Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Definition
The probability distribution or probability mass function (pmf)
of a discrete rv is defined for every number x by
p(x) = P(X = x) = P(all s ∈ S : X (s) = x).
In words, for every possible value x of the random variable, the
pmf specifies the probability of observing that value when the
experiment
is performed. (The conditions p(x) ≥ 0 and
P
all possible x p(x) = 1 are required for any pmf.)
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Definition
The cumulative distribution function (cdf) F (x) of a discrete rv
X with pmf p(x) is defined for every number x by
X
F (x) = P(X ≤ x) =
p(y )
y :y ≤x
For any number x, F(x) is the probability that the observed value
of X will be at most x.
Probability Distributions for Discrete RV
Definition
The cumulative distribution function (cdf) F (x) of a discrete rv
X with pmf p(x) is defined for every number x by
X
F (x) = P(X ≤ x) =
p(y )
y :y ≤x
For any number x, F(x) is the probability that the observed value
of X will be at most x.
F (x) = P(X ≤ x) = P(X is less than or equal to x)
p(x) = P(X = x) = P(X is exactly equal to x)
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
pmf =⇒ cdf:
F (x) = P(X ≤ x) =
X
y :y ≤x
p(y )
Probability Distributions for Discrete RV
pmf =⇒ cdf:
F (x) = P(X ≤ x) =
X
y :y ≤x
It is also possible cdf =⇒ pmf:
p(y )
Probability Distributions for Discrete RV
pmf =⇒ cdf:
F (x) = P(X ≤ x) =
X
p(y )
y :y ≤x
It is also possible cdf =⇒ pmf:
p(x) = F (x) − F (x−)
where “x−” represents the largest possible X value that is strictly
less than x.
Probability Distributions for Discrete RV
Probability Distributions for Discrete RV
Proposition
For any two numbers a and b with a ≤ b,
P(a ≤ X ≤ b) = F (b) − F (a−)
where “a−” represents the largest possible X value that is strictly
less than a. In particular, if the only possible values are integers
and if a and b are integers, then
P(a ≤ X ≤ b) = P(X = a or a + 1 or . . . or b)
= F (b) − F (a − 1)
Taking a = b yields P(X = a) = F (a) − F (a − 1) in this case.
Expectations
Expectations
Definition
Let X be a discrete rv with set of possible values D and pmf p(x).
The expected value or mean value of X , denoted by E (X ) or
µX , is
X
E (X ) = µX =
x · p(x)
x∈D
Expectations
Definition
Let X be a discrete rv with set of possible values D and pmf p(x).
The expected value or mean value of X , denoted by E (X ) or
µX , is
X
E (X ) = µX =
x · p(x)
x∈D
e.g (Problem 30)
A group of individuals who have automobile insurance from a
certain company is randomly selected. Let Y be the number of
moving violations for which the individual was cited during the last
3 years. The pmf of Y is
y
0
1
2
3
Then the expected value of
p(y) 0.60 0.25 0.10 0.05
moving violations for that group is
µY = E (Y ) = 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05 = 0.60
Expectations
Expectations
y
0
1
2
3
p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
Expectations
y
0
1
2
3
p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ=
0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
= 0.60
100
Expectations
y
0
1
2
3
p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ=
0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
= 0.60
100
60
25
10
5
+1·
+2·
+3·
100
100
100
100
= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
µ=0·
= 0.60
Expectations
y
0
1
2
3
p(y) 0.60 0.25 0.10 0.05
Assume the total number of individuals in that group is 100, then there
are 60 individuals without moving violation, 25 with 1 moving violation,
10 with 2 moving violations and 5 with 3 moving violations.
The population mean is calculated as
µ=
0 · 60 + ·1 · 25 + ·2 · 10 + 3 · 5
= 0.60
100
60
25
10
5
+1·
+2·
+3·
100
100
100
100
= 0 · 0.60 + 1 · 0.25 + 2 · 0.10 + 3 · 0.05
µ=0·
= 0.60
The population size is irrevelant if we know the pmf!
