Chapter 4
4.1
Random Variable = the numerical value of an outcome (outcomes happen by chance)
Discrete Random Variable = “whole” numbers, the count of something, things you can count and
cannot be divided into parts of things.
Example: number of cars, number of sales calls, number of people
Continuous Random Variable = random variable that can have fraction amounts like on a number
line
Example: Height of person, weight of backpack, temperature, distance traveled, time, volume
Discrete Probability Distribution = list of all outcomes a discrete random variable can have and the
probability of each.
Remember: All probabilities are between 0 and 1, inclusive. 0 ≤ P(X) ≤ 1
If a probability is less than 0 or greater than 1, it is invalid.
Sum of all probabilities = 1.0
If the sum doesn’t =1, then it isn’t a complete probability distribution.
Example:
Sales per Day
X
0
1
2
3
4
TOTAL
Mean of Discrete Random Variable
μ = Σ X ∙ P(x)
Frequency
7
10
15
5
3
40
Probability
P(X)
7/40 = 0.175
10/40 = 0.25
15/40 = 0.375
5/40 = 0.125
3/40 = 0.075
1.00
μ
Example:
μ = 0 ∙ 0.175 + 1 ∙ 0.25 + 2 ∙ 0.375 + 3 ∙ 0.125 + 4 ∙ 0.075 = 1.675
The average number of sales per day is 1.675
Expected Value of Discrete Random Variable
E(X) = μ
E(X)
Example:
E(X) = μ = 1.675
The expected value is 1.675 sales per day.
Variance of Discrete Random Variable
σ2 = Σ (X – μ)2 ∙ P(x)
σ2
σ2 = (0 – 1.675)2 ∙ 0.175 + (1 – 1.675)2 ∙ 0.25 + (2 – 1.675)2 ∙ 0.375 + (3 – 1.675)2 ∙ 0.125
+ (4 – 1.675)2 ∙ 0.075 = 1.269
Standard Dviation of Discrete Random Variable
σ=
√𝝈𝟐
σ = √1.269 = 1.127
σ
4.2
Binomial Distribution ( A type of discrete probability distribution)
Binomial Experiment = experiment or event where there are only TWO outcomes, success or failure,
yes or no, make a purchase or not make a purchase, making a basket or not making a basket.
n = number of times you repeat the experiment, number of trials, this number of trials is FIXED, you
know how many you will be doing.
p = probability of success, this probability must be the same for all trials
q = probability of failure
Remember, there are only two outcomes so
p+q=1
q=1-p
x = the number of successes in n trials. x = 0, 1, 2, 3, 4, … , n
Example:
Randomly select one card from a deck of cards. If it is a club it is a success; if it is not a club it
is a failure. Do this 5 times.
n=5
p = 13/52 = ¼
q=1–p=1-¼=¾
x can be 0, 1, 2, 3, 4, 5 times that you get a club in the 5 trials.
The FOUR conditions for a Binomial Experiment
1. The experiment is repeated for a fixed number of trials and each trial is independent (not
affecting the other trials).
2. There are only two possible outcomes for each trial. The outcomes can be classified as success
(S) or failure (F).
3. The probability of a success P(S) is the same for each trial.
4. The random variable x is the number of successful trials in the n trials that you do (x will be a
number from 0 to n).
Find the probability of x successes in n trials in binomial experiment:
𝑛!
P(X) = nCx px qn-x = (𝑛−𝑥)!
𝑥!
𝑝 𝑥 𝑞 𝑛−𝑥
Binomial Distribution
Find the probability of all the x’s possible from 0 to n.
P(0)
P(1)
P(2)
P(3)
…
P(n)
Mean of a Binonmial Distribution μ
μ = np
Variance of a Binonmial Distribution σ2
σ2 = npq
Standard Deviation of a Binonmial Distribution σ
σ = √𝝈𝟐 = √𝒏𝒑𝒒
Example:
Binomial Experiment
Probabilility of success = 41%. Do the trial 4 times. Find the probability that there are at least 2
successes out of the 4 trials.
n= 4
p = 0.41
q = 1 – p = 1 – 0.41 = 0.59
Find P(2 or more) = P(2) + P(3) + P(4)
P(2) = 4C2 p2 q4-2 = 6 (0.41)2(0.59)2 =0.351094
Or
4!
P(2) = (4−2)!
4!
2!
𝑝2 𝑞 4−2 = (2)!
2!
(0.41)2 (0.59)2 = 0.351094
P(3) = 4C3 p3 q4-3 = 4 (0.41)3(0.59)1 =0. 162654
Or
4!
P(3) = (4−3)!
4!
3!
