Licture 4 - Random Variables and Probability Distributions

A lecture in mathematics

1. Lecture 4: Random variables and probability
distributions (Random variables, Discrete
probability distributions)

2. Random variable
Statistics is concerned with making inferences about popu-
lations and population characteristics.
It is often important to allocate a numerical description to
the outcome, e.g., the number of defectives that occur.
Recall previous examples, e.g., throwing a pair of dice, the
two numbers are the outcome, but their sum is a random
variable ...
Denition 3.1
Arandom variableis a function that associates a real number
with each element in the sample space.
These values are, of course, random quantities determined
by the outcome of the experiment.
We shall use a capital letter, sayX, to denote a random
variable and its corresponding small letter,x, for one of its
values.
FECU cWafaa S. Sayed 2 / 21

3. Random variable
Statistics is concerned with making inferences about popu-
lations and population characteristics.
It is often important to allocate a numerical description to
the outcome, e.g., the number of defectives that occur.
Recall previous examples, e.g., throwing a pair of dice, the
two numbers are the outcome, but their sum is a random
variable ...
Denition 3.1
Arandom variableis a function that associates a real number
with each element in the sample space.
These values are, of course, random quantities determined
by the outcome of the experiment.
We shall use a capital letter, sayX, to denote a random
variable and its corresponding small letter,x, for one of its
values.
FECU cWafaa S. Sayed 2 / 21

4. Random variable
Statistics is concerned with making inferences about popu-
lations and population characteristics.
It is often important to allocate a numerical description to
the outcome, e.g., the number of defectives that occur.
Recall previous examples, e.g., throwing a pair of dice, the
two numbers are the outcome, but their sum is a random
variable ...
Denition 3.1
Arandom variableis a function that associates a real number
with each element in the sample space.
These values are, of course, random quantities determined
by the outcome of the experiment.
We shall use a capital letter, sayX, to denote a random
variable and its corresponding small letter,x, for one of its
values.
FECU cWafaa S. Sayed 2 / 21

5. Illustrative examples
In an experiment, two balls are drawn in succession without re-
placement from an urn containing 4 red balls and 3 black balls.
What are the possibilities of the number of red balls obtained?
LetYbe the number of red balls.
The possible outcomes and
the valuesyof the random variableYare
LetXbe the random variable dened by the waiting time, in
hours, between successive speeders spotted by a radar unit.
What are its possible values?
The random variableXtakes on all valuesxfor whichx0.
FECU cWafaa S. Sayed 3 / 21

6. Illustrative examples
In an experiment, two balls are drawn in succession without re-
placement from an urn containing 4 red balls and 3 black balls.
What are the possibilities of the number of red balls obtained?
LetYbe the number of red balls.
The possible outcomes and
the valuesyof the random variableYare
LetXbe the random variable dened by the waiting time, in
hours, between successive speeders spotted by a radar unit.
What are its possible values?
The random variableXtakes on all valuesxfor whichx0.
FECU cWafaa S. Sayed 3 / 21

7. Illustrative examples
In an experiment, two balls are drawn in succession without re-
placement from an urn containing 4 red balls and 3 black balls.
What are the possibilities of the number of red balls obtained?
LetYbe the number of red balls.
The possible outcomes and
the valuesyof the random variableYare
LetXbe the random variable dened by the waiting time, in
hours, between successive speeders spotted by a radar unit.
What are its possible values?
The random variableXtakes on all valuesxfor whichx0.
FECU cWafaa S. Sayed 3 / 21

8. Illustrative examples
In an experiment, two balls are drawn in succession without re-
placement from an urn containing 4 red balls and 3 black balls.
What are the possibilities of the number of red balls obtained?
LetYbe the number of red balls.
The possible outcomes and
the valuesyof the random variableYare
LetXbe the random variable dened by the waiting time, in
hours, between successive speeders spotted by a radar unit.
What are its possible values?
The random variableXtakes on all valuesxfor whichx0.
FECU cWafaa S. Sayed 3 / 21

