
\magnification=\magstep1
\centerline{MONT 105N -- Analyzing Environmental Data}
\centerline{Answers for Midterm Exam Review Questions}
\centerline{March 13, 2020}
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\noindent
\item{1)}  A random sample of 1000 college first-year students in the US is selected and their
ages are determined.  
\itemitem{a)}  What would you expect was the sample mean age, to the nearest whole number?  
\itemitem{} {\it Solution:}  The sample mean is probably somewhere between 17 and 18, and 
I would suspect it's closer to 18.  Note:  This takes into account ``nontraditional students'' 
too because they form a relatively small fraction of the total college first-year population.
\itemitem{b)}  Give the formula that would be used to compute the sample SD, calling the 
ages $x_1,\ldots,x_{1000}$.
\itemitem{} {\it Solution:}  Letting the sample mean be $\overline{x}$, the formula is
$$\eqalign{S &= \sqrt{{1 \over 999}\sum_{i = 1}^{1000}(x_i - \overline{x})^2}\cr
&=\sqrt{{1\over 999}\left((x_1 - \overline{x})^2 + (x_2 - \overline{x})^2 + \cdots + (x_{1000} - \overline{x})^2\right)}
}$$
\itemitem{c)} Would you expect the sample SD to be about $0.1$ year, $1$ year, or $10$ years?  
Explain.
\itemitem{} {\it Solution:}  It makes no sense for the SD to be as big as 10 years since that
would leave the possibililty of a significant number of students being 8 years old or less.
I would accept either $0.1$ year or $1$ year as a reasonable estimate.  Almost all first year 
students would be 17, 18, or 19 years old (at least at liberal arts colleges).   But I would expect the SD
is closer to $0.1$, since 18 would be the most common age.
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\item{2)}
Find the 5-number summary, and draw the ``box and whisker plot'' for the data set 
$$\matrix{5 & 3 & 7 & 8 & 8 & 10 & 1 & 2 & 4\cr}$$
(Use the inclusive method for the quartiles.)
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\noindent
{\it Solution:}  Rearranging the numbers to non-decreasing order we have
$$1 < 2 < 3 < 4 < 5 < 7 < 8 < 8 < 10$$
The min is $1$, the max is $10$, the median is $5$.  The first quartile is the median
of the first half of the list (including the median):  
$$1 < 2 < 3 < 4 < 5$$
so $Q_1 = 3$.  Similarly, $Q_3 = 8$.  The box extends from 3 to 8 with a vertical
bar at 5.  ``The whiskers'' extend to $1$ on the left and to $10$ on the right.
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\item{3)} Eight birds -- two robins, two woodpeckers, two thrushes, and two blackbirds
are placed in a covered cage. Three random
draws of a single bird are made from the cage {\it with replacement}.
\itemitem{a)} Are the three draws {\it independent} in this case?  Why or why not?
\itemitem{} {\it Solution:}  Yes, since the draws are done with replacement.  The 
probabilities for each of the birds on each of the draws are the same, and knowing
how the last draw turned out does not change those probabilities.
\itemitem{b)} What are the chances that all three birds are robins?
\itemitem{} {\it Solution:}  Since $2/8 = 1/4$ of the birds are robins and the 
draws are independent, this is $(1/4)(1/4)(1/4) = 1/64$.
\itemitem{c)} What are the chances that none of the birds is a robin?
\itemitem{} {\it Solution:}  We have a $6/8 = 3/4$ chance of picking a robin each
time so the probability is $(3/4)(3/4)(3/4) = (27/64)$.                                    
\itemitem{d)} What are the chances that there is a something other than a robin among the three
birds?
\itemitem{} {\it Solution:}  This is the complementary probability to the one computed
in part (a): It equals $1 - (1/64) = 63/64$.
\itemitem{e)} (More challenging)  What would change in your answers to a,b,c,d if the
draws were done {\it without replacement}?
\itemitem{} {\it Solution:} (a) No, the probabilities change from one draw to the 
next depending on the outcomes of the previous draws.  So without replacement the 
draws are {\it not independent}.  (b) Since there are only $2$ robins to start, there is no
way to pick 3 robins if the draws are done without replacement.  The probability is zero.
(c) This is $(6/8)(5/7)(4/6) = 5/14$.  Explanation: You have a $(6/8)$ chance of picking a 
non-robin the first time.  After that, there are 5 non-robins and 7 birds left, so on the second
draw, you have a $5/7$ chance of drawing a non-robin.  Then after those two draws there are
$4$ non-robins and $6$ birds left, so the chance of picking a non-robin on the last draw is 5/14.
