \doteq 8.78$$
We have $\overline{x} - 2S = 1.72$ and $\overline{x} + 2S \doteq 15.84$.  The interval
between these two endpoints contains {\it eight out of nine}, or about $88.9\%$ of the data values.  This is consistent
with Chebyshev's Theorem, which says {\it at least} $\left(1 - \frac{1}{2^2}\right) \times 100\% = 75\%$
of the data values must lie within 2 SD's of the mean.  
\end{enumerate}
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\noindent
II.  A weighted (biased) coin is tossed repeatedly. On any one toss, the probability
of a head is $p = .44$.
\begin{enumerate}    
\item[A)] (10) What is the probability of getting
$2$ heads and $8$ tails in $n = 10$ tosses?
\item[] {\it Solution:}  Using the binomial p.m.f., if $X$ is the number out of the 
10 tosses on which a head comes up,
$$P(X = 2) = \binom{10}{2}(.44)^2(.56)^8 \doteq .084$$
or about an $8.4\%$ chance.  
\item[B)] (10)  Is your answer in A the same as the probability of 
drawing 2 red and 8 black balls on 10 draws from an urn containing a total of 
$44$ red and $56$ black balls, if the draws are done {\it without replacement}?  
Explain.  
\item[] {\it Solution:}  No, it is not the same.  If the balls are drawn without
replacement, the draws are not independent and the binomial p.m.f. formula does
not apply.  (We need the hypergeometric formula then:
$$P(X = 2) = \frac{\binom{44}{2}\cdot \binom{56}{8}}{\binom{100}{10}} \doteq .0776$$ 
\item[C)] (5)  {\it Given that the first 3 tosses come up heads}, 
what is the probability that the first tails comes up on the 7th toss?  
(Note:  This is a conditional probability.)  
\item[] {\it Solution}  We want $\frac{(.44)^3(.56)^3(.44)}{(.44)^3} = (.56)^3(.44) \doteq .0773$  
\end{enumerate}
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\noindent
III.   Males over 20 years old in the U.S. at present have an average
height of $178$ cm (about 5' 10''), with $SD = 10$ cm.
Their average weight is $76$ kg and with $SD = 12$ kg.  
\begin{enumerate}
\item[A)] (10)  If a male over 20 years old is selected at random, 
what is the probability that he has height between $178$ and $190$ cm (between about
5' 10'' and about 6' 3'')?
\item[] {\it Solution:}  We standardize using the mean and SD for the height and use the normal table:
$$P(178 < X < 190) = P\left(0 < Z < \frac{190-178}{10}\right) = P(0 < Z < 1.2) = .3849$$
(about $38.5\%$ chance).  
\item[B)] (10)  What is the probability that a male over 20 years old, selected at random, 
has a weight between $70$ and $90$ kg (between about 154.5 lb and about 198 lb)?
\item[] {\it Solution:}  Now we standardize using the mean and SD for the weight:
$$P(70 < X < 90) = P\left({70 - 76}{12} < Z <  \frac{90 - 76}{12}\right) \doteq P(-.5 < Z < .86)$$
By the symmetry of the normal curve, this equals
$$P(0 < Z < .5) + P(0 < Z < .86) = .1915 + .3051 = .4966$$    
\item[C)]  (5) (True/False and Explain): It would be reasonable to compute the
probability that a male 20 years old, selected at random has a height between $178$ and $190$ cm
{\it and} a weight between $70$ and $90$ kg by {\it multiplying your answers} from A and B. 
\item[] {\it Solution:}  False: It would not be reasonable because doing that is 
assuming that the height and the weight are {\it independent}, whereas this is 
clearly not true.  Larger heights are {\it correlated} with larger weights in the 
sense that taller bodies also tend to have higher weights.    
\end{enumerate} 
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\noindent
{\it Essay} (20)
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One very common everyday situation where we are given information phrased
in probabilistic terms is when weather forecasts for a given geographical
area include an estimated {\it probability of precipitation} for some
time period in the future.   But what does one of those probabilities mean, really?  What is the 
exact formula used to calculate a probability of precipitation?  Explain by 
indicating exactly what it means, for instance, if a weather forecast says 
there is an 80\% chance of rain in Worcester tomorrow.  Can it mean more than 
one thing?  What does the National Oceanographic and Atmospheric 
Administration (NOAA) say about this and what is the justification they give 
for doing it this way?  Do you think just the number gives an adequate indication 
to an average person?  Would it help to educate people better about how these 
probabilities are actually computed?  Is there a better way to indicate how likely 
certain weather conditions might be?  
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\noindent
{\it Model Answer:}  The probability of precipitation, PoP, is defined as a product:
$${\rm PoP} = C \times A,$$ 
where $C = $ the confidence or estimated probability that precipitation will occur 
{\it somewhere} in the forecast area, and where 
$A = $ the percent of the area that will receive measureable precipitation, 
if it occurs at all.  For instance if the forecast said there was an 80\% chance
of rain in Worcester tomorrow, that could mean:
\begin{itemize}
\item The forecaster estimated $C = .8$ or was $80\%$ certain that rain would occur, and if it does,
the rain will happen over $A = 1$ or $100\%$ of the city, or  
\item The forecaster estimated $C = 1$ or was $100\%$ certain that rain would occur, but if 
it does, it would only rain over $A = .8\%$ or $80\%$ of the city, or
\item (Among infinitely many other possibilities), the forecaster estimated
$C = .9$ or was $90\%$ certain rain would occur and if it does, 
it would cover about $A \doteq .889$ or about $89\%$ of the area.
\end{itemize}  
The justification offered for doing it this way is that forecasts are usually made for
rather large geographical areas and weather can vary a lot from point to point within 
these areas.   The forecaster is expressing a combination of degree of confidence 
in the assertion that the given precipitation event will happen, and an estimate of the
area it will cover if it does.   (Personal opinion):  Unless they are specifically told how this works,
though, I really doubt that most people understand this, and many (``non-math-people'')
might not understand it even if they are told(!)  
It might be better (i.e. more informative and clearer) to report the $C$ and $A$
values separately rather than multiplying them!
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