Probability and statistics for engineering

1. Probability and Statistics
Chapter 3
Random Variables and
Probability Distributions
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2. • Random Variable (RV):
A random variable is a function that assigns a real number
to each outcome in the sample space of a random
experiment
•
‫العشوائي‬ ‫المتغير‬
:
‫العينة‬ ‫فضاء‬ ‫على‬ ‫معرفه‬ ‫حقيقية‬ ‫قيمة‬ ‫ذات‬ ‫دالة‬ ‫هو‬
)
(S
Random Variables are denoted by upper case letters (X)
Individual outcomes for an RV are denoted by lower
case letters (x)
Concept of a Random Variable
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3. Types of Random Variables
 Number of sales
 Number of calls
 Shares of stock
 People in line
 Mistakes per page
• Length
• Depth
• Volume
• Time
• Weight
Random Variables RV
Qualitative RV
Quantitative RV
Discrete RV
Continuous RV Ordinal
Non Ordinal
Can be arranged
•Excellent
•Very Good
•Good
•
Can't be
arranged
Green
Blue
Red
Black
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4. Discrete Probability Distribution
The probability distribution of a discrete
random variable is called a probability
function if
1
)
(
0
)
(
:
Function
(Mass)
y
Probabilit
all




x
x
f
x
x
f
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5. Continuous Probability Distribution
1
)
(
0
)
(
:
Function
(Density)
y
Probabilit
all



x
x
f
x
x
f
The probability distribution of a continuous
random variable is called a probability
function if
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6. Cumulative Probability Function
• The cumulative probability function, denoted
F(xn), shows the probability that X is less than or
equal to xn
• In other words,
)
x
f(X
)
F(x n
n 




n
x
x
n f(x)
)
F(x



x
n dx
f(x)
)
F(x
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7. Example
• Determine the value of K so that each of the
following functions can serve as a probability
distribution of the discrete random variable X:


 


Otherwise
0
3;
2,
1,
0,
X
for
)
4
(
)
(
2
X
K
x
f
X 0 1 2 3 Sum
f(x) 4k 5K 8K 13K 30K
Solution:
30K=1 , then K=1/30
3.;
2,
1,
0,
X
for
30
)
4
(
)
(
2



X
x
f
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8. Example
• 1. Find the constant a
• 2. Find the probability















elsewhere
0
3
X
2
3a
aX
-
2
X
1
a
1
X
0
)
(
aX
x
f
Solution:
1
)
3
(
1
)
(
)
(
)
(
3
2
2
1
1
0
3
2
2
1
1
0














dx
a
ax
dx
a
dx
ax
dx
x
f
dx
x
f
dx
x
f
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9. Solution:
  1
3
2
2
3
2
2
2
1
1
0
2

















aX
x
a
x
a
x
a
2a=1, then a=0.5

















elsewhere
0
3
X
2
2
3
2
X
-
2
X
1
2
1
1
X
0
2
)
(
X
x
f
5
.
0
)
1
5
.
1
(
2
1
4
1
2
1
2
1
)
5
.
1
(
5
.
1
1
1
0






 
 dx
xdx
x
P
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10. Example
• Required:
(a) P(x=4) (b)
(c) (d) P(x>-6)








otherwise
0
4
3,
2,
1,
0,
x
25
1
2
)
(
x
x
f
)
4
2
( 
 x
P
)
1
( 
x
P
Solution:
x 0 1 2 3 4 sum
f(x) 0.04 0.12 0.2 0.28 0.36 1
Cumulative F(x) 0.04 0.16 0.36 0.64 1
(a) P(x=4)=0.36
or P(x=4) = F(4) – F(3) = 1- 0.64 = 0.36
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11. Solution:
0.2+0.28+0.36=0.84
or = F(4) – F( 1) = 1 – 0.16 = 0.84





 )
4
(
)
3
(
)
2
(
)
4
2
(
)
( p
p
p
x
P
b
16
.
0
)
1
(
)
( 

x
P
c
(d) P(x>- 6) =1
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12. Example
The probability mass function of the random variable X is given
as
Find the distribution function F(x) and graph this distribution
function.
Solution:
3
2
1
x
1/6
1/3
1/2
f(x)
 
















3
3
2
2
1
1
1
6
5
2
1
0
x
x
x
x
x
F
Graph of F(x) (step function)
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13. Discrete Probability Distribution
Experiment: for toss 4 coins, Calculate the mean
and variance
Let X = # heads.
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14. H
H
H
H
H
H
H
H
H
H
H
T
T
T
H
H
H
H
T
T
T
T
T
T
T
T
T
T
T
T
Solution:
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15. H
H
H
H
H
H
H
H
H
H
H
T
T
T
H
H
H
H
T
T
T
T
T
T
T
T
T
T
T
T
4
3
3
2
3
2
2
1
3
2
2
1
2
1
1
0
No.
of
Heads
Solution:
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16. 4
3
2
1
0
X
1
4
6
4
1
f
1/16
4/16
6/16
4/16
1/16
f(x)
16/16
15/16
11/16
5/16
1/16
F(x)
4/16
12/16
12/16
4/16
0
x.f(x)
Let X is a R.V represent the number heads
Mean =µ = x.f(x) = 32/16 =2
Solution:
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17. Probability mass function plot
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18. Probability histogram
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19. Discrete cumulative distribution function
1/16
5/16
11/16
15/16 16/16
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20. Joint Probability Distributions
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21. Joint Probability Distributions
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22. Example
Two ballpoint pens are selected at random from a box that
contains 3 blue pens, 2 red pens, and 3 green pens. If X is
the number of blue pens selected and Y is the number of
red pens selected, find
(a) the joint probability function f(x, y),
(b) P[(X, Y ) ∈ A], where A is the region {(x, y)|x + y ≤ 1}.
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23. solution
X (blue) Y (red)
0 0
0 1
1 0
1 1
2 0
0 2
The
possible
pairs
of
values
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24. solution
• f(0,0) = this represent the probability of getting
0 blue and 0 red means the two ballpoint pens
are green can be selected by 3 ways
• f(0,0 ) = 3/28
• The denominator (28) represents the number
of ways of selecting two ballpoint pens from
eight ballpoint pens
• f(0,1) = 0 blue, 1 red and 1 green (2*3 = 6)
• f(0,1 ) = 6/28
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25. Solution
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26. Solution
(b) The probability that (X, Y ) fall in the region A
is
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27. Solution
(b) The probability that (X, Y ) fall in the region A
is
P[(X, Y ) ∈ A] = P(X + Y ≤ 1) =
f(0, 0) + f(0, 1) + f(1, 0)
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28. Example
Verify the following function is probability function, then
find
P[(X, Y ) ∈ A],
where A = {(x, y) | 0 < x < 0.25, 0.25< y < 0.5}.
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29. Solution
(a) The integration of f(x, y) over the whole region
is
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30. Solution
(b) To calculate the probability, we use
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31. Solution
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32. 32
32