Note that the factors after the first one here are technically {\it conditional probabilities}.
For instance the $5/7 = P(non-robin\ on\ second\ |\ non-robin\ on\ first)$ and the $4/6$ is
$P(non-robin\ on\ third\ |\ non-robins\ on\ first\ and\ second)$
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\item{4)}  A  coin is tossed repeatedly.  
\itemitem{a)} What is the chance of getting
$7$ heads and $3$ tails in ten tosses if the heads and tails are 
equally likely?
\itemitem{} {\it Solution:}  We should assume the tosses are independent, so 
this is a binomial situation.  The probability is 
$${10 \choose 7} (1/2)^7(1/2)^3 = {120\over 1024} = {15\over 128} \doteq .117$$
(that is, about a $12\%$ chance).  
\itemitem{b)} {\it Given that the first 3 tosses come up heads}, 
what is the chance that the first tails comes up on the 7th toss if
heads and tails are equally likely?  (Note:  This is a conditional probability.)
\itemitem{} {\it Solution:}  The chance that the first three tosses come up heads are 
$$(1/2)(1/2)(1/2) = (1/2)^3 = 1/8$$  
The chance that the first three tosses come up heads {\it and} then 
three more tosses come up heads before the first tails are $(1/2)^6\cdot (1/2) = 1/128$. 
Hence the conditional probability we are asking for is 
$${(1/2)^7\over (1/2)^3} = (1/2)^4 = 1/16.$$
\itemitem{c)} How do your answers to (a) and (b) change if you know that on each
toss, the probability of a head is $.6$ and the probability of a tail is $.4$ (that is 
the coin is weighted)?
\itemitem{} {\it Solution:}  In (a), the probability changes to 
$${10 \choose 7} (.6)^7 (.4)^3 \doteq .215$$
In (b) the probability changes to 
$${(.6)^6\cdot (.4)\over (.6)^3} = (.6)^3\cdot (.4) = .0864$$
In both (b) and here, note that the form is ${q^6 p\over q^3} = q^3 p$, where $p$ is the probability
of a tail and $q$ is the probability of a head.  In other words, this is the same as ignoring the 
first three tosses, starting from the fourth, and asking what is the probability that the first
tails comes on the fourth toss after the first three.   
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\item{5)}  Assume that everyone starts school at age 5 and stays in school at least until age 16.
True or False and explain: 
\itemitem{a)}  The median years in school would be greater than 11.
\itemitem{} {\it Solution:}  True:  the number of years in school would be at least 11 for 
everyone, so the median years in school would have to be larger.  Recall the median 
is the number such that half the people have that number of years in school, and half
the people have that number of years in school or more.
\itemitem{b)}  The average years in school would be greater than the median. 
\itemitem{} {\it Solution:}  Almost certainly true.  The median is probably somewhere
around 13 or 14 (that is, finished high school and some college).  But there are
some people who finish college and get one or more graduate degrees, ending up 
with $12 + 4 + 5 = 21$ years in school or more.  The presence of those people
in the average would tend to pull the average up larger than the median.  (This is 
positively skewed distribution.)
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\item{6)}  One ticket is drawn at random from each of the two boxes:
(i)  $1\ 2\ 3\ 4\ 5\ $ and  (ii)  $1\ 2\ 3\ 4\ 5\ 6\ $. 
The tickets in the first box are blue and those in the second box are green, so you 
can tell which box the ticket came from.  
\itemitem{a)}  What is the probability that one of the numbers is $2$ and the other is $5$?
\itemitem{} {\it Solution:}  There are $5 \cdot 6 = 30$ pairs $(m,n)$ representing
the draws from the two boxes, with $1\le m \le 5$ and $1 \le n \le 6$.  ``At random'' should
mean that all of those are equally likely.  There are two pairs as stated: $(2,5)$ and $(5,2)$
so the probability is $2/30 = 1/15 \doteq .067$.
\itemitem{b)}  What is the probability that the sum of the numbers is $7$?
\itemitem{} {\it Solution:}  That can happen in these 5 ways: 
$(1,6), (2,5), (3,4), (4,3), (5,2)$.  So the probability is $5/30 = 1/6 \doteq .167$.  
