Chp 5: Discrete Probability Distribution
Lengchivon Kou
Middlesex Community College
March 6, 2023
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Motivation: Suppose tossing a coin twice
We get:
HH
HT
TT
P (0 head) =
1
= .25
4
P (1 head) =
2
= 0.5
4
P (2 head) =
1
= 0.25
4
TH
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Motivation: Suppose tossing a coin twice
We get:
HH
HT
TT
P (0 head) =
1
= .25
4
P (1 head) =
2
= 0.5
4
P (2 head) =
1
= 0.25
4
TH
Today, we study further: Probability Distribution
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
A Random Variable:
a variable that has a single numerical value
numerical value determined by chance
typically, x = Random Variable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
A Random Variable:
a variable that has a single numerical value
numerical value determined by chance
typically, x = Random Variable
Ex: Tossing a coin twice : HH
P (0 head) =
1
4
P (1 head) =
Lengchivon Kou
HT
2
4
TT
TH
P (2 head) =
1
4
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
A Random Variable:
a variable that has a single numerical value
numerical value determined by chance
typically, x = Random Variable
Ex: Tossing a coin twice : HH
P (0 head) =
1
4
P (1 head) =
HT
2
4
TT
P (2 head) =
Let x = # of head =⇒ Random Variable
Lengchivon Kou
TH
x=0
1
1
4
2
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Two Kinds Random Variable
Discrete random variable: a finite or countable values
Ex:
1
2
3
... =⇒ finite or countable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Two Kinds Random Variable
Discrete random variable: a finite or countable values
Ex:
1
2
3
... =⇒ finite or countable
Continuous random variable: infinite or uncountable values
Ex: weight, height: 3.175 kg
Lengchivon Kou
5.9 in =⇒ continuous scale
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
c. Numbers of students in statistics classes
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
c. Numbers of students in statistics classes
=⇒ Discrete random variable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
c. Numbers of students in statistics classes
=⇒ Discrete random variable
d. Foot length of humans
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
c. Numbers of students in statistics classes
=⇒ Discrete random variable
d. Foot length of humans
=⇒ Continuous random variable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
c. Numbers of students in statistics classes
=⇒ Discrete random variable
d. Foot length of humans
=⇒ Continuous random variable
e. Eye color of students in statistics classes
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Identify Discrete RV, Continuous RV or Not RV:
a. Exact weight of the next 100 babies born in the U.S
=⇒ Continuous random variable
b. Response of survey ”which hand do you lean to?”
=⇒ Not a random variable
c. Numbers of students in statistics classes
=⇒ Discrete random variable
d. Foot length of humans
=⇒ Continuous random variable
e. Eye color of students in statistics classes
=⇒ not random variable
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Probability Distribution: 3 requirements
1
Random variable x = numerical value and its probability P (x)
2
0 ≤ P (x) ≤ 1
P
P (x) = 1
3
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Tossing a coin twice:
HH
HT
TT
TH
1
= .25
4
2
P (1 head) = = 0.5
4
P (0 head) =
P (2 head) =
1
= 0.25
4
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Tossing a coin twice:
HH
HT
TT
TH
1
= .25
4
2
P (1 head) = = 0.5
4
P (0 head) =
P (2 head) =
Let
1
= 0.25
4
x = # of head
=⇒ x = 0, 1, 2
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Tossing a coin twice:
HH
HT
TT
TH
1
= .25
4
2
P (1 head) = = 0.5
4
P (0 head) =
P (2 head) =
Let
1
= 0.25
4
x = # of head
=⇒ x = 0, 1, 2
Probability of head = P (x) =⇒ P (0)
Lengchivon Kou
P (1) P (2)
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Probability of Distribution: # of heads in two coins tossing:
x = # of head
0
1
2
P (x)
.25
.50
.25
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Probability of Distribution: # of heads in two coins tossing:
x = # of head
0
1
2
P (x)
.25
.50
.25
It satisfied 3 conditions:
2
x = 0 1 2 numerical
P
P (x) = .25 + .50 + .25 = 1
3
0 ≤ P (x) ≤ 1
1
=⇒ Probability of Distribution
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Determine whether it is a Probability Distribution?
1. # of girls in four births
x = # of girls
0
1
2
3
4
P (x)
.25
.50
.25
.50
.25
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Determine whether it is a Probability Distribution?
1. # of girls in four births
x = # of girls
0
1
2
3
4
=
0
P (x)
.25
.50
.25
.50
.25
1
x
1
2
0 ≤ P (x) ≤ 1
2
3
4
Lengchivon Kou
numerical
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Determine whether it is a Probability Distribution?
1. # of girls in four births
x = # of girls
0
1
2
3
4
1
x
2
0 ≤ P (x) ≤ 1
P
P (x) = .25 + .50 + .25 + .50 + .25 = 1.75
P
P (x) = 1.75 6= 1 =⇒ Not a Probability of Distribution
3
=
0
P (x)
.25
.50
.25
.50
.25
1
2
3
4
Lengchivon Kou
numerical
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Determine whether it is a Probability Distribution?
