Download Example: Give the sample space giving a detailed description of

Examples: Random Variables and Probability Distributions
RANDOM VARIABLES and PROBABILITY DISTRIBUTIONS
Example: Give the sample space giving a detailed description of each
possible outcome when three electronic components are tested, where
N - denotes “nondefective”
D - denotes “defective”
Therefore, S = { NNN, NND, NDN, DNN, NDD, DND, DDN, DDD}
The number of defectives that occur will be assigned a numerical
value of 0, 1, 2, or 3, in the sample space. These values are random
quantities determined by the outcome of the experiment.
The random variable X - the no. of defective items when three
electronic components are tested would be
Sample Space
NNN
NND
NDN
DNN
NDD
DND
DDN
DDD
S
0
1
1
1
2
2
2
3
The possible values of x of X and their probabilities are given by
x
P(X = x)
0
1/8
1
3/8
2
3/8
3
1/8
Total
1
Note that the values of x exhaust all possible cases and hence the
probabilities add to 1. Representing all the probabilities by a formula,
f(x) = P(X = x), example: f(2) = P(X = 2).
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
The set of ordered pair (x, f(x)) is called the probability function or
probability distribution of the discrete random variable X.
Example: Discrete probability distribution
11/60) A shipment of 7 television sets contains 2 defective sets. A hotel makes a
random purchase of 3 of the sets. If X is the number of defective sets purchased by the
hotel, find the probability distribution of X. Express the results graphically as a
probability histogram.
Sol’n:
2 5
0 3
10
2
f(0) = --------------- = --------- = ------7
35
7
3
2 5
1 2
20
4
f(1) = --------------- = --------- = ------7
35
7
3
2 5
2 1
5
1
f(2) = --------------- = --------- = ------7
35
7
3
Therefore, the probability distribution of x is
x
f(x)
0
2/7
ENGSTAT Notes of AM Fillone
1
4/7
2
1/7
Examples: Random Variables and Probability Distributions
4/7
2/7
1/7
0
1
2
Probability histogram
Example: Discrete Probability Distributions
4.1/117)(M/S) The director of marketing for a small computer
manufacturer believes the discrete probability distribution shown in
the accompanying figure characterizes the number, y, of new
computers the firm will lease next year.
f(y)
.20
.15
.10
y
7
8
9
10
11
12
a) Is this a valid probability distribution? Explain
ENGSTAT Notes of AM Fillone
13
Examples: Random Variables and Probability Distributions
b) Display the probability distribution in tabular form.
c) What is the probability that exactly 9 computers will be leased?
d) What is the probability that fewer than 12 computers will be
leased?
Sol’n:
a) Yes
b)
y
f(y)
The Probability Distribution
8
9
10
11
12
.10
.15
.20
.20
.15
7
.10
13
.10
Total
1.0
c) f(9) = .15
d) f(y<12) = .75
Example: Discrete Probability Distributions
5/60) Determine the value c so that each of the following functions
can serve as a probability distribution of the discrete random variable
X:
for x = 0, 1, 2, 3;
a)
f(x) = c(x2 + 4)
b)
 2  3 
f(x) = c  

 x 3 – x 
for x = 0, 1, 2.
Sol’n:
a) for x = 0; f(x) = c(0 + 4) = 4c
for x = 1; f(x) = c(12+4) = 5c
for x = 2; f(x) = c(22+4) = 8c
for x = 3; f(x) = x(32+4) = 13c
3
Σ f(x) = f(0) + f(1) + f(2) + f(3) = 1
x=0
= 4c + 5c + 8c + 13c = 1
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
c = 1/30
b)
2  3 
for x = 0: f(0) = c 0 3 – 0  = c
2  3 
for x = 1: f(1) = c 1 3 – 1  = 6c
2  3 
for x = 2: f(2) = c 2 3 – 2  = 3c
2
Σ f(x) = f(0) + f(1) + f(2) = 1
x=0
c + 6c + 3c = 1
c = 1/10
13/61) The probability distribution of X, the number of imperfections
per 10 meters of a synthetic fabric in continuous rolls of uniform
width, is given by
X
f(x)
0
0.41
1
0.37
2
0.16
3
0.05
Construct the cumulative distribution of X.
