Last Lecture: • Histograms: – Definition – Interpretation in terms of probability – Estimate of distribution function • Sample Means, Sample Medians, and Sample Variances / Standard Deviations (also known as “statistics”) – Definitions – Interpretations – Estimates of “true” values • This Thursday, see www.math.umass.edu/~jstauden / for homework 2. • In the HW, you’ll also learn about percentiles and boxplots (from Chapters 2 and 3). • We’ll learn a lot more about the stuff in the first 3 chapters later in the semester…
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Last Lecture: Histograms: –Definition –Interpretation in terms of probability –Estimate of distribution function Sample Means, Sample Medians, and Sample.
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Last Lecture:• Histograms:
– Definition– Interpretation in terms of probability– Estimate of distribution function
• Sample Means, Sample Medians, and Sample Variances / Standard Deviations (also known as “statistics”)– Definitions– Interpretations– Estimates of “true” values
• This Thursday, see www.math.umass.edu/~jstauden/ for homework 2.
• In the HW, you’ll also learn about percentiles and boxplots (from Chapters 2 and 3).
• We’ll learn a lot more about the stuff in the first 3 chapters later in the semester…
Example with dicePr( rolling a 1 on 1 die in one roll) = 1/6Pr( rolling a 2 on 1 die in one roll) = 1/6Pr( rolling a 3 on 1 die in one roll) = 1/6Pr( rolling a 4 on 1 die in one roll) = 1/6Pr( rolling a 5 on 1 die in one roll) = 1/6Pr( rolling a 6 on 1 die in one roll) = 1/6
Pr( rolling less than 4 in one roll) = Pr( rolling 1 or 2 or 3 in one roll)= Pr( rolling 1 in one roll) + Pr( rolling 2 in one roll)
+ Pr( rolling 3 in one roll) - Pr( rolling a and a 2 and a 3 in one
roll) = 1/6 + 1/6 + 1/6 – 0 = 1/2
Example with 2 dice
1
2
Outcome 3
on die 2 4
5
6
Outcome on die 11 2 3 4 5 6 Each square is
a possibleevent
Pr( any specificevent) =
1/6 * 1/6 = 1/36
Pr( rolling a seven in total) =
6/36 = 1/6
(squares w/ xs inthem are 7s)
x
x
x
x
x
x
Example with 2 dice (a related interpretation)
Pr( rolling a seven in total ) = number of ways to roll a seven number of possible outcomes
In general when ways the event could occur are equally probable, Pr( event ) =
number of ways that the event could occurnumber of possible outcomes
What’s a simple expression for Pr( event doesn’t occur?)(hint: it involves Pr( event )…)
Aside: Odds• “Odds” are related to probabilities. More specifically, the odds of an
event are:
“Pr( event does not occur ) / Pr (event occurs) to 1”
• At start of last football season, “the odds”* that the Patriots would win the Superbowl were: 250 to 1.
• Sample means and Sample variances are calculated from datasets.• True means and True variances are part of the theoretical model for
the data.
• KEY IDEA: as the size of the dataset becomes larger, the Sample means and variances get closer to the true means and variances…
Powerball Example:# Winners: 0 1 2 3 4 5 6Probability:8% 21% 26% 21% 13% 9% 2%(These are estimates based on historical data, but assume that they are the truthfor the sake of the example.)
Probability that I am a winner if I buy 1 ticket = 1/80 million (= 1/80M).Jackpot (pre tax) = $200 million. (if >1 person wins, jackpot is divided).
Assume whether or not I win is independent of the number of winners.
Let X = millions of dollars I win from one ticket.
PDF:x 0 200 100 66.7 50 40 33.3pr(x) ? .2283/80M .2826/80M .2283/80M .1413/80M .0978/80M .0217/80M1) What does “?” equal? (and how did I compute the other pr(x)’s)?(see next slide for ans)2) mu = E(X) = sum(x * p(x)). In dollars this is about $1.26. (you can confirm this)3) Var(X) = sum((x – mu)*pr(x))4) Interpretation: If I play powerball a lot when there is a $200 million jackpot, then I can expect to win $1.26 on average.5) If tickets are a dollar each, why doesn’t Powerball lose money? (These numbers are all based on real data.)