2. job interview mistake
x
Lateness
Bad attire
Eye contact
Texting
P (x)
.25
.15
.25
.35
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Determine whether it is a Probability Distribution?
2. job interview mistake
x
Lateness
Bad attire
Eye contact
Texting
1
2
P (x)
.25
.15
.25
.35
0 ≤ P (x) ≤ 1
P
P (x) = .25 + .15 + .25 + .35 = 1
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Determine whether it is a Probability Distribution?
2. job interview mistake
x
Lateness
Bad attire
Eye contact
Texting
P (x)
.25
.15
.25
.35
2
0 ≤ P (x) ≤ 1
P
P (x) = .25 + .15 + .25 + .35 = 1
3
x not numerical =⇒ Not a Probability of Distribution
1
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Parameter of a Probability Distribution:
µ = mean for a probability distribution
Formula:
µ = Σ[x · P (x)]
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Parameter of a Probability Distribution:
µ = mean for a probability distribution
Formula:
µ = Σ[x · P (x)]
σ = Standard deviation for Prob Distribution
p
p
Formula: σ = Σ[(x − µ)2 · P (x)] =
Σ[x2 · P (x)] − µ2
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Parameter of a Probability Distribution:
µ = mean for a probability distribution
Formula:
µ = Σ[x · P (x)]
σ = Standard deviation for Prob Distribution
p
p
Formula: σ = Σ[(x − µ)2 · P (x)] =
Σ[x2 · P (x)] − µ2
Variance = σ 2
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
µ=
Σ[x · P (x)] =
0(.25) + 1(.50) + 2(.25) =
Lengchivon Kou
1.0
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
µ=
Σ[x · P (x)] =
0(.25) + 1(.50) + 2(.25) =
1.0
b) Find standard deviation σ
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
µ=
Σ[x · P (x)] =
0(.25) + 1(.50) + 2(.25) =
1.0
b) Find standard deviation σ
p
σ = Σ[(x − µ)2 · P (x)]
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
µ=
Σ[x · P (x)] =
0(.25) + 1(.50) + 2(.25) =
1.0
b) Find standard deviation σ
p
σ = Σ[(x − µ)2 · P (x)]
p
σ = (0 − 1)2 · .25 + (1 − 1)2 · .50 + (2 − 1)2 · .25 = .7
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
µ=
Σ[x · P (x)] =
0(.25) + 1(.50) + 2(.25) =
1.0
b) Find standard deviation σ
p
σ = Σ[(x − µ)2 · P (x)]
p
σ = (0 − 1)2 · .25 + (1 − 1)2 · .50 + (2 − 1)2 · .25 = .7
c) Find the variance
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
a) Find expected value or mean µ
µ=
Σ[x · P (x)] =
0(.25) + 1(.50) + 2(.25) =
1.0
b) Find standard deviation σ
p
σ = Σ[(x − µ)2 · P (x)]
p
σ = (0 − 1)2 · .25 + (1 − 1)2 · .50 + (2 − 1)2 · .25 = .7
c) Find the variance
Variance = σ 2 =(.7)2 = .5
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
Find mean, standard deviation, variance
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
Find mean, standard deviation, variance
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
Find mean, standard deviation, variance
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
2
STAT =⇒ CALC =⇒ 1 enter : List = L1 Freqlist = L2
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex1: Probability of Distribution: # of heads in two coins tossing
x = # of head
0
1
2
P (x)
.25
.50
.25
Find mean, standard deviation, variance
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
2
STAT =⇒ CALC =⇒ 1 enter : List = L1 Freqlist = L2
We get:
µ = 1.0
σ = .7
Variance = σ 2 =(.7)2 = .5
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex2 Prob distribution: # of girls in eight births
x = # of girls P (x)
0
.004
1
.031
2
.109
3
.219
4
.273
5
.219
6
.109
7
.031
8
.004
Find mean, standard deviation, variance
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex2 Prob distribution: # of girls in eight births
x = # of girls P (x)
0
.004
1
.031
2
.109
3
.219
4
.273
5
.219
6
.109
7
.031
8
.004
Find mean, standard deviation, variance
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
2
STAT =⇒ CALC =⇒ 1 enter : List = L1 Freqlist = L2
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex2 Prob distribution: # of girls in eight births
x = # of girls P (x)
0
.004
1
.031
2
.109
3
.219
4
.273
5
.219
6
.109
7
.031
8
.004
Find mean, standard deviation, variance
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
2
STAT =⇒ CALC =⇒ 1 enter : List = L1 Freqlist = L2
We get:
µ = 4.0
σ = 1.4
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Range Rule of Thumb: Identify significant value
Significantly low value
Significantly high value
≤ (µ − 2σ)
≥ (µ + 2σ)
Value not Significant: between (µ − 2σ) and (µ + 2σ)
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
From Ex2 of # of girls in eight births,
We found: µ = 4.0
σ = 1.4
Is 3 girls a significant low? Is 7 girls a significant high?
