\magnification=\magstep1
\centerline{MONT 105Q -- Mathematical Journeys}
\centerline{Solutions for Practice Problems for Midterm Exam}
\centerline{March 17, 2016}
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\noindent
{\it Sample Mathematical Questions}
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\noindent
{\it Note:}  There are somewhat more questions and more 
parts to these questions than I could reasonably ask you to do in the 
actual exam (because of the limited time available).  This should
give you an idea of the range of possible topics and styles of questions I might 
include in the exam, though.  
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\noindent
A.  The following data set has $n = 9$:
$23,28,40,44,47,50,51,54,55$
\medskip
\item{1)}  Find the ``5-number'' summary for this
data set (min, 1st quartile, median, 3rd quartile, max).
\item{} {\it Solution:}  The numbers are already in increasing order 
so we can see min $= 23$, the median is the ``middle'' element $ = 47$, the
max $= 55$ and then (according to the convention we used in class), $Q_1 = {28 + 40\over 2} = 34$
(the median of the 4 data items strictly below the overall median) and 
$Q_3 = {51 + 54\over 2} = 52.5$ (from the 4 data items strictly above the median).
\item{2)}  Draw the corresponding ``box and whisker'' plot.
\item{}  {\it Solution:}  Should show a ``box'' extending from $Q_1 = 34$ to $Q_3 = 52.5$, 
with a vertical line at $47$, plus ``whiskers'' extending from $23$ to $34$ and
$52.5$ to $55$.
\item{3)}  Construct a frequency histogram for the data using 4 equal bins
on the range 20 to 60.  Explain how you are treating data that falls at 
a bin boundary.  (Any consistent method is OK.)
\item{} {\it Solution:}  Say we decide to place data that falls at a bin
boundary into the higher (right) bin each time.  Then the four bins yield counts:
$$\eqalign{[20,30) &: 2\cr 
           [30,40) &:  0\cr
           [40,50) &:  3\cr
           [50,60] &:  4\cr}$$
The histogram shows 4 boxes, the first of height 2, the second of height 0, the 
third of height 3, the last of height 4.
\item{4)}  One measure of ``skewness'' of a data set is the quantity
$$B = {Q_3 - 2*{\rm median} + Q_1\over Q_3 - Q_1}.$$ 
Compute that statistic for this data set.  What does it seem to indicate?
Does this seem reasonable from the box plot?
\item{} {\it Solution:}  We get
$$B = {52.5 - 2*47 + 34\over 52.5 - 34} = -0.41$$
This indicates a slight skew to the left (i.e. toward lower values).  That should
seem reasonable since the median is farther from $Q_1$ than it is from $Q_3$, and 
the left ``whisker'' is also quite a bit longer.
\item{5)}  Compute the SD of the data set.  How many of the 
data values lie within two SD's of the mean?  
\item{} {\it Solution:}  To compute the SD we also need the mean:
$$\overline{x} = {23 + 28 + 40 + 44 + 47 + 50 + 51 + 54 + 55\over 9} = 43.\overline{5} 
\doteq 43.6$$
(rounding).  
(Note that this also confirms what we were saying in part 4 -- the mean is 
smaller than the median which indicates a skew to the lower values.)  
Then from the basic formula 
$$SD \doteq \sqrt{{(23 - 43.6)^2 + (28 - 43.6)^2 + \cdots + (55-43.6)^2\over 9}} \doteq 
10.7$$
(rounded again).  This could also be computed by the equivalent, more efficient, formula we 
mentioned in class:
$$SD = \sqrt{ {23^2 + 28^2 + \cdots + 55^2\over 9} - (43.\overline{5})^2} \doteq 10.7$$
(Practical note:  If you use the rounded value $43.6$ here, you'll get
something a bit smaller than the $10.7$ -- a rounding artifact.)  The numbers within
two standard deviations of the mean are (roughly)
$43.6 - 2(10.7) \le x \le 43.6 + 2(10.7)$, or $22.2 \le x \le 65$.  {\it All nine} of the
numbers in the data set satisfy those inequalities.
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\noindent
B. The aces, kings, and queens from a standard deck of cards are removed and 
placed in a stack of twelve cards by themselves (there are three hearts, 
three spades, three diamonds and three clubs).
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\item{1)} If a single card is selected at random from the stack, what is the 
probability that it is a heart?  
\item{} {\it Answer:}  ${3\over 12} = {1\over 4}$. 
\item{2)}  If a single card is selected at from from the stack, what is the
conditional probability that the card is a queen, given that it is a spade?
\item{} {\it Answer:}  ${1\over 3} = {{1\over 12}\over {3\over 12}}$.
\item{3)}  If a single card is selected from the stack, consider the two 
events:  (1) the card is an ace, and (2) the card is a diamond.  Are these
independent?
\item{} {\it Answer:}  Yes.  $P({\rm ace}) = {4\over 12} = {1\over 3} = 
P({\rm ace}|{\rm diamond})$.
\item{3')} Same question as 3) but on the stack of 11 cards obtained
by removing the queen of diamonds. 
\item{} {\it Answer:} No.  $P({\rm ace}) = {4\over 11} \ne P({\rm ace}|{\rm diamond}) = {1\over 2}$.