Expectations
Expectations
Examples:
Let X be a Bernoulli rv with pmf


1 − p x = 0
p(x) = p
x =1


0
x 6= 0, or 1
Expectations
Examples:
Let X be a Bernoulli rv with pmf


1 − p x = 0
p(x) = p
x =1


0
x 6= 0, or 1
Then the expected value for X is
E (X ) = 0 · p(0) + 1 · p(1) = p
Expectations
Examples:
Let X be a Bernoulli rv with pmf


1 − p x = 0
p(x) = p
x =1


0
x 6= 0, or 1
Then the expected value for X is
E (X ) = 0 · p(0) + 1 · p(1) = p
We see that the expected value of a Bernoulli rv X is just the
probability that X takes on the value 1.
Expectations
Expectations
Examples:
Consider the cards drawing example again and assume we have
infinitely many cards this time. Let X = the number of drawings
until we get a ♠. If the probability for getting a ♠ is α, then the
pmf for X is
(
α(1 − α)x−1 x = 1, 2, 3, . . .
p(x) =
0
otherwise
Expectations
Examples:
Consider the cards drawing example again and assume we have
infinitely many cards this time. Let X = the number of drawings
until we get a ♠. If the probability for getting a ♠ is α, then the
pmf for X is
(
α(1 − α)x−1 x = 1, 2, 3, . . .
p(x) =
0
otherwise
The expected value for X is
E (X ) =
X
D
x · p(x) =
∞
X
x=1
xα(1 − α)x−1 = α
∞
X
d
[− (1 − α)x ]
dα
x=1
Expectations
Examples:
Consider the cards drawing example again and assume we have
infinitely many cards this time. Let X = the number of drawings
until we get a ♠. If the probability for getting a ♠ is α, then the
pmf for X is
(
α(1 − α)x−1 x = 1, 2, 3, . . .
p(x) =
0
otherwise
The expected value for X is
E (X ) =
X
x · p(x) =
D
∞
X
x=1
∞
E (X ) = α{−
xα(1 − α)x−1 = α
∞
X
d
[− (1 − α)x ]
dα
x=1
d X
d 1−α
1
[ (1 − α)x ]} = α{− (
)} =
dα
dα α
α
x=1
Expectations
Expectations
Examples 3.20
Let X be the number of interviews a student has prior to getting a
job. The pmf for X is
(
k
x = 1, 2, 3, . . .
2
p(x) = x
0 otherwise
P
2
wherePk is chosen so that ∞
x=1 (k/x ) = 1. (It can be showed
∞
2
that x=1 (1/x ) < ∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =
∞
X
x=1
∞
X1
k
= ∞!
x· 2 =k
x
x
The expected value is NOT finite!
x=1
Expectations
Examples 3.20
Let X be the number of interviews a student has prior to getting a
job. The pmf for X is
(
k
x = 1, 2, 3, . . .
2
p(x) = x
0 otherwise
P
2
wherePk is chosen so that ∞
x=1 (k/x ) = 1. (It can be showed
∞
2
that x=1 (1/x ) < ∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =
∞
X
x=1
∞
X1
k
= ∞!
x· 2 =k
x
x
The expected value is NOT finite!
Heavy Tail:
x=1
Expectations
Examples 3.20
Let X be the number of interviews a student has prior to getting a
job. The pmf for X is
(
k
x = 1, 2, 3, . . .
2
p(x) = x
0 otherwise
P
2
wherePk is chosen so that ∞
x=1 (k/x ) = 1. (It can be showed
∞
2
that x=1 (1/x ) < ∞, which implies that such a k exists.)
The expected value of X is
µ = E (X ) =
∞
X
x=1
∞
X1
k
= ∞!
x· 2 =k
x
x
x=1
The expected value is NOT finite!
Heavy Tail: distribution with a large amount of probability far
from µ
Expectations
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 dollars, would
you accept the guaranteed amount or would you gamble?
