Distributions
1
ICEBREAKER Binomial EXAMPLE
Let us suppose that you are planning to have three children. Would you like to have three boys? Or two
boys and then a girl? Think of all the possible gender arrangements that you could be fortunate enough
to have, and fill in the table below. We will assume that the probability of each child being a girl is 0.6.
Gender Arrangement
Probability (work)
Probability (calculated)
BBB
0.4 x 0.4 x 0.4
0.064
BBG
0.4 X 0.4 X 0.6
0.096
BGB
0.4 X 0.6 X 0.4
0.096
GBB
0.6 X 0.4 X 0.4
0.096
BGG
0.4 X 0.6 X 0.6
0.144
GBG
0.6 X 0.4 X 0.6
0.144
GGB
0.6 X 0.6 X 0.4
0.144
GGG
0.6 X 0.6 X 0.6
0.216
Total
1
Now let X = the number of girls you are blessed with. Construct a probability distribution table for X.
X
Probability
0
1 x 0.064 = 0.064
1
3 x 0.096 = 0.288
2
3 x 0.144 = 0.432
3
1 x 0.216 = 0.216
Total
1
And finally, plot a probability histogram for X by shading in squares in the grid below. Fill in the values of
X in the second row from the bottom, and then for each value, shade in a column where the height of the
column represents the probability of that value of X.
2010 Radha Bose FSU Department of Statistics
Binomial Distributions
27
18
Probability: 1 square = 0.016 probability
14
4
0
1
2
3
Values of X
Notice that the distribution is L E F T-skewed.
2010 Radha Bose FSU Department of Statistics
2
Binomial Distributions
Bernoulli Trials
Bernoulli trials are trials that satisfy the following three conditions:
~ there are only two possible outcomes at each trial (same two for all trials), called "success" and
"failure"
~ the probabilities of success and failure are the same for all trials
~ the trials are independent
THE BINOMIAL DISTRIBUTIONS
Bin(n , )
If X = the number of successes that occur in a fixed number of Bernoulli trials, then the appropriate
probability model for X is the Binomial model.
~ fixed number of trials
(n )
4 Binomial
~ only two possible outcomes at each trial (hence the name binomial)
Conditions
~ probability of success
( )
is the same at all trials,
hence probability of failure
(1 )
is also the same at all trials
~ trials are independent
~ model characterized by
n and
~ X is discrete with sample space S = {0, 1, 2, , n}
~ probabilities obtained by using binomial probability formula
P ( X = r ) = n Cr r (1 )
~ probability histograms are bell-shaped and exactly symmetric when
< 0.5
and left-skewed when
~ tail probabilities decrease as
nr
= 0.5 , right-skewed when
> 0.5 , skewness increases as
moves away from 0.5
n gets larger (tail area shrinks)
IDENTIFYING BINOMIAL SETTINGS
Identify which of the settings described below are Binomial. If a setting is not binomial, state why not
by choosing one or more of answer choices A, B, C or D. Answers on last page.
(A) the number of trials is not fixed
(B) there are more than two possible outcomes
(C) the probability of success is not the same for all trials
(D) the trials are not independent
2010 Radha Bose FSU Department of Statistics
3
Binomial Distributions
4
Note that, in some cases, even though a specific sample size, or a specific number of trials, is not
mentioned, the fact that it is a fixed number as opposed to a varying or infinite quantity is implied
in the story.
(i) You randomly select some cards (without replacement) from a full deck of cards and record the
number of hearts you get. This example credited to Sarah Wilson STA 2122-01 Su-A05.
(ii) Tallahassee Memorial Hospital keeps a record of the gender of the first baby born there each day.
We assume that gender determination occurs independently in humans, and the probability of a baby
being a boy is the same for all babies.
(iii) We have a stack consisting of several $1, $5, $10 and $20 bills. We draw a bill at random, record
whether it is $1 or not and put it back in the stack. We perform 30 such draws.
This example credited to Spencer Anderson, STA 2023-10 SP04.
(iv) You go to an orphanage one day and record whether or not each child has the flu, which is a contagious
infection.
(v) We have a stack consisting of several $1, $5, $10 and $20 bills. We draw a bill at random, record
what it is and put it back in the stack. We perform 30 such draws.
(vi) You toss 15 different coins (some fair, some biased) and record the number of heads that show up.
(vii) We have a big bowl full of assorted candy. We randomly select a piece of candy, see if it is Kit-Kat
or not, and then replace it in the bowl. We perform 15 such selections.