\item{4)} Three random draws {\it with replacement} are made from this 
stack of cards. What is the probability that all three cards are spades?
\item{} {\it Answer}:  With replacement, there is a ${3\over 12}$ chance of 
getting a spade on each draw.  So 
the probability of getting three spades is 
${3\over 12}\cdot {3\over 12} \cdot {3\over 12} = {1\over 64}$.
\item{5)}  Three random draws {\it without replacement} are made from 
the whole stack of 12 cards.  What is the probability that all three are clubs? 
\item{} {\it Answer}: ${3\over 12}\cdot {2\over 11}\cdot {1\over 10}$. 
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\noindent
C.  (Short answer) Suppose that a researcher collects 80 individuals of
the Atlantic surf clam.  These clams can be found at levels down to about a 
meter in the sand, and larger clams tend to live at deeper levels.  
The researcher finds an average shell width $10.2$ cm.  Think of this as a 
sampling process.
\item{1)}  What is the population?  What is the sample?
\item{}  {\it Answer:} The population might be all clams of this genus (or possibly
all the clams of this genus at this particular location, depending on what you
might want to be studying).  The sample is the 80 particular clams collected.   
\item{2)}  Is the $10.2$ a value derived from particular data (a ``statistic'')
or is it a property of the population as a whole (a ``parameter'')?
\item{} {\it Answer:}  It's a statistic since it's computed from the shell widths
of those 80 individual clams.
\item{3)}  Would the researcher {\it know} the population mean in
this circumstance?
\item{} {\it Answer:}  Most probably not.  The idea of doing the 
collection and measurements might be to estimate the population mean.
\item{4)}  What additional information would the researcher need in order
to understand the distribution of shell widths?  Describe
how that would be estimated and how that would be interpreted.
\item{} {\it Answer:}  In addition to the sample mean, the other big
statistics to look at would be the sample SD, or other measures of the
distribution such as the ``five number summary'' as in question A above.
\item{5)}  What could we say about the distribution of 
sample means of samples of size 80 (even if the distribution of the individual
shell widths was unknown)? 
\item{} {\it Answer:} With samples that large, the Central Limit Theorem
would ``kick in'' and say that the sample means would have an approximately
normal distribution with $\mu = $ population mean and SD $= {\sigma\over \sqrt{80}}$,
where $\sigma = $ population standard deviation.
\item{6)}   If you knew that reasearcher was being lazy about digging 
and the clams he collected were all taken from sand levels no deeper than
$10$cm, would that be a {\it simple random sample}?  What kind of bias
might this process introduce? 
\item{} {\it Answer:}  Not a simple random sample, because of a selection bias
including fewer larger clams.
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\noindent
D.  Suppose that a large data set of air temperature readings is normally 
distributed with  $\mu = 18.6^\circ$C and $\sigma = .2^\circ$ C.
The ``$z$-score'' of a temperature reading $x$ is computed by
$$z = {x - \mu \over \sigma}$$
\item{1)}  What would be the $z$-score of a reading of $x = 17.9^\circ$?
\item{} {\it Answer:}  
$$z = {17.9 - 18.6\over .2} = -3.5$$
(i.e. this $x$ value is $3.5$ standard deviations below the population mean.
It would be very unusual to observe a value this small!)
\item{2)}  What temperature reading would correspond to a $z$-score
of $1.4$?
\item{} Doing some easy algebra, 
$$z = {x -18.6\over .2} = 1.4\quad {\rm when}\quad x = (1.4)(.2) + 18.6 = 18.88$$ 
\item{3)}  Based on this information, if a temperature reading $x$ 
is selected at random from the data set, 
what is the probability that $18.2^\circ \le x \le 18.9^\circ$?
\item{} {\it Answer}:  This corresponds to the range of $z$-scores
$${18.2 - 18.6\over .2} \le z \le {18.9 - 18.6\over .2},$$
or $-2\le z \le 1.5$.  From the standard normal table, the probability
that $z$ is in this range is 
$$.5 - P(z > 2) + .5 - P(z > 1.5) = .5 - .0228 + .5 - .0668 = .9104$$
\item{4)}  Based on this information, if a temperature reading $x$ 
is selected at random from the data set, 
what is the probability that $x \ge 18.9^\circ$?
\item{} {\it Answer}:  $.0668$
\item{5)} Based on this information, if a temperature reading $x$ 
is selected at random from the data set, 
what is the probability that $x \le 18.2^\circ$?
\item{} {\it Answer}: $.0228$
\item{6)}  Based on this information, if $n = 10$ measurements
are taken, what is the probability that exactly 5 of them are greater
than $18.8$?
\item{} {\it Answer}:  The probability that any one $x > 18.8$ is 
$$P(x > 18.8) = P\left(z > {18.8 - 18.6\over .2}\right) = P(z > 1) = .1587$$
Out of the 10 measurements (assuming independence), probability that
exactly 5 of them are $> 18.8$ is given by the binomial probability
formula:
$${10\choose 5}(.1587)^5(.8413)^5$$
(It's OK to leave the answer in this form since it's clear where it comes 
from!  If you really need a single number, this works out to about $.01$
(a 1 in 100 chance).  

\bye
