Reference no: EM131725484

1. Z-tables - Elementary statistics textbooks typically have a table of the standard normal distributions. This is a formatted table with values of the cumulative distribution (CDF) that generally looks something like this:

x	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.5000	0.5040	0.5080	0.5120	0.5160	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
. . .

The table is read by locating the first and second decimals of the x value in the corresponding row and column. Thus F(0.13) = 0.5517. Using Python, produce a table of the normal CDF with rows starting at x = 0, 0.1, 0.2, . . . , 3.3, 3.4, and with columns 0.00, 0.01, . . . , 0.08, 0.09.

2. Central Limit Theorem - Read the review of the Central Limit Theorem before doing this exercise.

The purpose of this exercise is to verify the portion of the central limit theorem that says that the standard deviation of the sample means varies as σ/√n, where σ is the population standard deviation, and n is the sample size.

The output of this exercise will be a single plot. It will show "standard deviation of the sample mean" on the y-axis and "sample size" on the x-axis.

You will draw a smooth curve and a set of markers on the plot. The markers will NOT be on the smooth curve.

Pick any distribution you like and use it to produce random numbers with a mean µ and standard deviation σ. These numbers (µ and σ) will be fixed constants that you choose and will not change during the course of the program.

For each of the following sample sizes, you should generate one point on the plot: n = 3, 7, 10, 30, 70, 100. Generate 10 random samples from your distribution, and calculate the means x₁^-, x₂^-, . . . , x₁₀^- of the 10 random samples of size n. Then calculate the standard deviation of the set of 10 numbers {x₁^-, x₂^-, . . . , x₁₀^-}. This will give you a standard deviation of the means sn, that corresponds to a sample size n. Plot a marker at the location (n, s_n) and then repeat the process for each of the remaining values of n in the set of n's.

3. Monty Hall Problem - In a game show a prize is placed behind one of three closed doors, labeled A, B, and C. You then choose one of the doors. After you have made your choice, Monty opens one of the doors, and shows you that there is no prize behind it. Monty will NEVER open the door you pick, and he will NEVER open a door that has a prize behind it. He gives you the choice of staying with your original choice or changing. Write a program to simulate one round of the Monty Hall Problem, and then run it at least 100 times to determine the probability of winning if you change your choice and if you stay.

[Note: (Students who have not had a course in probability can skip this note; it is not needed to solve the problem. The remainder of this page is here to convince doubters as to the correct answer.) This problem can be solved exactly. Let P(A) = 1/3, P(B) = 1/3, and P(C) = 1/3 be the probabilities that the prize is behind doors A, B, or C. Let x be your choice. Suppose you initially choose x = A. Since your choice is independent of where the prize is, then the conditional probabilities are

P(A|x = A) = 1/3

P(B|x = A) = 1/3

P(C|x = A) = 1/3

Monty is only going to open a door that does not have a prize behind it, and will not open your door (GIVEN). Let y denote the door that Monty picks to open. If there is a prize behind your door, he can open either of the other doors. If there is a prize behind a door that you did not select, he must open the third door. Thus:

P(y = C|A, x = A) = 1/2

P(y = C|B, x = A) = 1

P(y = C|C, x = A) = 0

The question of switching then becomes one of finding the probability

P(B|y = C, x = A)

= P(y = C|B, x = A)P(B|x = A)/P(y = C, x = A)

= (1/3) × (1)/P(y = C, x = A)

But the denominator is

P(y = C, x = A)

= P(y = C|A, x = A)P(A|x = A)

+ P(y = C|B, x = A)P(B|x = A)

+ P(y = C|C, x = A)P(C|x = A)

= (1/2) × (1/3) + (1) × (1/3)

+ (0) × (1/3)

= 1/2

giving

P(B|y = C, x = A) = 2/3

Thus there is a 2/3 chance the prize is behind door B, and a 1/3 chance the door is behind your original choice, door A.]