9. Illustrative examples
In an experiment, two balls are drawn in succession without re-
placement from an urn containing 4 red balls and 3 black balls.
What are the possibilities of the number of red balls obtained?
LetYbe the number of red balls.
The possible outcomes and
the valuesyof the random variableYare
LetXbe the random variable dened by the waiting time, in
hours, between successive speeders spotted by a radar unit.
What are its possible values?
The random variableXtakes on all valuesxfor whichx0.
FECU cWafaa S. Sayed 3 / 21

10. Illustrative examples
In an experiment, two balls are drawn in succession without re-
placement from an urn containing 4 red balls and 3 black balls.
What are the possibilities of the number of red balls obtained?
LetYbe the number of red balls.
The possible outcomes and
the valuesyof the random variableYare
LetXbe the random variable dened by the waiting time, in
hours, between successive speeders spotted by a radar unit.
What are its possible values?
The random variableXtakes on all valuesxfor whichx0.
FECU cWafaa S. Sayed 3 / 21

11. Do you notice any difference?
FECU cWafaa S. Sayed 4 / 21

12. Discrete random variables (nite or countable)
Denition 3.2
If a sample space contains a nite number of possibilities or an
unending sequence with as many elements as there are whole
numbers, it is called adiscrete sample space.
A random variable is called a discrete random variable if its
set of possible outcomes is countable.
In most practical problems, discrete random variables rep-
resent count data, such as the number of defectives in a
sample of k items or the number of highway fatalities per
year in a given state.
FECU cWafaa S. Sayed 5 / 21

13. Discrete random variables (nite or countable)
Denition 3.2
If a sample space contains a nite number of possibilities or an
unending sequence with as many elements as there are whole
numbers, it is called adiscrete sample space.
A random variable is called a discrete random variable if its
set of possible outcomes is countable.
In most practical problems, discrete random variables rep-
resent count data, such as the number of defectives in a
sample of k items or the number of highway fatalities per
year in a given state.
FECU cWafaa S. Sayed 5 / 21

14. Continuous random variables (uncountable)
Denition 3.3
If a sample space contains an innite number of possibilities
equal to the number of points on a line segment, it is called a
continuous sample space.
A random variable is called a continuous random variable
if its possible outcomes fall in an entire interval of numbers
and are not discrete or whole numbers. They can take on
values on a continuous scale.
In most practical problems, continuous random variables
represent measured data, such as all possible heights, weights,
temperatures, distance, or life periods.
FECU cWafaa S. Sayed 6 / 21

15. Continuous random variables (uncountable)
Denition 3.3
If a sample space contains an innite number of possibilities
equal to the number of points on a line segment, it is called a
continuous sample space.
A random variable is called a continuous random variable
if its possible outcomes fall in an entire interval of numbers
and are not discrete or whole numbers. They can take on
values on a continuous scale.
In most practical problems, continuous random variables
represent measured data, such as all possible heights, weights,
temperatures, distance, or life periods.
FECU cWafaa S. Sayed 6 / 21

16. More discrete examples
A stockroom clerk returns three safety helmets at random to
three steel mill employees who had previously checked them. If
Smith, Jones, and Brown, in that order, receive one of the three
hats, list the sample points for the possible orders of returning
the helmets, and nd the valuemof the random variableMthat
represents the number of correct matches.
The possible arrangements in which the helmets may be
returned and the number of correct matches are
FECU cWafaa S. Sayed 7 / 21

17. More discrete examples
A stockroom clerk returns three safety helmets at random to
three steel mill employees who had previously checked them. If
Smith, Jones, and Brown, in that order, receive one of the three
hats, list the sample points for the possible orders of returning
the helmets, and nd the valuemof the random variableMthat
represents the number of correct matches.
The possible arrangements in which the helmets may be
returned and the number of correct matches are
FECU cWafaa S. Sayed 7 / 21

18. More discrete examples
A stockroom clerk returns three safety helmets at random to
three steel mill employees who had previously checked them. If
Smith, Jones, and Brown, in that order, receive one of the three
hats, list the sample points for the possible orders of returning
the helmets, and nd the valuemof the random variableMthat
represents the number of correct matches.
The possible arrangements in which the helmets may be
returned and the number of correct matches are
FECU cWafaa S. Sayed 7 / 21