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 dollars, would
you accept the guaranteed amount or would you gamble?
x
1 2 3 4 5 6
p(x) 16 16 16 16 16 16
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 dollars, would
you accept the guaranteed amount or would you gamble?
x
1 2 3 4 5 6
p(x) 16 16 16 16 16 16
1
1 12 13 14 15 16
x
Expectations
Example (Problem 38)
Let X = the outcome when a fair die is rolled once. If before the
1
die is rolled you are offered either 3.5
dollars or X1 dollars, would
you accept the guaranteed amount or would you gamble?
x
1 2 3 4 5 6
p(x) 16 16 16 16 16 16
1
1 12 13 14 15 16
x
Then the expected dollars from gambling is
6
E(
X1
1
1
)=
· p( )
X
x
x
x=1
1 1 1
1 1
+ · + ··· + ·
6 2 6
6 6
49
1
=
>
120
3.5
=1·
Expectations
Expectations
Proposition
If the rv X has a set of possible values D and pmf p(x), then the
expected value of any function h(X ), denoted by E [h(X )] or µhX ,
is computed by
X
E [h(X )] =
h(x) · p(x)
D
Expectations
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Let X denote the number of computers sold, and suppose that
p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.
Let h(X ) denote the profit associated with selling X units,
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Let X denote the number of computers sold, and suppose that
p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.
Let h(X ) denote the profit associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
Expectations
Example 3.23
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece.
Let X denote the number of computers sold, and suppose that
p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4.
Let h(X ) denote the profit associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
The expected profit is
E [h(X )] = h(0) · p(0) + h(1) · p(1) + h(2) · p(2) + h(3) · p(3)
= (−900)(0.1) + (−100)(0.2) + (700)(0.3) + (1500)(0.4)
= 700
Expectations
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + b.)
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X ) − 900 = 700
Expectations
Proposition
E (aX + b) = a · E (X ) + b
(Or, using alternative notation, µaX +b = a · µX + b.)
e.g. for the previous example,
E [h(X )] = E (800X − 900) = 800 · E (X ) − 900 = 700
Corollary
1. For any constant a, E (aX ) = a · E (X ).
2. For any constant b, E (X + b) = E (X ) + b.
Expectations
Expectations
Definition
Let X have pmf p(x) and expected value µ. Then the variance of
X, denoted by V (X ) or σX2 , or just σX2 , is
V (X ) =
X
(x − µ)2 · p(x) = E [(X − µ)2 ]
D
The stand deviation (SD) of X is
q
σX = σX2
Expectations
Expectations
Example:
For the previous example, the pmf is given as
0
1
2
3
x
p(x) 0.1 0.2 0.3 0.4
Expectations
Example:
For the previous example, the pmf is given as
0
1
2
3
x
p(x) 0.1 0.2 0.3 0.4
then the variance of X is
2
V (X ) = σ =
3
X
x=0
2
(x − 2)2 · p(x)
= (0 − 2) (0.1) + (1 − 2)2 (0.2) + (2 − 2)2 (0.3) + (3 − 2)2 (0.4)
=1
Expectations
Expectations
Recall that for sample variance s 2 , we have
Sxx
=
s2 =
n−1
P
xi )2
n
P
xi2 − (
n−1
Expectations
Recall that for sample variance s 2 , we have
Sxx
=
s2 =
n−1
P
xi )2
n
P
xi2 − (
n−1
Proposition
X
V (X ) = σ 2 = [
x 2 · p(x)] − µ2 = E (X 2 ) − [E (X )]2
D
Expectations
Recall that for sample variance s 2 , we have
Sxx
=
s2 =
n−1
P
xi )2
n
P
xi2 − (
n−1
Proposition
X
V (X ) = σ 2 = [
x 2 · p(x)] − µ2 = E (X 2 ) − [E (X )]2
D
e.g. for the previous example, the pmf is given as
x
0
1
2
3
p(x) 0.1 0.2 0.3 0.4
Then
V (X ) = E (X 2 ) − [E (X )]2 = 12 · 0.2 + 22 · 0.3 + 32 · 0.4 − (2)2 = 1
Expectations
Expectations
Proposition
If h(X ) is a function of a rv X , then
X
2
V [h(X )] = σh(X
{h(x)−E [h(X )]}2 ·p(x) = E [h(X )2 ]−{E [h(X )]}2
) =
D
If h(X ) is linear, i.e. h(X ) = aX + b for some nonrandom constant
a and b, then
2
2
2
V (aX + b) = σaX
+b = a · σX and σaX +b =| a | ·σX
