Fall Semester 2002

Lab Six, Friday November 15, 2002
Date
Nov 15, 2002

Due today
Tutorial Five
Last day to turn in Homework Three with no penalty.

Starting today
Excel Tutorial Six (from the lab manual).
There's a problem and an experiment listed below too.

Due next time
QuizSite exercises need to be turned in (for credit).
Excel Tutorial Six
Prepare an experiment like the one listed below.

Today's grading
You might receive questions of the following nature:
  1. How do you sort data in an Excel list?
  2. What are the elements of an Excel list?
  3. How can you use worksheet lables in formulas?
  4. How do you summarize a list using pivot tables?
  5. How do you create a pivot chart?
  6. Plot a normal distribution with a mean of 400 and a standard deviation of 12.5
  7. Define a z-score and explain its use.

Today's custom experiment
First, here's a possible experiment for the two gold coins problem: one can imagine 6000 cards, each 1000 marked with one of the 6 initial "initial choice of drawer and coin distribution in the drawer." We imagine putting these 6000 cards in a container, stirring thoroughly, and drawing a card to represent one experimental trial of selecting a drawer and then a coin, and then after looking at the card either discarding the trial if an S showed as the first coin, and if not then going on to see what second coin is marked on the card. We can then return the card, stir the cards and try again and again, until we have done enough trials to convince ourselves of the result. Try it.

The Birthday Problem

The famous birthday problem asks the question:

"What is the fewest number of people that can be assembled in a room so there is a probability greater than 1/2 of a duplicate birthday?"
We, of course, must make some assumptions about the distribution of birthdays throughout the year. For convenience it is natural to assume that there are exactly 365 days in a year (neglect the leap year effects) and assume that all birthdays are equally likely, namely each date has a probability 1/365. We also assume birthdays are independent (there are no known twins, etc.)

There are the cases of one pair of duplicate birthdays, two pairs of duplicate birthdays, triples, etc. - many different cases to be combined. This is the typical situation where you use the complement probability approach and compute the probability that there are no duplicates. But we won't go into the details of the calculation here and instead present the results, and ask you to put together an experiment to verify or calculate a particular probability.

The result is that for 23 people (and our assumptions) the probability of a duplicate birthday (first) exceeds 1/2. That it indeed does may seem surprising until you remember that any two people can have the same birthday, and it is not just a duplicate of your birthday.

Set up an experiment to verify this conclusion.

Then calculate the probability of a duplicate birthday for a group of 15 people.

Last updated: Oct 31, 2002 by Adrian German for A113