19. More discrete examples
Statisticians use sampling plans to either accept or reject
batches or lots of material. Suppose one of these sampling
plans involves sampling independently 10 items from a lot of 100
items in which 12 are defective. What are the possible values of
the number of items found defective in the sample of 10?
LetXbe the random variable dened as the number of items
found defective in the sample of 10.
In this case, the random
variable takes on the values0;1;2; : : : ;9;10.
FECU cWafaa S. Sayed 8 / 21

20. More discrete examples
Statisticians use sampling plans to either accept or reject
batches or lots of material. Suppose one of these sampling
plans involves sampling independently 10 items from a lot of 100
items in which 12 are defective. What are the possible values of
the number of items found defective in the sample of 10?
LetXbe the random variable dened as the number of items
found defective in the sample of 10.
In this case, the random
variable takes on the values0;1;2; : : : ;9;10.
FECU cWafaa S. Sayed 8 / 21

21. More discrete examples
Statisticians use sampling plans to either accept or reject
batches or lots of material. Suppose one of these sampling
plans involves sampling independently 10 items from a lot of 100
items in which 12 are defective. What are the possible values of
the number of items found defective in the sample of 10?
LetXbe the random variable dened as the number of items
found defective in the sample of 10.
In this case, the random
variable takes on the values0;1;2; : : : ;9;10.
FECU cWafaa S. Sayed 8 / 21

22. More discrete examples
When a die is thrown until a 5 occurs, we obtain a sam-
ple space with an unending sequence of elements,S=
fF;NF;NNF;NNNF; : : :g, whereFandNrepresent, respectively,
the occurrence and non-occurrence of a 5.
Even in this experiment, the number of elements can be equated
to the number of whole numbers so that there is a rst element,
a second element, a third element, and so on, and in this sense
can be counted.
Countable but not nite
FECU cWafaa S. Sayed 9 / 21

23. More discrete examples
When a die is thrown until a 5 occurs, we obtain a sam-
ple space with an unending sequence of elements,S=
fF;NF;NNF;NNNF; : : :g, whereFandNrepresent, respectively,
the occurrence and non-occurrence of a 5.
Even in this experiment, the number of elements can be equated
to the number of whole numbers so that there is a rst element,
a second element, a third element, and so on, and in this sense
can be counted.
Countable but not nite
FECU cWafaa S. Sayed 9 / 21

24. More discrete examples
When a die is thrown until a 5 occurs, we obtain a sam-
ple space with an unending sequence of elements,S=
fF;NF;NNF;NNNF; : : :g, whereFandNrepresent, respectively,
the occurrence and non-occurrence of a 5.
Even in this experiment, the number of elements can be equated
to the number of whole numbers so that there is a rst element,
a second element, a third element, and so on, and in this sense
can be counted.
Countable but not nite
FECU cWafaa S. Sayed 9 / 21

25. More discrete examples
Suppose a sampling plan involves sampling items from a pro-
cess until a defective is observed. The evaluation of the process
will depend on how many consecutive items are observed.
In that regard, letXbe a random variable dened by the number
of items observed till a defective is found. WithNa non-defective
andDa defective, sample spaces areS=fDggivenX=1,
S=fNDggivenX=2,S=fNNDggivenX=3, and so on.
Countable but not nite
FECU cWafaa S. Sayed 10 / 21

26. More discrete examples
Suppose a sampling plan involves sampling items from a pro-
cess until a defective is observed. The evaluation of the process
will depend on how many consecutive items are observed.
In that regard, letXbe a random variable dened by the number
of items observed till a defective is found. WithNa non-defective
andDa defective, sample spaces areS=fDggivenX=1,
S=fNDggivenX=2,S=fNNDggivenX=3, and so on.
Countable but not nite
FECU cWafaa S. Sayed 10 / 21

27. More discrete examples
Consider the simple condition in which components are arriving
from the production line and they are stipulated to be defective
or not defective.
Dene the random variableXby
X=

1;if the component is defective,
0;if the component is not defective.
Categorical or binary “Bernoulli random variable”
FECU cWafaa S. Sayed 11 / 21

28. More discrete examples
Consider the simple condition in which components are arriving
from the production line and they are stipulated to be defective
or not defective.
Dene the random variableXby
X=

