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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 1}
\centerline{September 10, 2009}
\vskip 10pt
\noindent
1.4.  (a) Using bins $[2,3),[3,4),$ etc. the histogram should
show boxes of heights
$$10/40, 12/40, 7/40, 5/40, 2/40, 3/40,$$
then three
boxes with height zero, and one final box of height $1/40$.

(b)  The proportion is $18/40 = 9/20$ or about $45\%$

(c)  The probability is the proportion $29/40$, or $72.5\%$.
\vskip 10pt
\noindent
1.8.  (a)  Using six bins $[10.1,12.1),[12.1,14.1),\ldots,[20.1,22.1)$ 
the histogram should show boxes of heights
$$7/26, 7/26, 4/26, 2/26, 5/26, 1/26.$$

(b)  The lower portion in the center (on the interval
$[16.1,18.1)$) and the two humps in the histogram
on either side of the central depression
is not like many other distributions.
One possibility is that the pottery found
at the first two sites (Llanederyn and Caldicott) has
some definitely different composition from the pottery
found at the other two sites (Island Thorns, Ashley Rails).
Some reasonable explanations:  the clay used at the two
groups of sites could have had different
chemical compositions, or there might have been some
different sort of glaze used on the pottery from
the second two sites that contained more aluminum oxide.
\vskip 10pt
\noindent
1.11.  We will show the desired equality by starting on the left.
Begin by squaring out: 
\[
(y_i - \overline{y})^2 = y_i^2 - 2y_i \overline{y} + \overline{y}^2.
\]
Then and using a, b, c
\begin{align*}
s^2 &= \frac{1}{n-1} \sum_{i=1}^n (y_i - \overline{y})^2\\
&= \frac{1}{n-1} \sum_{i=1}^n (y_i^2 - 2y_i \overline{y} + \overline{y}^2)\\
&= \frac{1}{n-1}\left(\sum_{i=1}^n y_i^2 - 2 \overline{y} \sum_{i=1}^n y_i + n \overline{y}^2\right),
\end{align*}
where we note that $\overline{y}$ is constant, so $\sum_{i=1} \overline{y} = n \overline{y}$.

