Transcript probability distributions

PROBABILITY DISTRIBUTIONS
Probability distributions → A listing of all the
outcomes of an experiment and the probability
assosiated with each outcome, the type :
a. Discrete Probability Distributions
b. Continuos Probability Distributions
How can we generate a probability distribution ?
Suppose we are interested in the number of
heads showing face up on the first tosses of a
coin. This is the experiment.the possible results
are: zero heads, one heads, two heads, and
three heads. What is the probability distribution
for the number of heads?
These results are listed below
Possible
Result
Coin Toss
First
Second
Number of
Third
Heads
1
T
T
T
0
2
T
T
H
1
3
T
H
T
1
4
T
H
H
2
5
H
T
T
1
6
H
T
H
2
7
H
H
T
2
8
H
H
H
3
Probability distribution for the events of zero,one,
two, and three heads showing face up on three of
a coin
Number of Heads
x
0
1
2
3
Total
Probability of Outcome
P (x)
1/8 = 0,125
3/8 = 0,375
3/8 = 0,375
1/8 = 0,125
8/8 = 1,000
Graphical presentation of the number of
heads resulting from three tosses of a coin
and the corresponding probability
0,4
0,35
0,3
0,25
0,2
0,15
0,1
0,05
0
0
1
2
3
The Mean, Variance, and Standard Deviation of a
Probability Distributions
1. Mean : μ = Σ [xP(x)]
2. Variance : σ2 = Σ [(x-μ)2 P(x)]
3. Stand Dev : σ = √σ2
Example:
John sells new car for Pelican Ford. John
usually sells the largest number of cars on
Saturday. He has the following probability
distributions for the number of cars he expects
to sell on a particular Saturday.
Number of Cars Sold Probability of Outcome
x
P (x)
0
0,1
1
0,2
2
0,3
3
0,3
4
0,1
Total
1,0
Questions:
1.What type of distributions is this?
This a discrete probability distribution
2.On a typical Saturday, how many cars does
John expect to sell?
μ = Σ [xP(x)]
= 0(0,10)+1(0,20)+2(0,30)+3(0,30)+4(0,10)
= 2,1
Number of
Cars Sold
X
0
1
2
3
4
Total
Probability
P(x)
x . P(x)
0,10
0,20
0,30
0,30
0,10
1,00
0,00
0,20
0,60
0,90
0,40
μ = 2,10
This value indicates that, over a large number
on Saturday expects to sell a mean 2,1 cars
a day. In a year he can expect to sell 50 x 2,1
= 105 cars.
3. What is the variance of the distribution?
Number Probability
of Cars
P(x)
Sold
0
0,10
1
0,20
2
0,30
3
0,30
4
0,10
(x-μ)
(x-μ)2
(x-μ)2 P(x)
0-2,1
1-2,1
2-2,1
3-2,1
4-2,1
4,41
1,21
0,01
0,81
3,61
0,441
0,242
0,003
0,243
0,361
σ2 = 1,290
Recall that the standard deviation, σ, is the
positive square root of the variance. In this
example = 1,136 cars.
If other salesperson, Rita, also sold a mean
of 2,1 cars on Saturday and have the
standard deviation in her sales was 1,91
cars. We would conclude that there is
more variability in Saturday sales of Rita
than john .
A. Discrete Probability Distributions
→ a discrete can assume only a certain
number of separated values. It there are
100 employees, then the count of the
number absent on Monday can only be
0,1,2,3,…,100. A discrete is usually the
result of counting something.
1. Binomial Probability Distribution
Characteristics :
a. An outcome on each trial of an experiment is
classified into one of two mutually exclusive
categories- success or failure
b. The random variable counts the number of
successes in a fixed number of trials
c. The probability of success and failure stay the
same for each trial
d. The trials are independent, meaning that the
outcome of one trial does not affect the
outcome of any other trial
Binomial PD : P(x) = nCx πx (1-π)n-x
Example:
There are five flights daily from pittsburgh
via US Airways into the Bradford,
Pennysylvania Regional Airport. Suppose
the probability that any flight arrives late is
0.20. What is the probability that none of
the flights are late today? What is the
probability that exactly one of the flights is
late today?
