Transcript Distribution - The Toppers Way

Theoretical Distributions
Frequency Distribution:
•
•
Observed Frequency Distribution
Theoretical Frequency Distribution
Theoretical Distributions:
1.
2.
3.
4.
5.
6.
Binomial Distributions
Multinomial Distributions
Negative Binomial Distributions
Poisson Distribution
Hyper Geometric Distribution
Normal Distributions
A Discrete distribution
is based on random
variables which can
assume only clearly
separated values.
A Continuous
distribution usually
results from measuring
something.
Discrete distributions
studied include:
o Binomial
o Hypergeometric
o Poisson.
Continuous distributions
include:
o Uniform
o Normal
o Others
Discrete and continuous distributions
Binomial Distribution:
“It is a discreet probability distribution which
expresses the probability of one set of
dichotomous alternative, ie. Success (p)
and failure (q).”
Binomial Distribution:
•
•
Also Known as Bernoulli Distribution
Given By Swiss Mathematician Jacob Bernoulli
Assumptions:
1.
2.
3.
4.
5.
An experiment is performed under same condition
for a fixed no. of trials.
In each trial, there are only two possible outcomes of
experiment i.e.. Success or Failure.
The probability of Success is denoted by “p”and it remains
constant from trial to trial.
The probability of Failure is denoted by “q” which is equal to
(1-p)
The trials are statistically independent i.e. the outcome of any
trial do not effect the outcome of subsequent trials.
The Binomial Distribution:
P(r) = nCr q n-r p r
Where
p = probability of success
q = Probability of failure (1-p)
n = no. of trials
r = no. of success in ‘n’ trials
Constants of Binomial Distribution:
Mean = np
Standard Deviation = npq
Measures of Skew ness = 1 = (q-p)2 / npq
Measures of Kurtosis = 2 = 3 + (1-6pq/npq)
Examples:
1. A coin is tossed 6 times. What is the probability of
obtaining 4 or more heads.
2. Assuming that half the population is vegetarian so
that the chance of an individual being vegetarian is
½ and assuming that 100 investigators make a sample
of 10 individuals to see whether they are vegetarian.
How many investigators would you expect to report
that three or less were vegetarian.
3. If p = 0.4 and n = 50,
Calculate the constants of Binomial Distribution.
Seven coins are tossed and the number of heads were noted.
This experiment was repeated 128 times and the following
observations were made.
No. of
head
0
1
2
3
4
5
6
7
Throws
7
6
19
35
30
23
7
1
Fit a binomial distribution assuming the coin to be unbiased.
Also find its mean and standard deviation.
Solution:
Given:
N = 128
n=7
p=½
q = (1-1/2)= ½
For fitting a Binomial distribution:
No. of Heads
Expected Frequencies [ N x (nCr q n-r p r)
0
128 x 7C0 q 7-0 p 0 = 1
1
128 x 7C1 q 7-1 p 1 = 7
2
128 x 7C2 q 7-2 p 2 = 21
3
128 x 7C3 q 7-3 p 3 = 35
4
128 x 7C4 q 7-4 p 4 = 35
5
128 x 7C5 q 7-5 p 5 = 21
6
128 x 7C6 q 7-6 p 6 = 7
7
128 x 7C7q 7-7 p 7 = 1
Poisson Distribution:
• Discrete Probability Distribution
• Given By French Mathematician Simeon Denis Poison
• It is expected where chance of an individual event being
success is very small.
• It is a distribution used to describe the behavior of rare
events i.e. no. of printing errors in books, serious floods.
It is defined as:
P(r) = e-m mr / r !
Where
r = 0, 1, 2………
e = 2.7183
m = mean of Poisson Distribution = np
If we want to know the expected no. of occurrences for
different success, we have to multiply each term by N
Which is the total no. of observations
Constants of Poisson Distribution:
Mean = np
Standard Deviation = np
1 = 1/ mean
2 = 3 + 1/Mean
“ In General, the Poisson Distribution explains the
behavior of those Discrete variables where the probability
of occurrence of the event is small and the total no. of
possible cases is sufficiently large.”
Examples:
1. The mean of Poisson Distribution is 2.25.
Find other constants.
2. Suppose on an average, one house in thousand in
a certain district has a fire during a year. If there are
two thousand houses in that district, what is the
probability that exactly five houses will have a fire
during the year.
(Take e-2 = 0.1352)
3. Ten percent of the tools produced in a certain
manufacturing process turn out to be defective. Find
the probability that in a sample of 10 tools chosen at
random, exactly two will be defective by using:
• The Binomial Distribution
• The Poisson Distribution
Normal Probability Distribution:
• An important Continuous Probability Distribution
• The Graphical shape of normal Distribution called
normal curve, is bell shaped smooth symmetrical
curve.
• The normal curve or distribution is a theoretical model
which may be used to describe the frequency
distribution of vast variety of continuous variables.
• A continuous random variable is said to be normally
distributed if:
P(x) = 1
2
e
–1/2 (x- / )
2
r
a
l
i
t r
b
u
i o
n
:

