Download STAT 380 Some Discrete Probability Distributions I. Binomial

STAT 380
Some Discrete Probability Distributions
I. Binomial Distribution
1. Properties of a binomial experiment
(a) The experiment consists of n repeated trials or attempts.
(b) Each trial results in an outcome that may be classified as “success” or
“failure”.
(c) The probability of success, denoted by p, remains constant from trial to
trial.
(d) The trials are independent of each other.
2. Random variable of interest: X = number of “successes” out of n trials
3. Probability distribution:
Ã
b(x; n, p) = P[X = x] =
Ã
where
n
x
n
x
!
px q n−x , x = 0, 1, 2, . . . , n,
!
=
n!
x!(n−x)!
and q = 1 − p
4. Cumulative distribution function: Given in Table A.1, pages 742–747
5. Mean and Variance:
µ = E(X) = np
σ 2 = Var(X) = npq
6. Example: A civil engineering professor assigns a bridge building project using
Popsicle sticks each semester. To get a passing grade, the structure must
support at least 20 lbs. Historically, 10% of the student bridges fail to support
20 lbs. A random sample of 15 student bridges was taken.
(a)
(b)
(c)
(d)
Determine the probability that everyone passes the project.
Determine the probability that at least ten passes the project.
Determine the probability that between 2 and 5 (inclusive) fail the project.
Determine the mean and standard deviation of the number of bridges that
pass the project.
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II. Geometric Distribution
1. Properties of a geometric experiment
(a) Each trial results in an outcome that may be classified as “success” or
“failure”.
(b) The probability of success, denoted by p, remains constant from trial to
trial.
(c) The trials are independent of each other.
2. Random variable of interest: X = number of the trial on which the first “success” occurs or X = number of trials until the first “success”
3. Probability distribution:
g(x; p) = P[X = x] = pq x−1 , x = 1, 2, 3, . . . ,
where q = 1 − p
4. Cumulative distribution function: P[X ≤ x] =
x
X
P(X = t) = 1−(1−p)x , x =
t=1
1, 2, 3, . . .
5. Mean and Variance:
µ = E(X) =
σ 2 = Var(X) =
1
p
1−p
p2
6. Example: A car manufacturer gives a 5-year/60,000-mile warranty on its drive
train. Historically, 7% of this manufacturer’s cars have required service under
this warranty. Consider a new dealer.
(a) Determine the probability that the first claim under this warranty is the
tenth car sold.
(b) Determine the expected number of cars sold until the first claim.
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III. Negative Binomial Distribution
1. Properties of a negative binomial experiment
(a) Each trial results in an outcome that may be classified as “success” or
“failure”.
(b) The probability of success, denoted by p, remains constant from trial to
trial.
(c) The trials are independent of each other.
2. Random variable of interest: X = number of the trial on which the k th “success” occurs or X = number of trials until the k th “success”
3. Probability distribution:
Ã
∗
b (x; k, p) = P[X = x] =
Ã
where
x−1
k−1
x−1
k−1
!
pk q x−k , x = k, k + 1, k + 2, . . . ,
!
=
(x−1)!
(k−1)![(x−1)−(k−1)]!
and q = 1 − p
4. Cumulative distribution function: P[X ≤ x] =
x
X
P(X = t)
t=k
5. Mean and Variance:
µ = E(X) =
σ 2 = Var(X) =
k
p
k(1 − p)
p2
6. Example: A car manufacturer gives a 5-year/60,000-mile warranty on its drive
train. Historically, 7% of this manufacturer’s cars have required service under
this warranty. Consider a new dealer.
(a) Determine the probability that the twentieth car sold is the third to require
service under this warranty.
(b) Determine the expected number of cars sold until the third claim.
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IV. Poisson Distribution
1. Properties of a Poisson process
(a) The number of outcomes occurring in a given time interval or specified
region is independent of the number of occurrences in any other disjoint
time interval or region.
(b) The probability of a single outcome occurring in a short time interval or a
small region is proportional to the length of the time interval or the size
of the region, and is independent of the number of occurrences outside the
time interval or region.
(c) The probability that more than one outcome will happen in a short time
interval or region is negligible.
2. Random variable of interest: X = number of outcomes occurring in a given
time interval or region denoted by t
3. Probability distribution:
p(x; λt) = P[X = x] =
e−λt (λt)x
, x = 0, 1, 2, . . . ,
x!
where λ is the average number of outcomes per unit time, distance, area or
volume.
4. Cumulative distribution function: Given in Table A.2, pages 748–750
5. Mean and Variance:
µ = E(X) = λt
σ 2 = Var(X) = λt
6. Example: Historically, a process has averaged 2.5 flaws per 1000-m length of
wire.
(a) Determine the probability that a 1000-m length of wire has more than two
flaws.
(b) Determine the standard deviation of the number of flaws in a 1000-m
length of wire.
(c ) Determine the probability that a 3000-m length of wire will have more
than 3 flaws.
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