http://onlinestatbook.com/stat_sim/sampling_dist/index.html

We already know that:
A range from -1 std.dev. to 1 std.dev. contains 68.3% of outcomes.
A range from -2 std.dev. to 2 std.dev. contains 95.4% of outcomes.
A range from -3 std.dev. to 3 std.dev. contains 99.7% of outcomes.

So the question is, how many Std.Dev's do we have to move away from the mean in both directions on the graph to contain 98% of outcomes. Not 95.4%, Not 99.7%, exactly 98%. Right away you know the answer will be between 2 and 3 std.dev's, as 98% is between 95.4% and 99.7%

To

Learning statistics can be a little strange. It almost seems like you're trying to lift yourself up by your own bootstraps. Basically, you learn about populations working under the assumption that you know the mean/stdev, which is silly, as you say, but later you begin to drop these assumptions and learn to make inferences about populations based on your samples.

Once you have some version of the Central Limit Theorem, you can start answering some interesting questions, but it takes a lot of study just to get there!

There is a difference. Your "samples" (random selections of values "x") that are made up of "instances" (referred to as the variable "n") provide what will essentially be the building blocks of your Sampling Distribution of the Sample Mean. Because your "instances" determine the value of the mean of "x", your size of "n" determines the value of "x"'s mean, and the Sampling Distribution of the Sample Mean's standard deviation (Defined as The original dataset's standard deviation divided by the square root of "n").
For example: If you were to take 1 "sample" with 100 "instances", you would get only one piece of data regarding the mean of 100 items [1,1,3,6,3,6,3,1,1,1,1,1...] from your original data. Your sampling distribution of the Sample mean's standard deviation would have a value of ((The original sample's S.D.)/(The square root of 100)), but that wouldn't really matter, because your data will likely be very close to your original data's mean, and you'd only have one sample.
Now if you take 100 samples with 1 instance [3], you'll get many pieces of data, but no change in standard deviation from your first sample: ((The original sample's S.D.)/(The square root of 1)). Functionally, with enough samples taken like this, you'll re-create your original dataset! You won't be creating a useful sampling distribution of the sample mean because "x" will equal the mean of "x". With 100 "samples" of 1 "instance", you're randomly picking 100 values of "x" and re-plotting them.
I hope that helps.

This is answered in the next video in the series.

THe general population is known to have a mean IQ of 100. That means that the distribution of sample means also has a mean of 100.

As long as you know all the values in the sample, you can do the series of calculations described under "basic examples" here http://en.wikipedia.org/wiki/Standard_deviation to figure out what the sample's standard deviation is. Of course, you have to divide by N-1 with samples like the wikipedia article (as well as Sal's video on standard deviation) explains, otherwise it's exactly the same. Perhaps you are limiting your definition of "standard deviation" to "standard deviation of population", which you of course can't figure out with just one sample of values? If it's not specified that the population's SD is asked for in the exam question you're describing, it's safe to assume that they are asking for the sample's SD.

Computers can quite easily simulate uniform distributions (for example the rand() function in matlab that gives a number between 0 and 1 accordingly to an uniform distribution). With that number you can simulate all sorts of other distributions.
For example if you want to simulate a fair dice you do :
x = rand(1)
if (x<1/6) then y = 1
elseif (x<2/6) then y = 2
elseif (x<3/6) then y = 3
elseif (x<4/6) then y = 4
elseif (x<5/6) then y = 5
else y=6

This is how you can simulate easily discrete distributions.

@cnidoblast, selecting 5 types of animals invalidates the CLT. One of the assumptions of the most common CLT (there are actually many versions, this one is the most common) is that the observations, what Mr. Khan calls samples, are independent and identically distributed instances of a random variable. A random variable is a function that converts an observation from a random process in to a number. Your animals are not numbers, so it's meaningless to sum them much less find the mean. If you're talking about averaging their weights then it still fails the CLT assumptions because the weights that you're averaging do not come from an identical distribution. That is, the distribution of weights of zebras is very different from the distribution of weights of goats. Hope this helps! :)