This example credited to Gabrielle Rey STA 2023-07 F04.
(viii) You survey the students in your class to see what letter grade they got on the last test. Assume
that the students in the class did independent, individual studying and test-taking.
(ix) You survey your immediate family members to see if they will attend your cousin Clementine's
wedding or not.
(x) Tallahassee Hospital keeps a record of the gender of the first baby born there each day last year.
We assume that gender determination occurs independently in humans, and the probability of a baby
being a boy is different among ethnic groups.
Can you think of a Binomial setting? Can you think of a setting that might be mistaken for a Binomial setting? If so, I would love to
hear about it. Please email me your ideas at the address given in our syllabus. Thanks!
2010 Radha Bose FSU Department of Statistics
Binomial Distributions
Here's another example to help you get a feel for the binomial probability formula.
Toss a coin 6 times and count the number of heads landed. Assume P(H) = 0.8.
X=
number
of
heads
in 6
tosses
Permutations of
Heads and Tails
0
TTTTTT
Calculating the probability
1 of 0.26 = 6C0 * (0.80 * 0.26)
HTTTTT
0.2 * 0.2 * 0.8 * 0.2 * 0.2 * 0.2
TTTHTT
0.2 * 0.2 * 0.2 * 0.8 * 0.2 * 0.2
TTTTHT
0.2 * 0.2 * 0.2 * 0.2 * 0.8 * 0.2
TTTTTH
1
0.2 * 0.8 * 0.2 * 0.2 * 0.2 * 0.2
TTHTTT
0.000064
0.8 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2
THTTTT
Probability
(do this
part on
your own)
0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.8
0.001536
6 of (0.8 * 0.25) = 6C1 * (0.81 * 0.25)
HTHTTT
2
15 of (0.82 * 0.24) = 6C2 * (0.82 * 0.24)
0.01536
20 of (0.83 * 0.23) = 6C3 * (0.83 * 0.23)
0.08192
15 of (0.84 * 0.22) = 6C4 * (0.84 * 0.22)
0.24576
etc
THTHTH
3
etc
THTHHH
4
etc
HTHHHH
5
6 of (0.85 * 0.2) = 6C5 * (0.85 * 0.21)
0.393216
1 of 0.86 = 6C6 * (0.86 * 0.20)
0.262144
etc
6
HHHHHH
Total
2010 Radha Bose FSU Department of Statistics
1
5
Binomial Distributions
6
NOTATION
Denotation / Interpretation
Symbol
n
the fixed number of trials
the probability of success at a single trial
r
a particular value of the binomial random variable X
Cr
binomial coefficient (the number of different ways we can arrange a total of
n
have r of one type and
n-r of another type)
BINOMIAL FORMULAS
(i) For r = 0, 1, ..., 2, n
P ( X = r ) = n Cr r (1 )
(ii) Mean or Expected Value
(iii) Standard deviation
n objects when we
= n
or
nr
E ( X ) = n
= n (1 )
If you use the nCr function on
your calculator, you will save on
keystrokes if you keep in mind
that the following hold for any
positive integer n:
nCn = nC0 = 1
nCn-1 = nC1 = n.
Most calculators have the nCr function, for example, on the TI-83/84:
10C7
= 10
Math
PRB
3:nCr
7
Enter = 120.
Try this now: compute the number of ways in which you can have 6 girls among 15 children. ________
If we wanted to write down the possible arrangements, they might look like this:
GGGGGGBBBBBBBBB
GGGGGBGBBBBBBBB
GGGGGBBGBBBBBBB
GGGGGBBBGBBBBBB
GGGGGBBBBGBBBBB
GGGGGBBBBBGBBBB
GGGGGBBBBBBGBBB
GGGGGBBBBBBBGBB
GGGGGBBBBBBBBGB
GGGGGBBBBBBBBBG
2010 Radha Bose FSU Department of Statistics
Binomial Distributions
7
1. It is estimated that about 24% of all drivers do not wear their seatbelts while driving. Assume
that drivers make their decisions independently. We randomly select 15 drivers and let X = the
number of drivers who are unbuckled.
X is Binomially distributed because:
~ there are a fixed number of trials,
n = _______________,
~ there are only two possible outcomes, __________________ and _________________,
~ the probability of success is the same for all drivers,
= __________,
~ trials are independent because drivers _______________________________________.
(a) Write down the distribution of X. ______________________________
(b) What is the sample space of X?
What is the probability we find:
(c) exactly 6 drivers unbuckled?
(d) 2 or 3 drivers unbuckled?