In particular,
σaX =| a | ·σX , σX +b = σX
Expectations
Expectations
Example 3.23 continued
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece. Let X denote the number
of computers sold, and suppose that p(0) = 0.1,
p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(X ) denote the profit
associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
Expectations
Example 3.23 continued
A computer store has purchased three computers of a certain type
at $500 apiece. It will sell them for $1000 apiece. The
manufacturer has agreed to repurchase any computers still unsold
after a specified period at $200 apiece. Let X denote the number
of computers sold, and suppose that p(0) = 0.1,
p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(X ) denote the profit
associated with selling X units, then
h(X ) = revenue − cost
= 1000X + 200(3 − X ) − 1500 = 800X − 900.
The variance of h(X ) is
V [h(X )] = V [800X − 900]
= 8002 V [X ]
= 640, 000
And the SD is σh(X ) =
p
V [h(X )] = 800.
Binomial Distribution
Binomial Distribution
1. The experiment consists of a sequence of n smaller experiments
called trials, where n is fixed in advance of the experiment;
Binomial Distribution
1. The experiment consists of a sequence of n smaller experiments
called trials, where n is fixed in advance of the experiment;
2. Each trial can result in one of the same two possible outcomes
(dichotomous trials), which we denote by success (S) and failure
(F );
Binomial Distribution
1. The experiment consists of a sequence of n smaller experiments
called trials, where n is fixed in advance of the experiment;
2. Each trial can result in one of the same two possible outcomes
(dichotomous trials), which we denote by success (S) and failure
(F );
3. The trials are independent, so that the outcome on any
particular trial dose not influence the outcome on any other trial;
Binomial Distribution
1. The experiment consists of a sequence of n smaller experiments
called trials, where n is fixed in advance of the experiment;
2. Each trial can result in one of the same two possible outcomes
(dichotomous trials), which we denote by success (S) and failure
(F );
3. The trials are independent, so that the outcome on any
particular trial dose not influence the outcome on any other trial;
4. The probability of success is constant from trial; we denote this
probability by p.
Binomial Distribution
1. The experiment consists of a sequence of n smaller experiments
called trials, where n is fixed in advance of the experiment;
2. Each trial can result in one of the same two possible outcomes
(dichotomous trials), which we denote by success (S) and failure
(F );
3. The trials are independent, so that the outcome on any
particular trial dose not influence the outcome on any other trial;
4. The probability of success is constant from trial; we denote this
probability by p.
Definition
An experiment for which Conditions 1 — 4 are satisfied is called a
binomial experiment.
Binomial Distribution
Binomial Distribution
Examples:
1. If we toss a coin 10 times, then this is a binomial experiment
with n = 10, S = Head, and F = Tail.
Binomial Distribution
Examples:
1. If we toss a coin 10 times, then this is a binomial experiment
with n = 10, S = Head, and F = Tail.
2. If we draw a card from a deck of well-shulffed cards with
replacement, do this 5 times and record whether the outcome is ♠
or not, then this is also a binomial experiment. In this case, n = 5,
S = ♠ and F = not ♠.
Binomial Distribution
Examples:
1. If we toss a coin 10 times, then this is a binomial experiment
with n = 10, S = Head, and F = Tail.
2. If we draw a card from a deck of well-shulffed cards with
replacement, do this 5 times and record whether the outcome is ♠
or not, then this is also a binomial experiment. In this case, n = 5,
S = ♠ and F = not ♠.