Sol’n:
Therefore, the cumulative distribution of X is
F(X) =
 0
 0.41
 0.78
 0.94
 0.99
 1.00
ENGSTAT Notes of AM Fillone
for x < 0
for 0 ≤ x < 1
for 1 ≤ x < 2
for 2 ≤ x < 3
for 3 ≤ x < 4
for x ≥ 4
4
0.01
Examples: Random Variables and Probability Distributions
1.0
0.8
0.6
0.4
0.2
0
1
2
3
4
Discrete cumulative distribution
2/78) If the joint probability distribution of X and Y is given by
(x + y)
f(x,y) = ----------,
30
for x = 0, 1, 2, 3; y = 0, 1, 2
find,
(a) P(X ≤ 2, Y = 1);
(b) P(X > 2, Y ≤ 1);
(c) P(X > Y);
(d) P(X + Y = 4).
The Joint Probability Distribution
f(x,y)
x
0
1
2
0
0
1/30
2/30
y
1
1/30
2/30
3/30
2
2/30
3/30
4/30
Column
3/30 = 6/30 =
9/30 =
Totals
1/10
1/5
3/10
ENGSTAT Notes of AM Fillone
Row Totals
3
3/30
4/30
5/30
12/30 =
2/5
6/30 = 1/5
9/30 = 3/10
14/30 = 7/15
30/30 = 1
Examples: Random Variables and Probability Distributions
a)
b)
c)
d)
P(X ≤ 2, Y = 1)
= f(0,1) + f(1,1) + f(2,1) =1/30 + 2/30 + 3/30
= 6/30
= 1/5
P(X > 2, Y ≤ 1)
= f(3,0) + f(3,1) = 3/30 + 4/30
= 7/30
P(X > Y)
= f(1,0) + f(2,0) + f(3,0) + f(2,1) + f(3,1) + f(3,2)
= 1/30 + 2/30 + 3/30 + 3/30 + 4/30 + 5/30
= 18/30
= 3/5
P(X + Y = 4) = f(2,2) + f(3,1) = 4/30 + 4/30
= 8/30
= 4/15.
Example: Joint probability distribution
6.4/202)(M/S) From a group of three data-processing managers, two
senior analysts, and two quality control engineers, three people are to
be randomly selected to form a committee that will study the
feasibility of adding computer graphics at a consulting firm. Let y1
denote the number of data-processing managers and y2 the number of
senior systems analysts selected for the committee.
a. Find the joint probability distribution of y1 and y2.
b. Find the marginal distribution of y1.
Sol’n:
a.
y1
y2
0
1
2
0
0
2/35
2/35
1
3/35
12/35
3/35
2
6/35
6/35
0
b.
P(y1)
0
4/35
ENGSTAT Notes of AM Fillone
1
18/35
2
12/35
3
1/35
3
1/35
0
0
Examples: Random Variables and Probability Distributions
7/79) Let X denote the reaction time, in seconds, to a certain
stimulant and Y denote the temperature (oF) at which a certain
reaction starts to take place. Suppose that two random variables X and
Y have the joint density given by
0 < x, y < 1
 4xy,
f(x,y) = 
elsewhere
 0,
Find
a) P(0 ≤ X ≤ 1/2 and 1/4 ≤ Y ≤ 1/2);
b) P(X < Y).
Soln:
½ ½
a) P(0 ≤ X ≤ 1/2 and 1/4 ≤ Y ≤ 1/2) = ∫
∫ 4xy dx dy
0
¼
½
x=1/2
2
= ∫ 4y (x /2) dy
¼
x=0
½
y=1/2
= ∫ ydy/2
= y2/4
y=1/4
¼
= 3/64
x 1
b) P(X < Y) = ∫ ∫ 4xy dx dy
0 y
1
1
x=y
2
= ∫ 4y dy x /2
= ∫
y
0
y
4
= 1/2 (1 – 0) = 1/2
ENGSTAT Notes of AM Fillone
3
4
1
2y dy = 2y /4
y=0
Examples: Random Variables and Probability Distributions
Example: Independence
24/80) Determine whether the two random variables of Exercise 7 are
dependent or independent.