1;if the component is defective,
0;if the component is not defective.
Categorical or binary “Bernoulli random variable”
FECU cWafaa S. Sayed 11 / 21

29. More discrete examples
Consider the simple condition in which components are arriving
from the production line and they are stipulated to be defective
or not defective.
Dene the random variableXby
X=

1;if the component is defective,
0;if the component is not defective.
Categorical or binary “Bernoulli random variable”
FECU cWafaa S. Sayed 11 / 21

30. Yet, one more continuous example
Interest centers around the proportion of people who respond to
a certain mail order solicitation. LetXbe that proportion.
Xis a random variable that takes on all valuesxfor which
0x1.Don't confuse with probability!
FECU cWafaa S. Sayed 12 / 21

31. Yet, one more continuous example
Interest centers around the proportion of people who respond to
a certain mail order solicitation. LetXbe that proportion.
Xis a random variable that takes on all valuesxfor which
0x1.Don't confuse with probability!
FECU cWafaa S. Sayed 12 / 21

32. Yet, one more continuous example
Interest centers around the proportion of people who respond to
a certain mail order solicitation. LetXbe that proportion.
Xis a random variable that takes on all valuesxfor which
0x1.Don't confuse with probability!
FECU cWafaa S. Sayed 12 / 21

33. Discrete probability distributions
Again, recall throwing a pair of dice, this is about computing
the probability that the sum of the numbers obtained equals
2, or 3, ... or 12. We have done it before, we are just formu-
lating it ...
Denition 3.4
The set of ordered pairs(x;f(x))is aprobability function,
probability mass function, orprobability distributionof the
discrete random variableXif, for each possible outcomex,
1
f(x)0, (0f(x)1)
2
P
x
f(x) =1,
3
P(X=x) =f(x).
FECU cWafaa S. Sayed 13 / 21

34. A shipment of 20 similar laptop computers to a retail outlet con-
tains 3 that are defective. If a school makes a random purchase
of 2 of these computers, nd the probability distribution for the
number of defectives.
Solution: LetXbe a random variable whose valuesxare the
possible numbers of defective computers purchased by the
school. Thenxcan only take the numbers 0, 1, and 2.
Similar to the poker hand problem in Lectures 2-3,
f(0) =P(X=0) =

3
0

17
2


20
2
=
68
95
,
f(1) =

3
1

17
1


20
2
=
51
190
andf(2) =

3
2

17
0


20
2
=
3
190
:
x 0 1 2
f(x)
68
95
51
190
3
190
FECU cWafaa S. Sayed 14 / 21

35. A shipment of 20 similar laptop computers to a retail outlet con-
tains 3 that are defective. If a school makes a random purchase
of 2 of these computers, nd the probability distribution for the
number of defectives.
Solution: LetXbe a random variable whose valuesxare the
possible numbers of defective computers purchased by the
school. Thenxcan only take the numbers 0, 1, and 2.
Similar to the poker hand problem in Lectures 2-3,
f(0) =P(X=0) =

3
0

17
2


20
2
=
68
95
,
f(1) =

3
1

17
1


20
2
=
51
190
andf(2) =

3
2

17
0


20
2
=
3
190
:
x 0 1 2
f(x)
68
95
51
190
3
190
FECU cWafaa S. Sayed 14 / 21

36. A shipment of 20 similar laptop computers to a retail outlet con-
tains 3 that are defective. If a school makes a random purchase
of 2 of these computers, nd the probability distribution for the
number of defectives.
Solution: LetXbe a random variable whose valuesxare the
possible numbers of defective computers purchased by the
school. Thenxcan only take the numbers 0, 1, and 2.
Similar to the poker hand problem in Lectures 2-3,
f(0) =P(X=0) =

3
0

17
2


20
2
=
68
95
,
f(1) =

3
1

17
1


20
2
=
51
190
andf(2) =

3
2

17
0


20
2
=
3
190
:
x 0 1 2
f(x)
68
95
51
190
3
190
FECU cWafaa S. Sayed 14 / 21

37. A shipment of 20 similar laptop computers to a retail outlet con-
tains 3 that are defective. If a school makes a random purchase
of 2 of these computers, nd the probability distribution for the
number of defectives.
Solution: LetXbe a random variable whose valuesxare the
possible numbers of defective computers purchased by the
school. Thenxcan only take the numbers 0, 1, and 2.
Similar to the poker hand problem in Lectures 2-3,
f(0) =P(X=0) =

