1. A

2.  The 72% because of selection bias (explain what
selection bias is for full credit) 
While the newspaper sample is considerably smaller in size, 
it is arguably more scientific because respondents were 
selected using a random probability method.

3.  

The standard error is root(361) * root(.75 * .25)
                      ---------------------------
                                361
multiplied by 100 you should get 2.279%

4.  It is possible and it should be 75% +/- the standard
error calculated in #3. (4.558) 

5.  This is a SE of a sum.  The box average is

(+9 * .1) + (-5 * .3) +( 0 * .6) = -.60  

so the expected value is -.60 * 100 = -60 dollars

the standard deviation is the square root of:

  (9 - (-.60))^2 + 3*(-5 - (-.60))^2 + 6*(0 - (-.60))^2
  ------------------------------------------------------
                           10

or $3.9038 so  the standard error is 10*3.9038 or $39.038

6.  You lost $158, this is -158 and you were only expecting
to lose $60 (-60).  Need to calculate Z to find the chance
of seeing this kind of outcome:

Z = -158 - (-60)
    ------------
       39.038

= -2.51, round down to -2.5, the area to left of -2.5 is
.62%.  So there was about a .6% or 6 in 1000 chance of losing $158
or more

7. Here, the population is known and you're being asked about the
chance of seeing a particular sample outcome -- specifically
between 5% and 9%.

Need a Z score

Z =          9% - 10%
    ------------------------------------- =  -.366 or about -.35
    (root 121) * (root .10 * .90)
      ------------------------    * 100
             121

Z =          5% - 10%
    ------------------------------------- =  -1.83 or about -1.80
    (root 121) * (root .10 * .90)
      ------------------------    * 100
             121

The area associated with a Z = -1.80 is 92.81%
The area associated with a Z = -.35 is  27.37%

Take the difference in this case and divide by 2 to give
32.72%