3. Again we draw a card from a deck of well-shulffed cards but
without replacement, do this 5 times and record whether the
outcome is ♠ or not. However this time it is NO LONGER a
binomial experiment.
Binomial Distribution
Examples:
1. If we toss a coin 10 times, then this is a binomial experiment
with n = 10, S = Head, and F = Tail.
2. If we draw a card from a deck of well-shulffed cards with
replacement, do this 5 times and record whether the outcome is ♠
or not, then this is also a binomial experiment. In this case, n = 5,
S = ♠ and F = not ♠.
3. Again we draw a card from a deck of well-shulffed cards but
without replacement, do this 5 times and record whether the
outcome is ♠ or not. However this time it is NO LONGER a
binomial experiment.
P(♠ on second | ♠ on first) =
12
= 0.235 6= 0.25 = P(♠ on second)
51
We do not have independence here!
Binomial Distribution
Binomial Distribution
Examples:
4. This time we draw a card from 100 decks of well-shulffed cards
without replacement, do this 5 times and record whether the
outcome is ♠ or not. Is it a binomial experiment?
Binomial Distribution
Examples:
4. This time we draw a card from 100 decks of well-shulffed cards
without replacement, do this 5 times and record whether the
outcome is ♠ or not. Is it a binomial experiment?
1299
= 0.2499 ≈ 0.25
5199
1295
P(♠ on sixth draw | ♠ on first five draw) =
= 0.2492 ≈ 0.25
5195
1300
P(♠ on tenth draw | not ♠ on first nine draw) =
= 0.2504 ≈ 0.25
5191
...
P(♠ on second draw | ♠ on first draw) =
Although we still do not have independence, the conditional
probabilities differ so slightly that we can regard these trials as
independent with P(♠) = 0.25.
Binomial Distribution
Binomial Distribution
Rule
Consider sampling without replacement from a dichotomous
population of size N. If the sample size (number of trials) n is at
most 5% of the population size, the experiment can be analyzed as
though it wre exactly a binomial experiment.
Binomial Distribution
Rule
Consider sampling without replacement from a dichotomous
population of size N. If the sample size (number of trials) n is at
most 5% of the population size, the experiment can be analyzed as
though it wre exactly a binomial experiment.
e.g. for the previous example, the population size is N = 5200 and
the sample size is n = 5. We have Nn ≈ 0.1%. So we can apply the
above rule.
Binomial Distribution
Binomial Distribution
Definition
The binomial random variable X associated with a binomial
experiment consisting of n trials is defined as
X = the number of S’s among the n trials
Binomial Distribution
Definition
The binomial random variable X associated with a binomial
experiment consisting of n trials is defined as
X = the number of S’s among the n trials
Possible values for X in an n-trial experiment are x = 0, 1, 2, . . . , n.
Binomial Distribution
Definition
The binomial random variable X associated with a binomial
experiment consisting of n trials is defined as
X = the number of S’s among the n trials
Possible values for X in an n-trial experiment are x = 0, 1, 2, . . . , n.
Notation
We use X ∼ Bin(n, p) to indicate that X is a binomial rv based on
n trials with success probability p.
We use b(x; n, p) to denote the pmf of X , and B(x; n, p) to denote
the cdf of X , where
B(x; n, p) = P(X ≤ x) =
x
X
y =0
b(x; n, p)
Binomial Distribution
Binomial Distribution
Example:
Assume we toss a coin 3 times and the probability for getting a
head for each toss is p. Let X be the binomial random variable
associated with this experiment. We tabulate all the possible
outcomes, corresponding X values and probabilities in the
following table:
Outcome
HHH
HHT
HTH
HTT
X
3
2
2
1
Probability
p3
p 2 · (1 − p)
p 2 · (1 − p)
p · (1 − p)2
Outcome
TTT
TTH
THT
THH
X
0
1
1
2
Probability
(1 − p)3
(1 − p)2 · p
(1 − p)2 · p
(1 − p) · p 2
Binomial Distribution
Example:
Assume we toss a coin 3 times and the probability for getting a
head for each toss is p. Let X be the binomial random variable
associated with this experiment. We tabulate all the possible
outcomes, corresponding X values and probabilities in the
following table:
Outcome
HHH
HHT
HTH
HTT
X
3
2
2
1
Probability
p3
p 2 · (1 − p)
p 2 · (1 − p)
p · (1 − p)2
Outcome
TTT
TTH
THT
THH
X
0
1
1
2
Probability
(1 − p)3
(1 − p)2 · p
(1 − p)2 · p
(1 − p) · p 2
e.g. b(2; 3, p) = P(HHT ) + P(HTH) + P(THH) = 3p 2 (1 − p).