2010 Radha Bose FSU Department of Statistics
Binomial Distributions
(e) at least one driver unbuckled?
(f) at least two drivers unbuckled?
(g) (i) If we kept on randomly selecting groups of 15 drivers and recording how many in each
group were unbuckled, what would be the long run mean?
(g) (ii) What would be the long run standard deviation?
2010 Radha Bose FSU Department of Statistics
8
Binomial Distributions
2. It is estimated that about 15% of college students cohabit (live together in an emotional and
sexual relationship). Let us assume that, apart from yourself, there are 40 students in your
recitation class, and they form a random sample of college students. We will also assume that
the decision to cohabit or not was made independently by each student.
In the questions that follow, exclude yourself from the mix.
X = the number of students in your recitation class who cohabit is a Binomial random variable.
What is the probability that
(a) a quarter of your recitation class are cohabiters?
(b) at least one person in your recitation class is a cohabiter?
(c) at most two persons in your recitation class are cohabiters?
2010 Radha Bose FSU Department of Statistics
9
Binomial Distributions
10
APPROXIMATING BINOMIAL PROBABILITIES USING A NORMAL MODEL
~ a Binomial model may be approximated by a Normal model when both n 10 and
~ it is sufficient to check that the smaller of
n (1 ) 10
n and n (1 ) is greater than or equal to 10
~ the continuity correction is used because Normal is continuous whereas Binomial is discrete
~ the Normal approximation becomes more accurate as n gets larger and as
~ the appropriate Normal model is
(
N = n , = n (1 )
~ z-scores are calculated in the usual way
points
z=
x
, with
moves closer to 0.5
)
and
as given above, and with the cutoff
x continuity-corrected first
Continuity Correction: an integer
r+0.5
r occupies the entire interval from r0.5 to r+0.5
(i)
Binomial P( X r ) Normal P( X < r + 0.5)
(ii)
Binomial P( X r ) Normal P( X > r 0.5)
(iii)
Binomial P(r1 X r2 ) Normal P r1 0.5 < X < r2 + 0.5
(
)
3. It is believed that 62% of all Americans listen to National Public Radio (NPR) on a regular
basis. We randomly select 500 Americans and let X = the number of Americans who listen to
NPR on a regular basis.
(a) Write down the Binomial distribution of X. ______________________________
(b) Show why, in this setting, it is appropriate to approximate Binomial probabilities using a
Normal model.
2010 Radha Bose FSU Department of Statistics
Binomial Distributions
(c) Write down the appropriate Normal model.
(d) Use your Normal model to find the probability that there would be between 325 and 335
(inclusive) who listen to NPR on a regular basis in a random sample of 500 Americans.
(e) Use your Normal model to find the probability that if we select a random sample of 500
Americans, we will find that 280 or more of them listen to NPR on a regular basis.
(f) Use your Normal model to find the probability that if we select a random sample of 500
Americans, we will find that three-fifths or less of them listen to NPR on a regular basis.
2010 Radha Bose FSU Department of Statistics
11
Binomial Distributions
12
ANSWERS
(i) C , D
C because after you remove a card from the deck, the total number of cards remaining decreases
and the number of cards of the same type as the one that was removed also decreases, so the
probability of drawing any particular type of card changes with successive drawings.
D because probabilities change, so trials are not independent.
(ii) A
The hospital is recording every day, forever, so the number of trials is not fixed.
(iii) Binomial
(iv) C , D
C because some children may be more susceptible than others so the probability of getting the flu
may change from child to child.
D because the flu is contagious - if someone you spend a lot of time with has it, you are more
likely to catch it than if you did not spend time around someone who has it.
(v) B
There are four possible outcomes if you are recording exactly what bill you draw.
(vi) C
Since you are tossing different coins, the probability of getting a Head is not the same for all
coins.
(vii) Binomial
(viii) B , C
B because there are more than two possible letter grades.
C because some students may be stronger than others in that subject, so the probabilities of
getting each type of letter grade are likely to be different from student to student.
(ix) C , D
C because schedules are likely to be different from person to person, so some family members
may have a better opportunity (example, more free time) to attend the wedding than others.
Also because some family members may be closer to Clementine and so they are more likely to
attend the wedding than others who may not be that close to her.
D because a family member's plans may depend on another family member's plans (example,
children may have no choice but to accompany their parents).
(x) C
The probability of being a boy (hence also the probability of being a girl) is different for babies of
different ethnic groups, so the probabilities of the two outcomes are not the same for all trials.
2010 Radha Bose FSU Department of Statistics