Transcript Probability Distributions

Probability Distributions
W&W Chapter 4
Discrete Random Variables
Suppose a couple plan to have 3 children
and are interested in the number of girls
they might have. This is an example of a
random variable, and is denoted by a
capital letter:
X = the number of girls
The possible values of X are 0, 1, 2, and 3;
however they are not equally likely.
Discrete Random Variables
We need to calculate
X
Pr(X)
0
1
2
3
Discrete Random Variables
Using Pr(boy)=.52 and Pr(girl)=.48
e
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
Pr(e)
.14
.13
.13
.12
.13
.12
.12
.11
x
0
p(x)
.14
1
.39 (.13+.13+.13)
2
.36 (.12+.12+.12)
3
.11
Pr(x)=1
Discrete Random Variables
A discrete random variable takes on
various values x with probabilities
specified by its probability
distribution, p(x).
Graphical Representation
p(x)
.4
.3
.2
.1
0
1
x
2
3
Example
What is the probability of fewer than
two girls?
Pr(X<2) = p(0) + p(1)
= .14 + 39 = .53
Notation
X: random variable
x: a specific value that X may take
p(0), p(1),..p(x) are the probabilities of x
Example: the probability of having one girl
in a family of three,
Pr(X=1) or just p(1)
A random variable can be discrete or
continuous.
Mean and Variance
Previously we learned how to
calculate the mean and variance for
a sample as follows:
Xbar = X/N
s2 = (X- Xbar)2/(N-1)
Population Mean and
Variance
We can calculate the mean and variance of
a random variable from its probability
distribution, p(x):
Mean:  = xp(x)
Variance: 2 = (x- )2p(x)
Remember that Greek letters denote
population statistics!
Variance
We can rewrite the formula for variance as follows:
2 = x2p(x) - 2
Start with 2 = (x- )2p(x)
=  (x2 - 2x + 2) p(x)
and noting that  is a constant:
2 =  x2p(x) - 2 xp(x) + 2 p(x)
Since xp(x) =  and p(x) = 1,
2 =  x2p(x) - 2() + 2(1)
=  x2p(x) - 2
Example
Let’s calculate the mean and variance of
the random variable X, the number of
girls
Mean
 = xp(x) = (0)(.14) + (1)(.39) + (2)(.36)
+ (3)(.11) = 1.44
2 = (x- )2p(x) = (0-1.44)2(.14)+ (11.44)2(.39) + (2-1.44)2(.36) + (31.44)2(.11) = 0.7464
Interpretation
The mean number of girls in a family of 3 is
1.44 and the variance is about .75.
Notice that 1.44/3 = .48, which is the
relative frequency (f/n) for girls!
 and 2 have similar interpretations to the
sample mean and variance.
 is a weighted average using probability
weights rather than relative frequency
weights, and the standard deviation () is
the typical deviation
Factorials
Question: Suppose you have 3 shirts,
2 sweaters, and 2 pairs of pants.
How many outfits can you form?
If we imagine a decision tree, we will
find that the answer is 12.
This can be derived by 322 = 12
Factorials (continued)
Rule of counting: A number of
multiple choices are to be made.
There are m1 possibilities for the
first choice, m2 for second, and so
on. If these choices can be
combined freely, then the total
number of possibilities for the whole
set of choices is m1m2m3…
Factorials (continued)
Suppose you have a survey questionnaire
with n questions. How many ways are
there to order the n questions?
There are n ways to choose the first
question, but after deciding this one,
there are only n-1 ways to choose the
second, n-2 ways to choose the third and
so on.
Thus the number is n(n-1)(n-2)21, which
we call n factorial, or n! for short.
The Binomial Distribution
There are many types of discrete
random variables and the most
common is called the binomial. The
classical example of a binomial
variable is:
S = number of heads in several tosses
of a coin
Assumptions of the
Binomial Distribution
1) We suppose there are n trials (tosses of
the coin)
2) In each trial, a certain event of interest
can occur or fail to occur; then we say a
success (head) or failure (tail) has
occurred. Their respective probabilities
are  and 1 - .
Assumptions (Continued)
3) We assume the trials are
statistically independent (remember
this means that the chances of
getting a head on one flip are not
influenced by getting a head or tail
on a previous flip).
4) S is the total number of successes
in n trials, and is called a binomial
variable.
Examples of Binomial
Variables
Trial
Success
Tossing a coin
Head
Failure
Tail

½
n
# tosses
S
# heads
Birth of a child
Girl
Boy
.48 # children
Multiple Choice
Correct
Wrong
1/5 # questions # correct
Drawing a voter Rep.
Dem/Other
# girls
f/N # surveyed # Rep.
Probability Distribution for a
Binomial Variable
p(s) =
( n ) s (1 - )n-s
(s)
where ( n ) = n!/[s!(n-s)!]
(s)
and the factorial n! is given by
n! = n(n-1)(n-2)1
Example of the Binomial
Recall that the probability of 1 girl, or
p(1) in a family of 3 children was
.39. We can demonstrate that the
binomial produces the same result.
p(1) = (3)(.48)1(.52)3-1 =
(1)
p(1) = (321)/[1(21)](.48)(.2704)
= 3(.129792) = .39
Another Example
Suppose we want to know if the chances for
women receiving tenure at FSU are fair, so in
this case S = number of women who receive
tenure at FSU in a given year. We assume that if
everyone has an equal chance for tenure, then
the proportion of women that have tenure
should be close to the proportion of women
hired as assistant professors. We collect this
information for 15 years and determine that:
 = .4 and 1 -  = .6
Example (continued)
We count the number of tenured faculty by gender and come
up with the following data: #female = 25 and #male = 75.
What is the probability that the tenure process is fair?
S = 25 females tenured
p(s) = (100)(.4)25(.6)75
(25)
p(s) = .0006
We conclude that if the process were fair, the chances of
getting only 25% of women tenured given hiring rates is
highly unlikely.
Sampling from a large
population
Recall the example of light bulbs which
demonstrated how sampling without
replacement can change the probability
for successive draws. If we draw one
card out of a deck of cards, the
probability for getting a particular card on
the second draw changes because we
have removed the first card. But in really
large populations, we can act as though
the removal does not matter.
Example
Suppose that a production run of
40,000 microwave ovens includes
32,000 (80%) that are flawless. But
the quality control department, not
knowing this figure, takes a random
sample of 10 to estimate the overall
quality. What is the chance that the
sample will be evenly split, 5
flawless and 5 not?
Example
Each of the 10 successive ovens in the sample can be
considered a trial, so n = 10. Now in this case, removing
one good oven will change the probability of getting a
good one on the next draw (even though the binomial
assumes independence). For the first oven, the probability
of success (flawless) is 32,000/40,000 = .8. If the first
oven was a success, then the probability of success is
31,999/39,999; if it was a failure, then the probability of
success on the second draw is 32,000/39,999. But this
comes out very close to .8. So the second trial is
practically independent of the first, and we can use the
binomial.
Example
p(5) = (10)(.80)5(.20)5
(5)
= 252(.000105) = .026
That is, in a random sample of 10 ovens, there is close to a
3% chance that 5 will be flawless and 5 will not.
We must emphasize the most important assumption of the
binomial distribution, which is that the trials are
independent. For smaller samples where the trials are
dependent on each other, the binomial would not be
appropriate.