Binomial Distribution
Binomial Distribution
More generally, for the binomial pmf b(x; n, p), we have
number of sequences of
probability of any
b(x; n, p) =
·
length n consisting of x S’s
particular such sequence
Binomial Distribution
More generally, for the binomial pmf b(x; n, p), we have
number of sequences of
probability of any
b(x; n, p) =
·
length n consisting of x S’s
particular such sequence
n
number of sequences of
=
and
length n consisting of x S’s
x
probability of any
= p x (1 − p)n−x
particular such sequence
Binomial Distribution
More generally, for the binomial pmf b(x; n, p), we have
number of sequences of
probability of any
b(x; n, p) =
·
length n consisting of x S’s
particular such sequence
n
number of sequences of
=
and
length n consisting of x S’s
x
probability of any
= p x (1 − p)n−x
particular such sequence
Theorem
( n
b(x; n, p) =
x
0
p x (1 − p)n−x
x = 0, 1, 2, . . . , n
otherwise
Binomial Distribution
Binomial Distribution
Example: (Problem 55)
Twenty percent of all telephones of a certain type are submitted
for service while under warranty. Of these, 75% can be repaired,
whereas the other 25% must be replaced with new units. if a
company purchases ten of these telephones, what is the probability
that exactly two will end up being replaced under warranty?
Binomial Distribution
Example: (Problem 55)
Twenty percent of all telephones of a certain type are submitted
for service while under warranty. Of these, 75% can be repaired,
whereas the other 25% must be replaced with new units. if a
company purchases ten of these telephones, what is the probability
that exactly two will end up being replaced under warranty?
Let X = number of telephones which need replace. Then
p = P(service and replace) = P(replace | service)·P(service) = 0.25·0.2 =
Binomial Distribution
Example: (Problem 55)
Twenty percent of all telephones of a certain type are submitted
for service while under warranty. Of these, 75% can be repaired,
whereas the other 25% must be replaced with new units. if a
company purchases ten of these telephones, what is the probability
that exactly two will end up being replaced under warranty?
Let X = number of telephones which need replace. Then
p = P(service and replace) = P(replace | service)·P(service) = 0.25·0.2 =
Now,
10
P(X = 2) = b(2; 10, 0.05) =
0.052 (1 − 0.05)10−2 = 0.0746
2
Binomial Distribution
Binomial Distribution
Binomial Tables
Table A.1 Cumulative Binomial
Probabilities (Page 664)
P
B(x; n, p) = xy =0 b(x; n, p) . . .
b. n = 10
p
0.01 0.05 0.10 . . .
0
.904 .599 .349 . . .
1
.996 .914 .736 . . .
2 1.000 .988 .930 . . .
3 1.000 .999 .987 . . .
...
...
... ...
Binomial Distribution
Binomial Tables
Table A.1 Cumulative Binomial
Probabilities (Page 664)
P
B(x; n, p) = xy =0 b(x; n, p) . . .
b. n = 10
p
0.01 0.05 0.10 . . .
0
.904 .599 .349 . . .
1
.996 .914 .736 . . .
2 1.000 .988 .930 . . .
3 1.000 .999 .987 . . .
...
...
... ...