\itemitem{c)}  What is the probability that one numbers is (strictly) 
bigger than twice the other?  (Hint: List all the cases where that is true and count!)
\itemitem{} {\it Solution:}  This happens for (and only for) 
$$(1,3),(1,4),(1,5),(1,6),(2,5),(2,6),(3,1),(4,1),(5,1),(5,2),(6,1),(6,2)$$
(I listed these by thinking of all possible values of the first entry of the 
pair--the draw from the first box, then seeing which draws from the second
box would satisfy the condition that one number was strictly more than twice the other.)
The probability is $12/30 = 2/5 = .4$.  
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\item{7)}  Body lengths of a population of immature lobsters are {\it uniformly distributed}
on the interval 15 to 22 cm.  
\itemitem{a)}  If a single lobster is selected at random, what is the probability 
that its body length is between 19 and 20 cm?
\itemitem{} {\it Solution:}  The corresponding uniform pdf has value $1/7$ for lengths
between 15 and 22 inclusive, but zero for all other lengths.  The probability
is 
$${20 - 19\over 22 - 15} = {1\over 7}$$
(This is the area under the graph of the pdf for $19 \le x \le 20$.
\itemitem{b)}  If 6 lobsters are selected with replacement from the population, 
what is the probability that 4 of them have body lengths between 19 and 20?
\itemitem{} {\it Solution:}  Binomial:
$${6\choose 4} (1/7)^4 (6/7)^2 \doteq .0046$$
(about $.5 \%$, so very unlikely).  
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\item{8)}  In a large class, the average score on a test was 75 out of 100 and the SD
was 10.  The scores followed a normal distribution.
\itemitem{a)}  A single test paper is selected randomly out of the pile.  What is the probability
that the score was between 75 and 85?
\itemitem{b)}  Using the standard normal table and standardizing as we discussed in class, 
$$P(75 < X < 85) = P\left({75 - 75\over 10} < Z < {85 - 75\over 10}\right) = P(0 < Z < 1) = .3413.$$
\itemitem{b)}  Same question as a) but the score is between 60 and 80?
\itemitem{} {\it Solution:}  Same method:
$$\eqalign{
P(60 < X < 80) &= P\left({60 - 75\over 10} < Z < {80 - 75\over 10}\right)\cr
&= P(-1.5 < Z < .5)\cr
\&= .4332 + .1915 = .6247}$$
\itemitem{c)}  What was the 80th percentile score (the score such that 80\% had scores
less than or equal to that, and only 20\% had scores greater than or equal to that)?
Round to the nearest whole number.  
\itemitem{} {\it Solution:}  We are looking for the score $c$ such that 
$$P(X \le c) = P\left(Z \le {c - 75\over 10}\right) = .80$$  
From the standard normal table the closest $z$ to having $P(Z < c) = .8$ comes from $c = .84$ 
(note we are including the $.5$ for negative values of $Z$ too, so we are looking for an area 
equal to $.8 - .5 = .3$ and $z = .84$ gives an area $.2995$ which is close enough to $.3$ for our
purposes.)  Then ${c - 75\over 10} = .84$ implies $c = 75 + 8.4 = 83.4$, or $83$ after rounding to the nearest 
whole number.   
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\item{9)} George is playing roulette (where, in the US version, there are 18 red numbers,
18 greens, and 2 additional slots, for a total of 38 possible ways for the ball to 
land).  Assume the roulette wheel is 
not biased or ``loaded'' so that each of the 38 outcomes is equally likely.  
George starts with $\$100$.  For his first 50 spins of the wheel, he bets \$1 on
         17, his favorite number (he can win if either 17 red or 17 green comes up). 
Unfortunately, he loses each time, losing \$50 in the
         process. Ever the optimist, George decides his luck must change on the next spin
         of the wheel because 17 is ``due'' (it hasn't come up 50 spins in a row, after all).
So he bets his remaining \$50 on 17 one more time.  Donald's significant other Laura, who
         always brings George back to earth, says he is a fool and that he's
         likely to lose again.
         Is George right, is Laura right, or are neither of them right? Explain your
         answer.
\item{} {\it Solution:}  George has fallen prey to the ``gambler's fallacy.''  Even
though 50 consecutive non-17's have come up, the probability of a 17 on 
the next roll is still the same as it is on every roll:  $2/38 = 1/19$.
There is no such thing as ``being due for a 17'' in this setting because 
the rolls are independent.
  
\bye