3
0

17
2


20
2
=
68
95
,
f(1) =

3
1

17
1


20
2
=
51
190
andf(2) =

3
2

17
0


20
2
=
3
190
:
x 0 1 2
f(x)
68
95
51
190
3
190
FECU cWafaa S. Sayed 14 / 21

38. A shipment of 20 similar laptop computers to a retail outlet con-
tains 3 that are defective. If a school makes a random purchase
of 2 of these computers, nd the probability distribution for the
number of defectives.
Solution: LetXbe a random variable whose valuesxare the
possible numbers of defective computers purchased by the
school. Thenxcan only take the numbers 0, 1, and 2.
Similar to the poker hand problem in Lectures 2-3,
f(0) =P(X=0) =

3
0

17
2


20
2
=
68
95
,
f(1) =

3
1

17
1


20
2
=
51
190
andf(2) =

3
2

17
0


20
2
=
3
190
:
x 0 1 2
f(x)
68
95
51
190
3
190
FECU cWafaa S. Sayed 14 / 21

39. A shipment of 20 similar laptop computers to a retail outlet con-
tains 3 that are defective. If a school makes a random purchase
of 2 of these computers, nd the probability distribution for the
number of defectives.
Solution: LetXbe a random variable whose valuesxare the
possible numbers of defective computers purchased by the
school. Thenxcan only take the numbers 0, 1, and 2.
Similar to the poker hand problem in Lectures 2-3,
f(0) =P(X=0) =

3
0

17
2


20
2
=
68
95
,
f(1) =

3
1

17
1


20
2
=
51
190
andf(2) =

3
2

17
0


20
2
=
3
190
:
x 0 1 2
f(x)
68
95
51
190
3
190
FECU cWafaa S. Sayed 14 / 21

40. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

41. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

42. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

43. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

44. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

45. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

46. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

47. If a car agency sells 50% of its inventory of a certain foreign
car equipped with side airbags, nd a formula for the probability
distribution of the number of cars with side airbags among the
next 4 cars sold by the agency.
Solution: Cars are either with or without airbags.
The total num-
ber of sample points can be obtained using
the generalized mul-
tiplication rule and equals
2
4
=16.
The number of ways of sellingxcars with side airbags among
the next 4 cars is equivalent to
combinations/partitioning.
)f(x) =P(X=x) =

4
x

16
; forx=0;1;2;3;4:
FECU cWafaa S. Sayed 15 / 21

48. There are many problems where we may wish to compute the
probability that the observed value of a random variableXwill
be less than or equal to some real numberx.
Denition 3.5
Thecumulative distribution functionF(x)of a discrete ran-
dom variableXwith probability distributionf(x)is
F(x) =P(Xx) =
X
tx
f(t);for 1<x<1:
Properties:
1
0F(x)1
2
F(1) =0andF(1) =1
3
P(a<Xb) =F(b)F(a)
4
P(X>x) =1F(x)
5
Can we getf(x)fromF(x)?
FECU cWafaa S. Sayed 16 / 21

49. Example
For the previous example, ndF(x)and verify thatf(2) =3=8
using it.
Recallf(x) =

4
x

16
; forx=0;1;2;3;4:
Solution: By direct substitution,
f(0) =1=16;f(1) =1=4;f(2) =3=8;f(3) =1=4;andf(4) =1=16:F(0) =f(0) =
1
16
;
F(1) =f(0) +f(1) =
5
16
;
F(2) =f(0) +f(1) +f(2) =
11
16
;
F(3) =f(0) +f(1) +f(2) +f(3) =
15
16
;
F(4) =f(0) +f(1) +f(2) +f(3) +f(4) =1:
FECU cWafaa S. Sayed 17 / 21