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
From Ex2 of # of girls in eight births,
We found: µ = 4.0
σ = 1.4
Is 3 girls a significant low? Is 7 girls a significant high?
Find
(µ − 2σ) = 4.0 − 2(1.4) = 1.2
Find
(µ + 2σ) = 4.0 + 2(1.4) = 6.8
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
From Ex2 of # of girls in eight births,
We found: µ = 4.0
σ = 1.4
Is 3 girls a significant low? Is 7 girls a significant high?
Find
(µ − 2σ) = 4.0 − 2(1.4) = 1.2
Find
(µ + 2σ) = 4.0 + 2(1.4) = 6.8
Since
3 ∈ [1.2, 6.8] =⇒ 3 is not significantly low value
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
From Ex2 of # of girls in eight births,
We found: µ = 4.0
σ = 1.4
Is 3 girls a significant low? Is 7 girls a significant high?
Find
(µ − 2σ) = 4.0 − 2(1.4) = 1.2
Find
(µ + 2σ) = 4.0 + 2(1.4) = 6.8
Since
3 ∈ [1.2, 6.8] =⇒ 3 is not significantly low value
Since
7 ≥ 6.8 =⇒ 7 is significantly high value
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex3: # of adults in group of five who reported sleepwalking.
x = # of adults
0
1
2
3
4
5
P (x)
.172
.363
.306
.129
.027
.002
a). Find mean, standard deviation
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex3: # of adults in group of five who reported sleepwalking.
x = # of adults
0
1
2
3
4
5
P (x)
.172
.363
.306
.129
.027
.002
a). Find mean, standard deviation
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
2
STAT =⇒ CALC =⇒ 1 enter : List = L1 Freqlist = L2
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Ex3: # of adults in group of five who reported sleepwalking.
x = # of adults
0
1
2
3
4
5
P (x)
.172
.363
.306
.129
.027
.002
a). Find mean, standard deviation
1
Enter Data L1, L2: STAT =⇒ Edit: L1 = x, L2 = P(x)
2
STAT =⇒ CALC =⇒ 1 enter : List = L1 Freqlist = L2
We get:
µ = 1.5 sleepwalkers
Lengchivon Kou
σ = 1.0 sleepwalkers
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
b). Use Range Rule of Thumb: To find Maximum and
Minimum value in the range
We found: µ = 1.5
σ=1
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
b). Use Range Rule of Thumb: To find Maximum and
Minimum value in the range
We found: µ = 1.5
σ=1
Minimum value in the range:
(µ − 2σ) = 1.5 − 2(1) = −.5
Maximum value in the range:
(µ + 2σ) = 1.5 + 2(1) = 3.5
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
b). Use Range Rule of Thumb: To find Maximum and
Minimum value in the range
We found: µ = 1.5
σ=1
Minimum value in the range:
(µ − 2σ) = 1.5 − 2(1) = −.5
Maximum value in the range:
(µ + 2σ) = 1.5 + 2(1) = 3.5
c) Use the range rule of thumb to determine 4 is a significantly
high number of sleepwalker?
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
b). Use Range Rule of Thumb: To find Maximum and
Minimum value in the range
We found: µ = 1.5
σ=1
Minimum value in the range:
(µ − 2σ) = 1.5 − 2(1) = −.5
Maximum value in the range:
(µ + 2σ) = 1.5 + 2(1) = 3.5
c) Use the range rule of thumb to determine 4 is a significantly
high number of sleepwalker?
since 4 > 3.5 =⇒ 4 is significantly high.
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
b). Use Range Rule of Thumb: To find Maximum and
Minimum value in the range
We found: µ = 1.5
σ=1
Minimum value in the range:
(µ − 2σ) = 1.5 − 2(1) = −.5
Maximum value in the range:
(µ + 2σ) = 1.5 + 2(1) = 3.5
c) Use the range rule of thumb to determine 4 is a significantly
high number of sleepwalker?
since 4 > 3.5 =⇒ 4 is significantly high.
d) Use the range rule of thumb to determine 3 is a significantly
high number of sleepwalker?
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
b). Use Range Rule of Thumb: To find Maximum and
Minimum value in the range
We found: µ = 1.5
σ=1
Minimum value in the range:
(µ − 2σ) = 1.5 − 2(1) = −.5
Maximum value in the range:
(µ + 2σ) = 1.5 + 2(1) = 3.5
c) Use the range rule of thumb to determine 4 is a significantly
high number of sleepwalker?
since 4 > 3.5 =⇒ 4 is significantly high.
d) Use the range rule of thumb to determine 3 is a significantly
high number of sleepwalker?
since 3 < 3.5 =⇒ 3 is not significantly high.
Lengchivon Kou
Chp 5: Discrete Probability Distribution
5.1 Probability Distribution
Lengchivon Kou
Chp 5: Discrete Probability Distribution