Download Random Phenomena (TRIBOS) Formulae

Random Phenomena (TRIBOS) Formulae - 2nd set
Probability mass function fX of a discrete random variable X with domain SX = {xi } has following
properties:
fX (xi ) = P (X = xi ) = p(xi ),
0 ≤ fX (xi ) ≤ 1,
fX (xi ) = 1.
(1)
xi ∈SX
Probability of X falling into interval [a, b]:
P (a ≤ X ≤ b) =
fX (xi ).
(2)
xi ∈[a,b]
Cumulative distribution function FX of a discrete random variable X is defined by:
fX (xj ).
FX (xi ) = P (X ≤ xi ) =
(3)
xj ≤xi
Probability density function fX of a continuous random variable X has following properties:
∞
−∞
fX (x) ≥ 0,
fX (x) dx = 1,
P (a ≤ X ≤ b) =
a
(4)
b
fX (x) dx.
Cumulative distribution function FX of a continuous random variable X is defined by:
x
fX (u) du
za
− ∞ < x < ∞.
FX (x) = P (X ≤ x) =
(5)
−∞
Also:
d
d
FX (x) =
dx
dx
x
−∞
fX (u) du = fX (x).
(6)
Mean of a discrete/continuous random variable X:
xk fX (xk ),
discrete:
mX = E [X] =
continuous:
mX = E [X] =
xk ∈SX
∞
−∞
(7)
xf (x) dx
(assuming SX = R).
Variance of a discrete/continuous random variable X:
2
= Var [X] = E (X − E [X])2 =
(xk − E [X])2 ,
discrete:
σX
continuous:
2
= Var [X] = E (X − E [X])2 =
σX
k ∞
−∞
(x − E [X])2 f (x) dx
(assuming SX = R).
(8)
Random Phenomena (TRIBOS) Homework Problems - 2nd set
1. The sample space of a random experiment is {a, b, c, d, e, f }, and each outcome is equally likely. A random
variable is defined as follows:
outcome
a
b
c
d
e
f
xi
0
0
1.5
1.5
2
3
Determine the probability mass function fX . Use it to determine the following probabilities: a) P (X =
1.5), b) P (0.5 < X < 2.7), c) P (X > 3), d) P (0 ≤ X < 2), e) P (X = 0 ali X = 2). Determine also the
corresponding cumulative probability function FX . R: f (0) = 2/6, f (1.5) = 2/6, f (2) = 1/6, f (3) = 1/6; a) 2/6, b)
3/6, c) 0, d) 4/6, e) 3/6
2. In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as
pass or fail. Assume that the probability that a wafer passes the test is 0.8 and that wafers are independent.
Determine the probability mass function of the number of wafers from a lot that pass the test. Determine
the corresponding cumulative probability function. R: f (0) = 0.008, f (1) = 0.096, f (2) = 0.384, f (3) = 0.512;
F (X < 0) = 0, F (0 ≤ X < 1) = 0.008, F (1 ≤ X < 2) = 0.104, F (2 ≤ X < 3) = 0.488, F (3 ≤ X) = 1
3. The probability density function of the time to failure of an electronic component in a copier (in hours)
is fX (x) = e−x/1000 /1000 for x > 0. Determine the corresponding cumulative probability function.
Determine the probability that
(a) A component lasts more than 3000 hours before failure.
(b) A component fails in the interval from 1000 to 2000 hours.
(c) A component fails before 1000 hours.
(d) Determine the number of hours at which 10 % of all components have failed.
R: FX (x) = 1 − e−x/1000 ; a) 0.0498, b) 0.233, c) 0.632, d) 105
4. Determine the mean and variance of the random variable in Exercises 1 and 2.
5. The thickness of a conductive coating in micrometers has a density function fX (x) = 600 x−2 for
100µm < x < 120µm.
(a) Determine the mean and variance of the coating thickness.
(b) If the coating costs $0.50 per micrometer of thickness on each part, what is the average cost of the
coating per part?