Sol’n:
The joint density function is given as
0 < x, y < 1
 4xy,
f(x,y) = 
elsewhere
 0,
1
1
2
g(x) = ∫ 4xy dy = 4x [y /2] = 2x [12 – 0] = 2x
0
0
1
1
2
h(y) = ∫ 4xy dx = 4y [x /2] = 2y [12 – 0] = 2y
0
0
Hence,
f(x,y) = g(x)h(y),
0 < x,y < 1
4xy = (2x)(2y)
4xy = 4xy
Therefore, x and y are independent.
Example: Continuous probability distribution
17/61) A continuous random variable X that can assume values
between x = 1 and x = 3 has a density function given by f(x) = 1/2.
a) Show that the area under the curve is equal to 1.
b) Find P(2 < X < 2.5).
c) Find P(X ≤ 1.6).
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
a)
f(x)
1/2
A = (1/2)(3-1) = 1
0
1
2
3
2.5
b) P(2 < X < 2.5) = ∫ f(x) dx
2
x
2.5
= ∫
2
2.5
(1/2) dx = x/2
2
= (1/2)(2.5 – 2) = 1/4
1.6
1.6
c) P( X ≤ 1.6) = ∫ (1/2) dx = x/2
= (1/2)(1.6 – 1) = 0.3
1
1
1/60) Classify the following random variables as discrete or continuous.
X:
Y:
M:
N:
P:
the number of automobile accidents per year in Virginia.
the length of time to play 18 holes of golf.
the amount of milk produced yearly by a particular cow.
the number of eggs laid each month by a hen.
the number of building permits issued each month in a certain
city.
Q:
the weight of grain produced per acre.
Sol’n:
X:
Discrete
Y:
Continuous
M: Continuous
N:
Discrete
P:
Discrete
Q:
Continuous
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
Example: Cumulative probability distribution
19/61) For the density function of Exercise 17, find F(x). Use it to
evaluate P(2 < X < 2.5).
Sol’n:
for 1 < x < 3.0
x
x
x
F(x) = ∫ f(t)dt = ∫ (1/2)dt = (t/2)
= (x –1 )/2
-∞
1
1
Therefore,
F(x) =
 0,
 (x –1)/2,
 1,
x≤1
1<x<3
x ≥3.
Then, P(2 < x < 2.5) = F(2.5) – F(2) = (2.5 –1)/2 – (2-1)/2 = 1/4
Example: Continuous probability distribution
21/61) Consider the density function
f(x) =
 k √ x,

 0,
0<x<1
elsewhere.
a) Evaluate k.
b) Find F(x) and use it to evaluate P(0.3 < X < 0.6).
Soln:
1
a) ∫ f(x) dx = 1
0
1
1
1
1/2
3/2
3/2
∫ k x dx = k [ (x )/(3/2) ] = 2k/3 [x ]
0
0
0
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
2k/3 [13/2 – 0] = 1
k = 3/2
x
b) F(X) = P (X ≤ x) = ∫ f(t) dt
-∝
x
x
x
1/2
3/2
3/2
= ∫ (3/2)t dt = (3/2) t / (3/2)
= t
= x3/2
0
0
0
F(x) =
 x3/2,
 x3/2,
 x3/2,
1
0 < x < 0.3
0.3 ≤ x < 0.6
0.6 ≤ x < 1
x ≥ 1.
F(x) = F(0.6) – F(0.3) = 0.464758 – 0.1643167 = 0.3004
Example: Conditional distribution
6/79) Let X and Y denote the lengths of life, in years, of two
components in an electronic system. If the joint density function of
these variables is given by
f(x,y) =
 e–(x+y),

 0,
x > 0, y > 0
elsewhere,
find P(0 < X < 1 | Y = 2).