Then using 
the definition of $\overline{y}$:
\begin{align*}
&=\frac{1}{n-1}\left(\sum_{i=1}^n y_i^2 - 2 n \overline{y}^2 + n \overline{y}^2\right)\\
&=\frac{1}{n-1}\left(\sum_{i=1}^n y_i^2 - n \overline{y}^2\right)\\
&=\frac{1}{n-1}\left(\sum_{i=1}^n y_i^2 - \frac{1}{n}\left(\sum_{i=1}^n y_i\right)^2\right).\\
\end{align*}
\vskip 10pt
\noindent
1.12.  Using the formula of 1.11, we have $\sum_{i=1}^6 y_i = 14$, and $\sum_{i=1}^6 y_i^2 = 40$.
So $s^2 = \frac{1}{5}(40 - \frac{1}{6} (14)^2) = 1.47$.
Then $s \doteq \sqrt{1.47} \doteq 1.21$.
\vskip 10pt
\noindent
1.15.  (a)  $\overline{y} = 4.39$, $s = 1.87$. (b) The
interval $\overline{y} \pm s$ is $(2.52,6.26)$  There are 
36 data points in this interval.  $36/40 = 90\%$ is much 
larger than one would expect by the empirical rule.  The
interval $\overline{y} \pm 2s$ is $(0.65,8.13)$.  This
contains $39/40 = 97.5\%$ of the data points -- again 
somewhat larger than we would expect.
The interval $\overline{y} \pm 3s$ is $(-1.22,10.0)$.  
This interval again contains $39/40$ of the data points.
(\emph{Note:}  This data set is somewhat unusual
because of the one very large ``outlier'' value $11.88$
at the start.  Because that number is so much larger
than the others, the value of $\overline{y}$ is 
pulled up and the the value of $s$ is larger
relative to the mean and standard deviation of the
remaining data points if the outlier is removed.
This explains the relatively poor agreement with
the empirical rule in this example.)
\vskip 10pt
\noindent
1.17.  The range of the measurements in 1.4 is $11.88 - 2.61 = 9.27$.
Since the interval $\overline{x} - 2s$ to $\overline{x} + 2s$ 
contains the large majority of the data, we can estimate $s$ by 
taking the range and dividing by $4$.
This yields an approximation to $s$: $s \doteq 9.27/4 \doteq 2.32$.
Using the formula from 1.11, the actual $s$ is approximately $1.87$.
Note that the one very large value has the effect of increasing
the range, and making one quarter of the range an overestimate.
\vskip 10pt
\noindent
1.21.  By the empirical rule, the number of measurements 
larger than $\overline{y} - s$ should be about $(\frac{1}{2}(68) + 50)\% = 84\%$.
So no, the manufacturer is just being slightly conservative with 
the $80\%$ figure.  
\vskip 10pt
\noindent
2.14.  Note: this problem is all about being able to interpret
the headings on the table properly!
 (a)  The probability that a single adult needs glasses is the sum
of the entries in the first row -- $0.58$.  (b) The probability that
an adult needs glasses but
does not use them -- $0.02$.  (c)  The probability that a single adult 
uses glasses whether or not he/she needs them is the sum
of the entries in the first column -- $0.46$.
\vskip 10pt
\noindent
2.15.  (a) The entry with the question mark is $1 - (.01 + .09 + .81) = .09$ since
these four alternatives are mutually exclusive.  (b)  The probability
that the company hits gas or oil on at least one drilling is
$(.01 + .09 + .09) = .19$.
\vskip 10pt
\noindent
2.16. (a) The probability that the randomly selected person has $O^+$ blood
is $\frac{1}{3} = .\overline{3}$.  (b)  The probability the person 
has type $O$ blood is $\frac{1}{3} + \frac{1}{15} = \frac{6}{15} = \frac{2}{5} = .4$.
(c)  The probability the person has type $A$ blood is 
$\frac{1}{3} + \frac{1}{16} = \frac{19}{48} \doteq .396$.  (d)  The probability
that the person has neither type $A$ nor type $O$ blood is 
$\doteq 1 - .4 - .396 = .204$ (using the answers from (b) and (c)). 
\vskip 10pt
\noindent
2.18.  (a) The possible outcomes are $S = \{HH,HT,TH,TT\}$. 
(b) If both coins are fair (balanced), it is reasonable to assign a probability
of $\frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}$ to each sample point.  They
are equally likely.  (c)  $A = \{HT,TH\}$ and $B = \{HT,TH,HH\}$.
(d) $P(A) = \frac{1}{2}$, $P(B) = \frac{3}{4}$, $P(A \cup B) = P(B) = \frac{3}{4}$
since $A \subset B$, $P(A\cap B) = \frac{1}{2}$ since $A\cap B = A$ whenever
$A \subset B$.  
\vskip 10pt
\noindent
2.19. (a) The sample space is $S = \{(V_1,V_1), (V_1,V_2), (V_1,V_3), (V_2,V_1), 
(V_2,V_2), (V_2,V_3), (V_3,V_1), (V_3,V_2), (V_3,V_3)\}$  (note that 
the same vendor could be used both days since that is not ruled out).  
(b) Each of the $9$ points in $S$ gets probability $\frac{1}{9}$.  
(c)  $A = \{(V_1,V_1),(V_2,V_2),(V_3,V_3)\}$ and $B = \{(V_2,V_1),(V_2,V_2),(V_2,V_3),
(V_1,V_2),(V_3,V_2)\}$.  Therefore $P(A) = \frac{1}{3}$, $P(B) = \frac{5}{9}$, 
$P(A\cup B) = \frac{7}{9}$ and $P(A\cap B) = \frac{1}{9}$.  
\end{document}