P(0) = nCx πx (1-π)n-x
= 5C0 (,20)0 (1-,20)5-0
= (1)(1)(,3277) = 0,3277
P(1) = nCx πx (1-π)n-x
=5C1 π1 (1-,20)5-1
= (5)(,20)(,4096) =n 0,4096
Binomial Probability Distribution for n=5,n=,20
Number of Late Flights
Probability
0
1
2
3
4
5
0,3277
0,4096
0,2048
0,0512
0,0064
0,0003
1,0000
Total
Mean of Binomial Distribution
μ = nπ
= (5)(,20) = 1,0
Variance of Binomial Distributions
σ2 = nπ (1- π)
= 5(,20)(1-,20) = 0,80
2. Hypergeometric Probability Distribution
Characteristics:
a. An outcome on each trial of an experiment is
classified into one of two mutually exclusive
categories- success or failure
b. The random variable is the number of
successes in a fixed number of trials
c. The trials are not independent
d. We assume that we sample from a finite
population without replacement. So the
probability of a success changes for each trial.
Hypergeometric Distribution
P(x) = (SCx)(N-SCn-x)
NCn
Example:
Play Time Toys, Inc. employs 50 people in the
assembly Departement. Forty of the employees
belong to a union and ten do not. Five
employees are selected at random to form a
commite to meet with management regarding
shift starting times. What is the probability that
four of the five selected for the committee belong
to a union?
The Answer:
P(x) = (SCx)(N-SCn-x)
NCn
P(4) = (40C4)(50-40C5-4)
50C5
= (91.390)(10) = 0,431
2.118.760
3.Poisson Probability Distribution
The probability Poisson describes the number
of times some event occurs during aspecified
interval
Characteristics :
1. The random variable is the number of times
some event occurs during a defined interval
2. The probability of the event is proportional to
the size of the interval
3. The intervals which do not overlap and are
independent
Poisson Distribution
P(x) = μxe-μ
x!
Where e = 2,71828
Mean of a Poisson Distribution
μ = nπ
B.Continuos Probability Distributions
→ A continuous probability distribution
usually results from measuring something,
such as the distance from the dormitory to
the classroom, the weight of an individual,
or the amount of bonus earned by Ceos.
Suppose we select five student and find
the distance, in miles, they travel to attend
class as 12.2, 8.9, 6.7 and 14.6.
We consider two families of Continuous
Distribution :
a. Uniform Probability Distribution
Uniform distribution : P(x) = 1
b–a
Mean : μ = a+b
2
Standar Deviasi : σ = √(b-a)2
12
b. Normal Probability Distribution
The number of normal distributions is unlimited,
each having a different mean, Standard
deviation, or both, While it is possible to provide
probability tables for discrete distributions such
asa the binomial and the poisson, providing
tables for infinite number of normal distributions
is impossible. Fortunately, one member of the
family can be used to determine the probabilities
for all normal distributions. It is called the
standard normal distribution, and it is unique
because it has a mean of 0 and a standard
deviation of 1.
Any normal distribution can be converted
into a standard normal distribution by
subtracting the mean from each observation
and dividing this difference by the standard
deviation. The results are called z values.
They are also referred to as z scores, the z
statistics, the standard normal deviates,the
standar normal values, or just the normal
deviate.
Z value → the signed distance between a
selected value, disigned x, and the
mean,divided by the standard deviation.
Formula :
Standard Normal Value : z = X –μ
σ
Where:
X is the value of any particular observation or
measurement.
μ is the mean of the distribution
σ is the standard deviation of the distribution
Example:
The weekly incomes of shift foremen in the glass
industry are normally distributed with a mean of
$1,000 and a standard deviation of $100. What is the
z value foe the income X of a foreman who earns
$1,100 per week? For a foreman who earns $900 per
week?
for X= $1,100
For X = $900
z=X–μ
z=X–μ
σ
σ
= $1,100 - $ 1,000
= $900 - $1,000
$100
$100
= 1,00
= - 1,00