View the step-by-step solution to:

SUPPLEMENTARY ASSIGNMENT 1 DUE FRI MARCH 14, 2014 (A Favorite Rule of Thumb Counterexample) In Chapter 16 homework problem 36d of Stats, Data and...

Could you please guide me through these 3 problems and show me how to do them?

SUPPLEMENTARY ASSIGNMENT 1 DUE FRI MARCH 14, 2014 (A Favorite Rule of Thumb Counterexample) In Chapter 16 homework problem 36d of Stats, Data and Models , you contem- plated 1000 plays of a $1 slot machine, where on average one play resulted in a $ . 08 casino profit with a standard deviation of $120 . The textbook doesn’t cover the CLT til several chapters later, but we have. And since 1000 trials are many more than 40, we encouraged you to “correctly” use the CLT while thinking about the average per play casino profit of N ( . 08 , 120 1000 ) to conclude that there was still a 49% chance the casino would lose money on that machine on a given day. Strike you as counterintuitive? This problem will show you that at least some- times, the calculation in 36d is not truly applicable. In particular, let’s assume the slot machine is “jackpot or nothing.” Meaning, when you put your $1 in, most of the time (i.e. with probability 1 - p ) you just lose, but with probability p , you get a jackpot of amount A . Let’s see what A and p should be to approximate the average $ . 92 gross payout with a standard deviation of $120 . Our probability model for the gross return (the random variable Y, say) to the gambler for one play is Amount Prob. 0 1-p A p If A were 1 this would just be a Bernoulli(p) (or Binomial(1,p)) random variable and we’d immediately know the mean and standard deviation. Since Y is just the constant A times such an X ( Y = AX ), the mean and standard deviation of Y are also quite clear. (The net return which keeps track of overall loss is Y - 1.) Problem 1. For the random variable Y described above, complete the following table for its mean and standard deviation: p A mean of Y st. dev. of Y .92 1 .92 .271 .04 600 24 117.58 .092 10 ? ? .00092 1000 ? ? .0000092 100000 ? ? Problem 2. The mean of Y is Ap . And when p is small, the standard deviation of Y is close to A p . Using this observation find values of A and p for which the mean and standard deviation of Y are within 1% of the values μ = . 92 and σ = 120 given in problem 36. Problem 3. With your value of p from problem 2, what is the probability that no jackpots are given out in 1000 plays of the slot machine? Your answer should be wildly different from the 51% = 100% - 49% that normal approximation in problem 36d gave. Try to give a simple explanation of why normal approximation of the random variable Y should not be expected to be accurate. 1
Background image of page 1
Sign up to view the entire interaction

Top Answer

Quite a tough concept. Anyway please check the answer and if you don't... View the full answer

8549638.docx

1) The required table is given below, P
0.92
0.04
0.092
0.00092
0.000009
2 A
1
600
10
1000
100000 Mean of Y St. Dev of Y
0.92
0.271
24
117.576
0.92
2.890
0.92
30.318
0.92 303.314 As,
Y = AX
So,...

Sign up to view the full answer

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

-

Educational Resources
  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question
Ask a homework question - tutors are online