=
0
,
2
=
1
Characteristics of a Normal Distribution
0
. 4
Normal
curve is
symmetrical
. 3
0
. 2
0
. 1
f ( x
0
Theoretically,
curve extends to
infinity
. 0
- 5
a
Mean, median, and
mode are equal
x
Properties of Normal Distribution:
• The normal curve is ‘bell shaped’ and symmetrical
in its appearance.
• If the curves were folded along its vertical axis, the two
halves would coincide.
• There is only one maximum point of normal curve which
occurs at mean.
• Since there is one maximum point, the normal curve is
unimodal.
• Because of Symmetry:
Mean = Median = Mode.
• Curve is asymptotic to X- axis.
• Total area under normal curve and above the horizontal
axis is 1.00
Area under normal curve is distributed as:
1. Mean  1 covers, 68.27% area
2. Mean  2 covers, 95.45% area
3. Mean  3 covers, 99.73% area
Area Relationship:
Distance from Mean
% of Total Area
0.5 
19.46%
1.0 
34.134%
1.96 
47.5%
2.00 
47.725%
2.5758 
49.5%
3.00 
49.865%
The various hypothesis are generally tested at 5% level
Or 1% level (ie. Taking account 95% and 99% of total area
Of the normal curve.
Area under normal Curve:
It is possible to transform any normal random variable X
With mean  and variance 2 to a new normal random
Variable Z with mean 0 and variance 1. This variable is
known as Standard Normal Variable (SNV).
The transformation of X to Z is given as
Z=X- /
Steps:
•
•
Use transformation as above for converting the given
normal random variable X to SNV Z.
Use the table to calculate area or probability in
the table areas under the standard normal curve.
z
X 

= 22000 - 20000
2000
= 1.00
The bi-monthly
starting salaries of
recent MBA
graduates follows
the normal
distribution with a
mean of Rs.20,000
and a standard
deviation of
Rs.2000. What is
the z-value for a
salary of
Rs.22,000?
MBA
Appendix: Area Under Normal Curve
Table of Area
z
0
1
2
3
4
5
6
7
8
9
0.0
.0000
.0040
.0080
.0120
.0160
.0199
.0239
.0279
.0319
.0359
0.1
.0398
.0438
.0478
.0517
.0557
.0596
.0636
.0675
.0714
.0753
0.2
.0793
.0832
.0871
.0910
.0948
.0987
.1026
.1064
.1103
.1141
1.0
.3413
.3438
.3461
.3485
.3508
.3531
.3554
.3577
.3599
.3621
2.7
.4965
.4966
.4967
.4968
.4969
.4970
.4971
.4972
.4973
.4974
2.8
.4974
.4975
.4976
.4977
.4977
.4978
.4979
.4979
.4980
.4981
Examples:
1. X is a normal variable with mean = 25 and standard
deviation = 5. Find the values of Z1 and Z2, such that
P(20<X<30) = P(Z1<Z<Z2)
2. Z is a standard normal variable, use table to determine
the following:
•
•
•
•
P(0<Z<1.2)
P(-1.2<Z<0)
P(-2<Z<2)
P(-<Z<1)
Example:
The average on a statistics test was 78 with a standard
deviation of 8. If the test scores are normally distributed,
find the probability that a student receives a test score
less than 90.
μ = 78
σ=8
z  x - μ = 90 - 78
σ
8
= 1.5
P(x < 90)
μ =78
90
μ =0
?
1.5
x
z
The probability that a
student receives a test
score less than 90 is
0.8531
P(x < 90) = P(z < 1.5) = 0.5 + .3531 = 0.8531
Example:
The average on a statistics test was 78 with a standard
deviation of 8. If the test scores are normally distributed,
find the probability that a student receives a test score
greater than than 85.
μ = 78
σ=8
P(x > 85)
μ =78 85
μ =0 0.88
?
z = x - μ = 85 - 78
σ
8
= 0.875  0.88
x
z
The probability that a
student receives a test
score greater than 85 is
0.1894.
P(x > 85) = P(z > 0.88) = .5  P(z < 0.88) = .5  0.3106 = 0.1894
Example:
The average on a statistics test was 78 with a standard
deviation of 8. If the test scores are normally distributed,
find the probability that a student receives a test score
between 60 and 80.
x - μ 60 - 78 = -2.25
z =
=
P(60 < x < 80)
μ = 78
σ=8
60
σ
8
x - μ 80 - 78
z2 
=
σ
8
1
μ =78 80
2.25
μ =0 0.25
?
?
x
z
= 0.25
The probability that a
student receives a test
score between 60 and 80
is 0.5865.
P(60 < x < 80) = P(2.25 < z < 0.25)
= .4878 +.0987 = 0.5865
3. Two thousand students appear in an examination.
Distribution of marks is assumed to be normal with
mean = 30 and standard deviation = 6.25. How many
students are expected to get
• Between 20 and 40 marks.
• Less than 35 marks.
4. Assuming mean height of soldiers to be 68.22 inches
with a variance of 10.8 inches. How many soldiers in
a regiment of 1000, would you expect to be over 6 feet
tall?
5. The local authorities in a certain district install
10000 electric lamps in the streets of the city. If these
lamps have an average life of 1000 burning hours with
standard deviation of 200 hours. Assuming normality,
what number of lamps might be expected to fall
• in the first 800 burning hours
• between 800 and 1200 burning hours