50. Example
For the previous example, ndF(x)and verify thatf(2) =3=8
using it.
Recallf(x) =

4
x

16
; forx=0;1;2;3;4:
Solution: By direct substitution,
f(0) =1=16;f(1) =1=4;f(2) =3=8;f(3) =1=4;andf(4) =1=16:F(0) =f(0) =
1
16
;
F(1) =f(0) +f(1) =
5
16
;
F(2) =f(0) +f(1) +f(2) =
11
16
;
F(3) =f(0) +f(1) +f(2) +f(3) =
15
16
;
F(4) =f(0) +f(1) +f(2) +f(3) +f(4) =1:
FECU cWafaa S. Sayed 17 / 21

51. Example
For the previous example, ndF(x)and verify thatf(2) =3=8
using it.
Recallf(x) =

4
x

16
; forx=0;1;2;3;4:
Solution: By direct substitution,
f(0) =1=16;f(1) =1=4;f(2) =3=8;f(3) =1=4;andf(4) =1=16:F(0) =f(0) =
1
16
;
F(1) =f(0) +f(1) =
5
16
;
F(2) =f(0) +f(1) +f(2) =
11
16
;
F(3) =f(0) +f(1) +f(2) +f(3) =
15
16
;
F(4) =f(0) +f(1) +f(2) +f(3) +f(4) =1:
FECU cWafaa S. Sayed 17 / 21

52. Example
For the previous example, ndF(x)and verify thatf(2) =3=8
using it.
Recallf(x) =

4
x

16
; forx=0;1;2;3;4:
Solution: By direct substitution,
f(0) =1=16;f(1) =1=4;f(2) =3=8;f(3) =1=4;andf(4) =1=16:F(0) =f(0) =
1
16
;
F(1) =f(0) +f(1) =
5
16
;
F(2) =f(0) +f(1) +f(2) =
11
16
;
F(3) =f(0) +f(1) +f(2) +f(3) =
15
16
;
F(4) =f(0) +f(1) +f(2) +f(3) +f(4) =1:
FECU cWafaa S. Sayed 17 / 21

53. Example (Continued)
Note thatF(x)is a staircase function unlikef(x), which is a set
of discrete values
F(x) =
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
0;forx<0
1
16
;for0x<1
5
16
;for1x<2
11
16
;for2x<3
15
16
;for3x<4
1;forx4:
Verication:f(2) =F(2)F(1) =
11
16

5
16
=
3
8
:
FECU cWafaa S. Sayed 18 / 21

54. Example (Continued)
Note thatF(x)is a staircase function unlikef(x), which is a set
of discrete values
F(x) =
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
0;forx<0
1
16
;for0x<1
5
16
;for1x<2
11
16
;for2x<3
15
16
;for3x<4
1;forx4:
Verication:f(2) =F(2)F(1) =
11
16

5
16
=
3
8
:
FECU cWafaa S. Sayed 18 / 21

55. In Smith, Jones, and Brown example, ndF(m).
m 0 1 3
f(m) =P(M=m)
1
3
1
2
1
6
F(m) =
8
>
>
<
>
>
:
0;form<0
1
3
;for0m<1
5
6
;for1m<3
1;form3:
FECU cWafaa S. Sayed 19 / 21

56. In Smith, Jones, and Brown example, ndF(m).
m 0 1 3
f(m) =P(M=m)
1
3
1
2
1
6
F(m) =
8
>
>
<
>
>
:
0;form<0
1
3
;for0m<1
5
6
;for1m<3
1;form3:
FECU cWafaa S. Sayed 19 / 21

57. In Smith, Jones, and Brown example, ndF(m).
m 0 1 3
f(m) =P(M=m)
1
3
1
2
1
6
F(m) =
8
>
>
<
>
>
:
0;form<0
1
3
;for0m<1
5
6
;for1m<3
1;form3:
FECU cWafaa S. Sayed 19 / 21

58. Instead of plotting the points(x;f(x)), we more frequently
construct rectangles with bases of equal width. They are
centered at each valuexwithout spaces in between and
their heights are equal to the corresponding probabilities
given byf(x).
P(X=x)is equal to the area of the rectangle centered atx,
where the bases usually have unit width.
FECU cWafaa S. Sayed 20 / 21

59. ThankYou