Fall Semester 2002

Lecture Notes Five: Experiments and the normal curve.
We are interested in two experiments.

In the first one you will plot the normal distribution. You will do it in two ways, in one plotting the distribution and in the other plotting the cumulative distribution. Read the section in the book for ways in which the cumulative distribution can be used to estimate probabilities. There will be more notes from Kirkup posted here too.

We've done this (already) last Thursday.

In the second one you will make an experiment. The experiment is actually a simulation. There is a theorem (called The Central Limit Theorem) that states the following about sampling:

If X is a simple random sample of n elements from a large (infinite) population, then the distribution of the mean (when you repeatedly sample the population, calculate the mean and the sample, and plot it) approaches the bell-shaped distribution of a normal random variable when n goes to infinity (and beyond, if you will).
In our example we will sample 500 times. Each time we take a small sample, 5 hot dogs (or breadsticks, if you want). We measure them. Their lengths are 12 inches plus or minus half an inch. The variation of the length is uniformly distributed in this interval from 11.5 to 12.5 inches. Well, in each sample we calculate the mean. Then we plot the histogram of the 500 means. We get a frequency distribution that is resembling the normal distribution. That's the way we verify the Central Limit Theorem. Again, use Kirkup to see if you can solve this problem, and how.

So in one experiment you

We will also analyze the following problem:

Sum of Two Dice.

What is the probability that the sum of the faces of two independently thrown dice total 7?

We have only six successful possibilities:

  1. 1 + 6
  2. 2 + 5
  3. 3 + 4
  4. 4 + 3
  5. 5 + 2
  6. 6 + 1
in the total sample space of 62 = 36 possible outcomes. Each possible outcome has probability 1/36, hence the probability of a total of 7 is 6/36 = 1/6. Let's look at the following table (of sums):

        1    2    3    4    5    6
1234567
2345678
3456789
45678910
567891011
6789101112
Given this table can you produce the probability distribution?

We now have the simple rule: count the number of different ways the compound event can occur in the original sample space of equally likely events (outcomes), then the probability of this compound event is the ration of the number of successes in the sample space to the total number of events in the original sample space. (See the example in lab tomorrow).

Last updated: Nov 11, 2002 by Adrian German for A113