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The Binomial Random Variable - Lecture Notes | STA 100, Study notes of Data Analysis & Statistical Methods

Material Type: Notes; Class: Statistical Methods; Subject: Statistics; University: SUNY Institute of Technology at Utica-Rome; Term: Unknown 1989;

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Pre 2010

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STA100
The Binomial Random Variable
Lecture 11
Weโ€™re just about done with the โ€œtheoryโ€ part of the course and are moving into the applications.
Most people find that the course gets much more understandable at this point, so
๏‚ท Good job making it this far!
๏‚ท Keep moving forward for the sorts of applications you came into the course for.
A Man and his Triangle
You may have noticed a pattern emerging from the previous examples. We would like a formula which
tells us, when sampling from a dichotomous population with a sample of size ๐‘› , what the probabilities
of getting 0,1, 2 , โ€ฆ ๐‘› successes are. Think of walking up to individuals in the Mall and asking
whether or not they have a match. The response will be Yes or No (Success or Failure) though some
people may also tell you to stop smoking.
๏‚ท If you sample with sample size ๐‘›= 1 then your sample space looks like S = ๐‘†,๐น and you
have exactly 1 outcome with 0 successes and exactly 1 outcome with 1 success. This is indicated
in the first row of the table.
๏‚ท If your sample is of size 2 then your sample space looks like S = ๐‘†๐‘†,๐‘†๐น,๐น๐‘†,๐น๐น and you have
exactly 1 outcome with 0 successes, 2 outcomes with 1 success, and 1 outcome with 2
successes. This is indicated in the second row of the table.
๏‚ท One last example here. If your sample is of size 3 then your sample space looks like S =
๐‘†๐‘†๐‘†,๐‘†๐‘†๐น,๐‘†๐น๐‘†,๐‘†๐น๐น,๐น๐‘†๐‘†,๐น๐‘†๐น,๐น๐น๐‘†,๐น๐น๐น and you have exactly 1 outcome with 0 successes, 3
outcomes with 1 success, 3 outcomes with 2 successes and 1 outcome with 3 successes.
๏‚ท The following triangle, named after Pascal, tells us the structure of our outcomes for any size
sample.
n=1
1 1
n=2
1 2 1
n=3
1 3 3 1
n=4
1 4 6 4 1
n=5
1 5 10 10 5 1
n=6
1 6 15 20 15 6 1
n=7
Look at the ๐‘›= 6 row. The way to construct any row in the table is as follows. First, if you
will sample with size 6 that means you will walk up to 6 people and say โ€œDo you have a match?โ€
There are 2*2*2*2*2*2=64 possible outcomes of the form, e.g. SFFFSF (corresponding to first
person says Yes, second person says No, third person says No, fourth person says No, fifth
pf3
pf4
pf5
pf8

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STA100 The Binomial Random Variable Lecture 11

Weโ€™re just about done with the โ€œtheoryโ€ part of the course and are moving into the applications. Most people find that the course gets much more understandable at this point, so ๏‚ท Good job making it this far! ๏‚ท Keep moving forward for the sorts of applications you came into the course for.

A Man and his Triangle

You may have noticed a pattern emerging from the previous examples. We would like a formula which tells us, when sampling from a dichotomous population with a sample of size ๐‘› , what the probabilities

of getting 0, 1, 2 , โ€ฆ ๐‘› successes are. Think of walking up to individuals in the Mall and asking whether or not they have a match. The response will be Yes or No (Success or Failure) though some

people may also tell you to stop smoking.

๏‚ท If you sample with sample size ๐‘› = 1 then your sample space looks like S = ๐‘†, ๐น and you have exactly 1 outcome with 0 successes and exactly 1 outcome with 1 success. This is indicated in the first row of the table. ๏‚ท If your sample is of size 2 then your sample space looks like S = ๐‘†๐‘†, ๐‘†๐น, ๐น๐‘†, ๐น๐น and you have exactly 1 outcome with 0 successes, 2 outcomes with 1 success, and 1 outcome with 2 successes. This is indicated in the second row of the table. ๏‚ท One last example here. If your sample is of size 3 then your sample space looks like S = ๐‘†๐‘†๐‘†, ๐‘†๐‘†๐น, ๐‘†๐น๐‘†, ๐‘†๐น๐น, ๐น๐‘†๐‘†, ๐น๐‘†๐น, ๐น๐น๐‘†, ๐น๐น๐น and you have exactly 1 outcome with 0 successes, 3 outcomes with 1 success, 3 outcomes with 2 successes and 1 outcome with 3 successes. ๏‚ท The following triangle, named after Pascal, tells us the structure of our outcomes for any size sample.

n=1 1 1 n=2 1 2 1 n=3 1 3 3 1 n=4 1 4 6 4 1 n=5 1 5 10 10 5 1 n=6 1 6 15 20 15 6 1 n=

Look at the ๐‘› = 6 row. The way to construct any row in the table is as follows. First, if you will sample with size 6 that means you will walk up to 6 people and say โ€œDo you have a match?โ€ There are 22222*2=64 possible outcomes of the form, e.g. SFFFSF (corresponding to first person says Yes, second person says No, third person says No, fourth person says No, fifth

person says Yes, sixth person says No). It gets tedious to list them all out, so we ask โ€œHow many of the 64 possibilities have 0 successes?โ€ Just 1, of course, the outcome FFFFFF. The we ask, โ€œHow many of the 64 possibilities have 6 successes?โ€ Just 1, of course, the outcome SSSSSS. Thus the first and last entries in the row must be 1โ€™s, and if you look at the table you will see this is so for any row. How do we do the other entries?