Answers to question on previous slide
• The ? = (80million-1)/80million– You know this is true since the probability that I do not win is one
minus the probability that I win (and the probability that I win is given to be 1/80 million).
• How did I compute the pr(x)’s:– The probability that I win $200million =
Pr(I win and there is only winner given that there is at least 1 winner)=Pr(I win)*Pr(there is only one winner given that there is at least one winner)=(1/80million) * (0.21/(.21+.26+.21+.13+.09+.02))Uses
independence
Uses the rule for conditional probability on page 141.
Cumulative Probability:
• A cumulative probability is the probability that X is less than or equal to a some number:
• Ex: powerball:• Pr(there are 3 or fewer winners)
=Pr(X<=3)=Pr(X=0 or X=1 or X=2 or X=3)=Pr(no winners)+Pr(1 winner)+Pr(2 winners)+Pr(3 winners)= 8%+21%+26%+21%
• Notation: F(3)=Pr(X <= 3) (F(k) = Pr(X<=k) is called the Cumulative Distribution Function or CDF)
• If this helps, think of F(k) as the integral of the PDF from 0 to k.
• Note: Pr(X > 3) = 1-Pr(X<=3) (careful about > and <=…)
Graphically:
6543210
0.2
0.1
0.0
Number of Winners (k)
Pr(
Win
ners
= k
)Pr( X <= 3 ) = sum of the areas of the shaded regions
= 1 – Pr( X>4 )= 1 – sum of the areas of the white regions
PDF for the randomvariable that representsthe number of winners
“Famous” PDFs• Binomial: “X~bin(n,p)”
– Setup• Let X = number of successes out of n identical trials• n identical independent trials• Each trial results in a success w/ probability p or failure with
probability q=1-p• X could possibly be 0,…,n
– PDF:• Pr(X = k) = (n choose k) pkqn-k
• (n choose k) = number of ways to choose k things from n things= (n) = n! / (k! (n-k)!)
One could show why the Poisson PDF is correct, but the math is more involved. If you’re interested, come talk to me sometime.
Example:
• Inspect an experimental rat’s brain for tumorous cells. You expect 10 tumorous cells in 60mm3 of brain. What’s the probability that you see either 2 or 3 tumorous cells in 10mm3?
• X = tumors found 10mm3 of brain. X~Poisson(5/3) (rate per 60mm3 is 10, so rate per 10mm3 is 10/6 = 5/3)
– Remember:(n choose k) = number of ways to choose k things from n things= (n) = n! / (k! (n-k)!)
(k)• Note that 0! = 1
– Note that binomial is like the hypergeometric, but the binomial is with replacement… (which results in a fixed p)
Hypergeometric Example• Cards: probability of being dealt a flush in hearts in a
hand of poker (flush=all cards of same suit)• X = number of hearts in the hand• N = 52• M = 52/4 = 13• n = 5• Want Pr(X=5)
(13 choose 5 ) (39 choose 0)/(52 choose 5)= 1287 * 1 / 25989600.0004951981 (NOTE THAT THIS NUMBER IS DIFFERENT FROM WHAT I WROTE ON THE BOARD IN THE CLASS)
What’s probabilty of getting a flush in any suit?
• (see minitab:calc:Probability Distributions: Hypergeometric)
• For each of the following:– What is the random variable?– What is it’s distribution and what are numbers
for its parameters?– What is the probability that is being asked for?– How can it be computed from the probability
density function.
More Examples:• There are 4 security checkpoints. The probability of being searched
at any one is 0.2. You may be searched more than once and all searches are independent. What’s the probability of being searched at least one time?
• 50 geese in a flock of 200 are tagged by a wildlife biologist. The next year, 10 ducks from the flock are captured. Assume the flock still has 200 ducks and no tags are lost. What’s the probability that at least 5 of the recaptured ducks have tags?
• Suppose a written test has 5 True/False questions. Passing = at least 3 correct answers and the test can be taken at most 3 times. (Assume no learning occurs between tests if one fails!)– If one randomly guesses what’s the probability of passing?
– What’s the probability that someone who randomly guesses will eventually pass?
• An overloaded server receives an average of 25 emails per second at 12:00PM. If it receives more than 30 emails in a second, it will crash. What’s the probability of a crash at 12:00PM on a given day (based on the traffic in the previous 1 second)?