Math 207 – Solutions to
Assignment 3
3.30
a.
The scatterplot is shown below.
b.
The
relationship shown in the scatterplot is generally
positive and linear (as Brett makes more completions, he tends to gain more
total yards, which makes sense). There is one outlier (outside the overall
pattern of the rest of the data), which took place in week 10 of the season. In
this game, Brett had both a small number of completions and small number of
total yards. (Did he get hurt that game? My sports-nuts students can fill me in
on that.)
c.
From
Minitab: Pearson correlation, r, of Completions and Total Yards = 0.850
d.
From
Minitab: The regression equation is Predicted Total Yards = 10.04 + 10.86
Completions
e.
Twenty pass
completions is within the range of the collected data, so we can use the
regression line to make this prediction: Predicted Total Yards = 
yards.
4.38
The order of
the questions on the exam doesn’t matter (it’s still the same exam, no matter
the order of the questions). To determine the total number of possible exams,
we need to count the number of combinations of ten questions taken five at
time. The number of combinations is 
. In order for the student to get all five
questions right, the instructor must pick five of the six questions that she
knows, and none of the questions that she doesn’t know. The number of ways to
do this is 
. Therefore, the probability that she can
solve all the problems is 
Note we can also solve this problem using the
multiplication rule (which itself is equivalent to using counting
methods—permutations, where order matters—in numerator and denominator):
P(1st
select a question she knows 
2nd
select a question she knows 3rd
select a question she knows 4th
select a question that she knows 5th
select a question that she knows) = P(1st select a question she
knows)P(2nd select a question she knows|1st select a
question she knows)P(3rd select a question she knows|1st
select a question she knows 2nd
select a question she knows)P(4th select a question that she knows|1st
select a question she knows 2nd
select a question she knows 3rd
select a question she knows)P(5th select a question that she knows|1st
select a question she knows 2nd
select a question she knows 3rd
select a question she knows 4th select a question that she
knows) = 
4.39
The ordering
of the shapes is important. To get the total number of permutations of the
shapes, we need to count the number of permutations of 12 objects taken 12 at a
time. The number of permutations is 
. Now we need to count the number of ways the
monkey can put the objects in order, grouping by shapes. Again, order is
important. There are 
ways
to order the shapes (e.g., circles, squares, triangles, rectangles is different
from circles, squares, rectangles, triangles)—this counts the “macro-level”
ordering. Furthermore, there are 
ways
to arrange each of the shape groupings (e.g. red circle, green circle, blue
circle is different from green circle, blue circle, red circle)—this counts
each “micro-level” ordering. Using the Basic Principle of Counting, there are 
ways
for the monkey to put the objects in order, grouping by shapes. Hence, the
probability of the monkey grouping by shape is 
This
is a very small probability! Which leads me to believe that the monkey wasn’t
simply ordering at random, but was actually grouping the shapes.
Another method of counting the number of ways
to group by shape:
Consider the
12 slots and think of filling them as a 12-stage experiment. There are 12
possibilities for the first slot, 2 possibilities for the second slot (since it
must be the same shape as the first slot), 1 possibility for the third slot, 9
possibilities for the fourth slot (since it can be any of the remaining
shapes), 2 possibilities for the fifth slot (since it must be the same shape as
the fourth slot), 1 possibility for the sixth slot, 6 possibilities for the
seventh slot (since it can be any of the remaining shapes), 2 possibilities for
the eighth slot (since it must be the same shapes as the seventh slot), 1
possibility for the ninth slot, 3 possibilities for the tenth slot (since it
can be any of the remaining shapes), 2 possibilities for the eleventh slot, and
1 possibility for the twelfth spot. Hence, there are 
ways
for the monkey to put the objects in order, grouping by shapes.
Note that we can also solve this problem
using the multiplication rule:
P(any object
picked first same
shape object picked second same
shape object picked third any of
the remaining objects picked fourth same
shape object picked fifth same
shape object picked sixth any of
the remaining objects picked seventh same
shape object picked eighth same
shaped object picked ninth any of
the remaining objected picked tenth same
shape object picked eleventh same
shape object picked twelfth) = 
(This is
actually equivalent to using the alternative-method of counting I mentioned
above.)
4.125
P(waiting 5 minutes or longer) = 0.2.
a.
P(man waits less than 5 minutes) = 1 – P(man
waits 5 minutes or longer) = 1 – 0.2 = 0.8.
b.
P(man waits less than 5 minutes woman
waits less than 5 minutes) = P(man waits less than 5 minutes)P(woman
waits less than 5 minutes) = (0.8)(0.8) = 0.64, since the events are independent.
c.
P(one or the other or both waits 5 minutes or
longer) = 1 – P(neither waits 5 minutes or longer) = 1 – 0.64 = 0.36.
Note that this probability can
also be found the “longer” way: P(man
waits less than 5 minutes)P(woman waits 5 minutes or longer) + P(man
waits 5 minutes or longer)P(woman waits less than 5 minutes) + P(man
waits 5 minutes or longer)P(woman waits 5 minutes or longer) =
(0.8)(0.2) + (0.2)(0.8) + (0.2)(0.2) = 0.36. (Again, because the events are
independent.)
Or it could be found as a union probability: P(man
waits 
woman waits)
= P(man waits) + P(woman waits) – P(man
waits woman
waits) = 0.2 + 0.2 – (0.2)(0.2) = 0.36. (Again, because the events are
independent.)
Additional Problem 1
a.
The fitted
line plot is shown below. The residual plot (needed for part b) is also shown.
b.
The residual
plot from this regression is shown above. There is a pattern of curvature in
the residuals. This indicates that a line is not the best summary of the
relationship in the data—a quadratic curve should be fit instead.
c.
The
quadratic curve fit is shown in the plot below. The 
improves to 83.5% (up from 70.8% with the
linear fit). Now 83.5% of the variation in TV price is explained by the model.
d.
A screen
size of 30 inches is well outside the range of the collected data. While we
could use the regression model in part c
to make a prediction, we shouldn’t do
this, as the prediction might be very inaccurate. (It’s dangerous to
extrapolate beyond the given data.)
Additional Problem 2
The sample
space for this experiment contains 8 outcomes:
S = {HHH,
HHT, HTH, HTT, THH, THT, TTH, TTT}
Because the
coin is fair, these simple outcomes are all equally likely. Note the outcomes
of individual coin tosses are independent, but
the events A and B are not necessarily independent. We need to use the
definition of independence to answer this question.
P(A) = P(Heads on first flip) = P[{HHH, HHT, HTH, HTT}] = 
P(B) = P(Exactly two heads in three flips) = P[{HHT, HTH, THH}] = 
P(A B) = P(Heads on first flip and exactly two
heads in three flips) = P[{HHT, HTH}]
= 
Then 
Note this conditional probability is not equal to the unconditional
probability, 
. Hence,
these events are dependent. (Knowing
there’s exactly two heads increases
the chance of a head on the first flip.)
Equivalently, you could have shown
Or equivalently,
you could have shown
(You need
not show all these results—just one suffices.)