You can think of the number 6 in the second entry of the row labeled โ€œn=6โ€ as the number of outcomes with exactly 1 success. We know this number is a 6 in two ways. One way is to work the table mechanically, adding the two numbers above our target, to the left and to the right, to get a 6 = 1+5. Then the next entry, corresponding to the number of outcomes with exactly 2 successes is obtained from the table mechanically by adding together the 5 and the 10 to get 15. In this way you can fill out the whole row and then move onto the next row.

But there is another way to think about all of these and, if youโ€™re a little ambitious, to see why the table โ€œworksโ€. As we have seen, the number of ways to get 2 successes on 6 trials is the combination of 6 items taken 2 at a time or

How about the entry to the right? The number 20 is (just working the table) the sum of the two entries above to the left and to the right. But also

You should be able to fill in the next row by yourself. Pascalโ€™s triangle helps us to calculate probabilities in our heads for simple sampling situations. Bringing back the ideas from the last lecture (and explained in your book in section 5.2) we have the following.

When an experiment is such that

๏‚ท you will perform a trial ๐’ times, and

๏‚ท each trial may come up as a success, with probability ๐’‘ , or a failure with probability

๐Ÿ โˆ’ ๐’‘ (think coin toss or polling dichotomous population) then,

๏‚ท if each trial is independent of all the others,

๏‚ท the probability of ๐’“ outcomes being successful on ๐’ trials is

๐’‘๐’“๐’๐’ƒ ๐’“ ๐’”๐’–๐’„๐’„๐’†๐’”๐’”๐’†๐’” = ๐‘ช๐’,๐’“ ๐’‘๐’“^ ๐Ÿ โˆ’ ๐’‘ ๐’โˆ’๐’“

b. What is the probability that I get at least 4 Hearts in my sample?

_r prob

10._

Since โ€œat least 4โ€ is the same as โ€œ4 or 5 or โ€ฆ or 10โ€ we can add the probabilities 0.146+0.058+ 0.016+0.003+0.000+0.000+0.000 to get 0.223.

c. What is the probability that I get at most 3 Hearts in my sample?

_r prob

10._

Since โ€œat most 3โ€ is the same as โ€œ0 or 1 or 2 o 3โ€ we can add the probabilities 0.056+0.188+0 .282+0 .250 to get 0.776. Note that these numbers add up to 1.

Some people like to draw a picture instead of a table, so take a look at the following.

Expected Value It turns out that we are interested in the โ€œaverageโ€ for these experiments

thought of in the following way. Suppose you walk up to 4 people in the Mall and ask โ€œDo you have a match?โ€ and then write down the number of people who say โ€œyesโ€. Your result will be one of 0, 1, 2, 3, or 4. Do this again and again, going from mall to mall. What do you think you will obtain on average? Assuming as before that 15% of people carry matches, I did this 100 times and got the following:

1 0 2 0 0 2 0 1 0 1 0 0 2 0 0 0 0 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1 1 4 0 1 1 1 0 0 0 0 0 0 1 1 2 1 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 1 1 1 2 0 1 0 1 1 0 1 1 1 1 0 1 0 0 1

How will we calculate an average?

0 1 2 3 4 5 6 7 8 9 10

0

Probabilities Of r Successes on 10 Trials

r

p

Pretty close. If I repeated the experiment in 1000 different malls Iโ€™ll bet Iโ€™d be even closer.

Take a second and, using the table in the appendix, fill in the following for an experiment where you take 6 foul shots (assume your results are independent from shot to shot) and suppose the probability you make any individual shot is 0.7 = 70%.

Number of Shots, r Probability, p 0 1 2 3 4 5 6

total 1

Iโ€™ve placed a plot summarizing this situation below.

Suppose you wanted to balance the stems in this figure. Where would you put the fulcrum?

-1 0 1 2 3 4 5 6 7

0

Probabilities for r=number of Successes on n=6 Attempted Foul Shots

Using the table we can calculate the mean, ๐œ‡. Hereโ€™s how I would set this up in Excel:

Number of Shots, r Probability, p rp=outcomeprobability 0 .001 0. 1 .010 0. 2 .060 0. 3 .185 0. 4 .324 1. 5 .303 1. 6 .118 0. total 1 4.

Our average, 4.2040, is just about where you would have to balance the plot above.

Summing up

At this point you should feel reasonably comfortable (assuming you are reading the text, working through practice examples from the text, etc.) with ๏‚ท Binomial Random Variables ๏‚ท Calculating the average outcome of a random variable.

In the next lecture we learn how to calculate standard deviations and move on to the Normal Distribution.