Then for b(2; 10, 0.05), we have
b(2; 10, 0.05) = B(2; 10, 0.05)−B(1; 10, 0.05) = .988−.914 = .074
Binomial Distribution
Binomial Distribution
Mean and Variance
Theorem
If X ∼ Bin(n, p), then E (X ) = np, V (X ) = np(1 − p) = npq, and
√
σX = npq (where q = 1 − p).
Binomial Distribution
Mean and Variance
Theorem
If X ∼ Bin(n, p), then E (X ) = np, V (X ) = np(1 − p) = npq, and
√
σX = npq (where q = 1 − p).
P
The idea is that X = ni=1 Y1 + Y2 + · · · + Yn , where Yi ’s are
independent Bernoulli random variable with probability p for
one outcome, i.e.
(
1, with probabilityp
Y =
0, with probability1 − p
Binomial Distribution
Mean and Variance
Theorem
If X ∼ Bin(n, p), then E (X ) = np, V (X ) = np(1 − p) = npq, and
√
σX = npq (where q = 1 − p).
P
The idea is that X = ni=1 Y1 + Y2 + · · · + Yn , where Yi ’s are
independent Bernoulli random variable with probability p for
one outcome, i.e.
(
1, with probabilityp
Y =
0, with probability1 − p
E (Y ) = p and V (Y ) = (1 − p)2 p + (−p)2 (1 − p) = p(1 − p).
Binomial Distribution
Mean and Variance
Theorem
If X ∼ Bin(n, p), then E (X ) = np, V (X ) = np(1 − p) = npq, and
√
σX = npq (where q = 1 − p).
P
The idea is that X = ni=1 Y1 + Y2 + · · · + Yn , where Yi ’s are
independent Bernoulli random variable with probability p for
one outcome, i.e.
(
1, with probabilityp
Y =
0, with probability1 − p
E (Y ) = p and V (Y ) = (1 − p)2 p + (−p)2 (1 − p) = p(1 − p).
Therefore E (X ) = np and V (X ) = np(1 − p) = npq.
Binomial Distribution
Binomial Distribution
Example: (Problem 60)
A toll bridge charges $1.00 for passenger cars and $2.50 for other
vehicles. Suppose that during daytime hours, 60% of all vehicles
are passenger cars. If 25 vehicles cross the bridge during a
particular daytime period, what is the resulting expected toll
revenue? What is the variance
Binomial Distribution
Example: (Problem 60)
A toll bridge charges $1.00 for passenger cars and $2.50 for other
vehicles. Suppose that during daytime hours, 60% of all vehicles
are passenger cars. If 25 vehicles cross the bridge during a
particular daytime period, what is the resulting expected toll
revenue? What is the variance
Let X = the number of passenger cars and Y = revenue. Then
Y = 1.00X + 2.50(25 − X ) = 62.5 − 1.50X .
Binomial Distribution
Example: (Problem 60)
A toll bridge charges $1.00 for passenger cars and $2.50 for other
vehicles. Suppose that during daytime hours, 60% of all vehicles
are passenger cars. If 25 vehicles cross the bridge during a
particular daytime period, what is the resulting expected toll
revenue? What is the variance
Let X = the number of passenger cars and Y = revenue. Then
Y = 1.00X + 2.50(25 − X ) = 62.5 − 1.50X .
E (Y ) = E (62.5−1.5X ) = 62.5−1.5E (X ) = 62.5−1.5·(25·0.6) = 40
Binomial Distribution
Example: (Problem 60)
A toll bridge charges $1.00 for passenger cars and $2.50 for other
vehicles. Suppose that during daytime hours, 60% of all vehicles
are passenger cars. If 25 vehicles cross the bridge during a
particular daytime period, what is the resulting expected toll
revenue? What is the variance
Let X = the number of passenger cars and Y = revenue. Then
Y = 1.00X + 2.50(25 − X ) = 62.5 − 1.50X .
E (Y ) = E (62.5−1.5X ) = 62.5−1.5E (X ) = 62.5−1.5·(25·0.6) = 40
V (Y ) = V (62.5−1.5X ) = (−1.5)2 V (X ) = 2.25·(25·0.6·0.4) = 13.5