1.  Let X_i represent the return from a single randomly selected stock.
a) We lose money if the return is negative.  We're asked to calculate P(X < 0).  Because X is normally distributed, we can use the normal table in the back, but first we need to standardize it.  The mean is .11 and the SD .28, so 0 is (0 - .11)/.28 = -.3929  (That is, 0 is .3929 SDs below the mean.)  So the table gives the area to the left of -.39 to be .3483.
b) A randomly selected average of 9 observations has a mean of .11 (same as the population mean) and an SD of (.28/sqrt(9))  (population SD divided by the square root of n.)  By the central limit theorem, the average also follows a normal distribution: N(.11, .09333).  So we're asked to find P(Xbar < 0), and in standard units, 0 is (0 - .11)/.09333 = -1.18.  From the table, P(Z < -1.18) = .1190.  So you're less likely to lose money with 9 stocks than with 1.
c) A randomly selected average of 25 stocks has a mean of .11 and an SD of .28/sqrt(25) = .056.  Again, we need to find P(Xbar < 0) and now 0 is
(0 - .11)/.056 = -1.96 SDs below mean.   So P(Z < -1.965) = 2.5% (approx).

2  Let X_i = winnings on a single toss of the two dice.
a) You win $10 if the spots add to 7.  There are 36 possible outcomes, and exactly  6 of them add to 7. ( 1 + 6, 6+1, 2+ 5, 5+2, 3+4, 4+3).  So the probability of winning is 6/36 = 1/6.
b)The game is fair if the mean winnings are 0.  That is: Mean(X)  = 0.  Mean(X) = 10*(1/6) + (-x)*(5/6) = 0.  Solving for x gives $2.00.
c) If you have to bet $3 to play, then Mean(X) = 10*(1/6) - 3(5/6) = -5/6 = - 0.833333  The game is no longer fair and is not to your advantage..
The SD is the square root of (10 - -.8333)^2 *(1/6) + (-3 - -.8333)^2 * (5/6) =  4.844814.  So although we typicall lose about 83 cents, the spread of possible winnings is so large that we stand a chance to make some money.
d) If you play 16 times, your total winnings would be X1 + X2 + ..... + X16.  The mean of this sum is -.833333 + .-8333... etc or 16*(-0.833333) = -13.3333.
The SD would be sigma*sqrt(n) = 4.844814*sqrt(16) =  19.37926.
How do we interpret this?  If infinitely many people were to play this game 16 times, their typical winnings would be -$13.33 and about 68% of them would win
-13.33 +- 19.379, 95% between the mean +- 2 SDs, etc.
e) The mean of the average winnings is also -0.83333, but the SD of the average is sigma/sqrt(n) = 4.844814/4 = 1.211.  So if you had infinitely many people play this game 16 times, and each person calculated their average winnings, the typical average winning would be -0.833 plus or minus 1.21.
f) By the Central Limit Theorem, the Average winnings (based on 16 spins of the wheel) is normally distributed with mean -0.8333 and SD 1.211.  So
P(Xbar >0) = P(Z > (0 - -.8333)/1.211) = P(Z > .688) = 1 - .7549 = .2451.  You you have about a 24% chance of having an average bigger than 0.

3.  Let X = number of heads.  The mean of X is 100*(1/2) = 50.  (Which should sound reasonable)  The SD is sqrt( 100 * 1/2 * 1/2) = sqsrt(25) = 5.
So in 100 tosses of a coin, we typically get 50 heads, plus or minus 5.
a. by the CLT, X is approximately N(50,5).  We want P(45<X<65).  In standard units, 45 is (45-50/5 = -1 SD below mean, and 65 is 3 SDs above mean. So we want to find P(-1 < Z < 3).  From the table we learn P(Z < 3) = .9987 and P(Z < -1) = .1587.  So the area in bewteen -1 and 3 is .9987 - .1587 = 0.84.
b. It's pretty unlikely: 70 is (70-50)/5 = 4 SDs above average. P(Z > 4) is so small its not even on the table, so this is approximately 0.
c. There are a couple of ways of doing this.  The easiest is to say: if there are 60% heads in 100 tosses, then there must have been 60 heads.  So I need to find P(X < 60) = P(Z < 2) = .9772.  Similarly, P(X < 40) = P(Z < -2) = 0.0228.  The other way is to say, well I know (X/n) is a random variable that is approximately N(.50, sqrt(1/2*1/2 divided by 100) = 1/sqrt(400) = 0.05.  So P(X/n < 60%) = P(Z < (60% - 50%)/5%) = P(Z < 2).