Soln:
f(x,y)
f(x|y) = ---------h(y)
∝
∝
α
-(x+y)
h(y) = ∫ f(x,y) dx = ∫ e
dx = - ∫ e-(x+y)(-dx)
-∝
-∝
0
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
-(x +y)
α
=-e
= - [e-(α+y) – e-y]
0
= e-y [ 1 – e-α] = e-y
e-xe-y
f(x,y)
e-(x+y)
1
f(x|y) = ----------- = ---------------- = --------------- = ∫ e-x dx
0
e-y
h(y)
e-y
1
-x
P(0 < X < 1 | Y = 2) = - e
= - [ e-1 – e0] = - [0.3679 – 1]
0
= 0.6321
Example: Conditional distribution/independence
11/79) The amount of kerosene, in thousands of liters, in a tank at the
beginning of any day is a random amount Y from which a random
amount X is sold during that day. Suppose that the tank is not
resupplied during the day so that x ≤ y, and assume that the joint
density function of these variables is given by
0 < x < y, 0 < y < 1
 2,
f(x,y) = 
elsewhere.
 0,
(a) Determine if X and Y are independent.
(b) Find P(1/4 < X < 1/2 | Y = 3/4 ).
Sol’n:
(a) f(x | y) = f(x,y)/h(y)
∞
∞
y
-∞
-∞
0
h(y) = ∫ f(x,y) dx = ∫ 2 dx = 2x ] = 2y
f(x | y) = 2/2y = 1/y,
0<x<y
Since f(x|y) depends on y, x and y are dependent.
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
1/2
1/2
(b) P(1/4<x<1/2| y=3/4) = ∫ (4/3) dx = 4x/3 ] = (4/3)(1/2 - 1/4) = 1/3
1/4
1/4
20/80) Determine whether the two random variables of Exercise 13
are dependent or independent.
Sol’n:
f(1,1) = .05
3
g(1) = Σ f(1,y) = .05 + .05 + 0 = .10
y=1
3
h(1) = Σ f(x,1) = .05 + .05 + .10 = .20
x=1
For the two variables to be independent, f(1,1) = g(1)h(1), however,
.05 ≠ .10(.20). Hence the two are dependent variables.
Example: Marginal distribution
13/79) Let X denote the number of times a certain numerical control
machine will malfunction: 1, 2, or 3 times on any given day. Let Y
denote the number of times a technician is called on an emergency
call. Their joint probability distribution is given as:
x
f(x,y)
1
2
3
1
0.05
0.05
0.1
Y 2
0.05
0.1
0.35
3
0
0.2
0.1
a) Evaluate the marginal distribution of X.
b) Evaluate the marginal distribution of Y.
c) Find P(Y=3 | X = 2).
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
Solution:
1
y 2
3
g(x)
Table
1
0.05
0.05
0
0.1
x
2
0.05
0.1
0.2
0.35
3
0.1
0.35
0.1
0.55
h(y)
0.2
0.5
0.3
1.0
a) The marginal distribution of x is
x
g(x)
1
0.1
2
0.35
3
0.55
b) The marginal distribution of y is
y
h(y)
1
0.2
2
0.5
3
0.3
c) P( Y= 3 | X = 2) = 0.2 (from the table)
Example: Continuous cumulative distribution
14/61) The waiting time, in hours, between successive speeders
spotted by a radar unit is a continuous random variable with
cumulative distribution
F(x) =
 0,

 1 – e-8x,
x≤0
x>0.
Find the probability of waiting less than 12 minutes between
successive speeders
a) using the cumulative distribution of X;
b) using the probability density function of X.
ENGSTAT Notes of AM Fillone
Examples: Random Variables and Probability Distributions
Sol’n:
12 minutes = 0.2 hrs.
-8(.2)
a) P(x < 0.2) = F(0.2) – F(0) = [1- e
0
-8(0)
] – [1 – e
] = 0.7981
b) f(x) = dF(x)/dx = (1-e-8x)(-8dx)/dx = (-8)( 1-e-8x)
0.2
0.2
-8x
-8x
= 0.7981
P(0 < x< 0.2) = ∫ (-8)( 1-e ) dx = [1-e ]
0
0
ENGSTAT Notes of AM Fillone