𝑝3 𝑞 4−3 = (1)!3! (0.41)3 (0.59)1 = 0.162654
P(4) = 4C4 p4 q4-4 = 4 (0.41)4(0.59)0 =0.028258
Or
4!
P(4) = (4−4)!
4!
4!
𝑝4 𝑞 4−4 = (0)!4! (0.41)4 (0.59)0 = 0.08258
P(2 or more) = P(2) + P(3) + P(4) = 0.351094 + 0.162654 + 0.08258 = 0.542
4.3
Geometric Distribution = ( A type of discrete probability distribution)
Experiment or event where there are only TWO outcomes, success or failure, yes or no, make a
purchase or not make a purchase, making a basket or not making a basket.
You repeat the trials UNTIL the first success.
n = number of times you repeat the experiment, number of trials until there is a success. This number
of trials is not fixed, you don’t know how many you will be doing.
p = probability of success, this probability must be the same for all trials
q = probability of failure
Remember, there are only two outcomes so
p+q=1
q=1-p
x = the number of trials needed til you get the first success. x = 0, 1, 2, 3, 4, 5, 6, 7, 8, … no end
The FOUR conditions for a Geometric Distribution
1. There are only two possible outcomes for each trial. The outcomes can be classified as success
(S) or failure (F).
2. The trials are repeated until a success occurs. The random variable x is the number of trials
needed until a success occurs.
3. The repeated trials are independent of each other.
4. The probability of a success P(S) is the same for each trial.
Find the probability of the first success occurring on trial number x:
P(X) = p qx-1
Example:
p = 0.23 q = 0.77
Find the probability of a success occurring on the 4th trial
P(4) = (0.23)(0.77)3 = 0.105003 (about 10%)
Poisson Distribution = ( A type of discrete probability distribution)
Experiment or event where you count the number of successes in a given interval of time, area,
volume.
Number of successes in a certain interval of time, area, volume.
Example: number of baskets made in a basketball game, number of cars through an intersection
between noon and 1 on any given day, number of sick days used in a year, number of accidents per
month.
There is no n, p, q in this type of distribution.
x = the number of success in a given interval of time, area, volume.
x = 0, 1, 2, 3, 4, 5, 6, 7, 8, … no end
𝜇 = mean number of successes in a certain interval of time, area, volume
The conditions for a Poisson Distribution
1. The random variable x is the number of success in a given interval of time, area, volume.
2. The probability of a success P(S) is the same for each interval.
3. The number of occurrences in one interval is independent of other intervals.
Find the probability of exactly x occurrences in a given interval of time, area, volume:
P(X) =
𝜇𝑥 𝑒 −𝜇
𝑥!
Example:
The mean number of accidents per month at a certain intersection is 3. What is the probability
that in any given month 4 accidents will occur at this intersection?
𝜇=3
P(4) =
x=4
𝜇𝑥 𝑒 −𝜇
𝑥!
=
34 𝑒 −3
4!
= 0.168
This is the probability that exactly 4 accidents will occur.
Example:
The mean number of accidents per month at a certain intersection is 3. What is the probability that in
any given month more than 4 accidents will occur at this intersection?
𝜇=3
x=4
We need to find out P(5), P(6), P(7), P(8), P(9), etc and add them all up … but that’s impossible to do.
So we will instead find the sum of all probabilities of 4 or fewer and subtract from 1.0.
P(more than 4) = 1.0 – P(4 or less)
P(4 or less) = P(0) + P(1) + P(2) + P(3) + P(4)
P(0) =
𝜇𝑥 𝑒 −𝜇
𝑥!
=
30 𝑒 −3
0!
= 0.049787
This is the probability that exactly 0 accidents will occur.
P(1) =
𝜇𝑥 𝑒 −𝜇
𝑥!
=
31 𝑒 −3
1!
= 0.149361
This is the probability that exactly 1 accident will occur.
P(2) =
𝜇𝑥 𝑒 −𝜇
𝑥!
=
32 𝑒 −3
2!
= 0.224042
This is the probability that exactly 2 accidents will occur.
P(3) =
𝜇𝑥 𝑒 −𝜇
𝑥!
=
33 𝑒 −3
3!
= 0.224042
This is the probability that exactly 3 accidents will occur.
P(4) =
𝜇𝑥 𝑒 −𝜇
𝑥!
=
34 𝑒 −3
4!
= 0.168031
This is the probability that exactly 4 accidents will occur.
P(4 or less) = P(0) + P(1) + P(2) + P(3) + P(4)
= 0.049787 + 0.149361 + 0.224042 + 0.224042 + 0.168031 = 0.815273
P(more than 4) = 1 – P(4 or less) = 1.0 – 0.815273 = 0.184727
The probability that more than four accidents will occur in any given month at the intersection is 0.185.