Transcript 2Prob Distn

PROBABILITY DISTRIBUTIONS
Probability distribution:
A listing of all the values the random
variable can assume with their
corresponding probabilities.
PROBABILITY DISTRIBUTIONS
A random variable does not mean that the
values can be anything ( a random number)
Random variables have a well defined set of
outcomes and well defined probabilities for
the occurrence of each outcome.
The random refers to the fact that the
outcomes happen by chance -- that is,
you don’t know which outcome will occur
next.
PROBABILITY DISTRIBUTIONS
Example : Rolling of a die
X
p(x)
1
2
3
4
5
1/6 1/6 1/6 1/6 1/6
6
1/6
sum
6/6=1
PROBABILITY DISTRIBUTIONS
There are two types of random variables:
that is , discrete and continuous.
DRV can assume a countable number
of values.
Example: The number of sales made by a
a salesperson in a given week.
The number of customers waiting
to be served in a restaurant at a
particular time.
PROBABILITY DISTRIBUTIONS
Two types of probability distributions
Discrete
Continuous
PROBABILITY DISTRIBUTIONS
Discrete variable can take only a limited
number of values, which can be listed.
Continuous variable can take on any
value within a given range, so we cannot
list all the possible values.
PROBABILITY DISTRIBUTIONS
The probability distribution of a discrete
random variable is a graph, table or formula
that specifies the probability associated with
each possible value the random variable can
assume.
PROBABILITY DISTRIBUTIONS
Example : Tossing two coins and let x be the
number of heads observed.
Solution:
X
p(x)
0
1/4
1
1/2
p(x)
p(x)
1/2
1/2
1/4
1/4
0 1 2
x
2
1/4
0 1 2
x
PROBABILITY DISTRIBUTIONS
The mean or expected value, of a discrete
random variable x is
 = E(x) =   x  p  x 
All x
The expected value is the mean of the
probability distribution, a measure of
its central tendency.
PROBABILITY DISTRIBUTIONS
The variance of a discrete random
variable x is
  E  x        x    p x 
2
2
2
All x
PROBABILITY DISTRIBUTIONS
Expected values of discrete random variables
Example:
Examine the probability distribution for
x ( the number of heads observed in the
tossing of two fair coins)
In a large number of experiments, 1/4 should
result in x=0, 1/2 in x=1 and 1/4 in x=2
heads.
Therefore, the average number of heads is
=0(1/4) + 1(1/2) + 2(1/4)
Example 1: Consider a random variable X with the
following probability distribution:
p(x) = 0.05x,
x = 2, 3, 4, 5, or 6
a) Express the probability distribution in tabular form.
b) Find the following probabilities:
P(X≥4)
P(X > 4)
P(3≤X≤5)
P(2 < X < 4)
P(X = 4.5)
Example 2
• Let X and Y be two independent random variables with
the following probability distributions:
X
0
1
2
p(x)
0.50
0.30
0.20
Y
p(Y)
0
0.40
• Calculate E(X) and E(Y)
Calculate V(X) and V(Y)
1
0.50
2
0.10
PROBABILITY DISTRIBUTIONS
Binomial Probabilities
A binomial experiment is an experiment
which satisfies these four conditions:
1. A fixed number of trials
2. Each trial is independent of the
others.
3. There are only two outcomes
4. The probability of each outcome
remains constant from trial to
trial.
PROBABILITY DISTRIBUTIONS
Binomial Probabilities
A binomial experiment has a fixed number
of independent trials, each with only two
outcomes.
Examples: 1. Asking 300 people if they
watch NDTV news.
2. Rolling a die to see if a 5
appears
PROBABILITY DISTRIBUTIONS
Binomial Probabilities
The Binomial Formula( Probability Distribution)
n!
pr  
pq
r ! n  r !
r
Where:
p = probability of a success
q = 1-p = probability of failure
r = number of success desired
n = number of trials
nr
PROBABILITY DISTRIBUTIONS
Binomial Probabilities
Mean:
  np
Standard Deviation:
  npq
Example 3: An insurance representative has
appointments with 4 prospective clients
tomorrow. From the past experience she
knows that the probability of making a sale on
any appointment is 1 in 5 or 0.20. What is the
probability that she will sell a policy to 3 out of
4 prospective clients?
Example 4: A company is planning to sell a
new product in four areas North, South, East
and West. The probability that the product
will be successful in an area is 0.3. Success
in one area will be independent of success or
failure in the other areas. What is the
probability of success in no areas, one area,
two areas, three areas, and four areas?
Example 5: Of the total output produced in a
factory, 8% is defective. In a batch of 500
units, what is the mean expected number of
defective items, and what is the standard
deviation?
Examples
6. Find the chance of getting 3 successes in 5
trials when the chance of getting a success in
one trial is 1/4.
7. For a binomial distribution the mean is 4 and
variance is 2. Find probability of getting
a) at least 2 successes
b) at the most 2 successes
PROBABILITY DISTRIBUTIONS
Poisson Distribution
A probability distribution used when a
density of items is distributed over a
period of time.
The sample size needs to be large and
the probability of success to be small.
PROBABILITY DISTRIBUTIONS
Poisson Distribution
Characteristics:
1. The experiment consists of counting the
number of times a particular event occurs
during a given unit of time or in a given
area or volume.
2. The probability that an event occurs in a
given unit of time, area or volume is the
same for all the units.
PROBABILITY DISTRIBUTIONS
3. The number of events that occur in one
unit of time, area or volume is independent
of the number that occur in other units.
4. The mean (or expected) number of events
in each unit will be denoted by

PROBABILITY DISTRIBUTIONS
Poisson Formula(Probability Distribution)
e
Px 
x!
x

Where:
P x 
e
= probability of exactly x
occurrences
= 2.71828, constant.
Example 8: Alden and Associates write
weekend trip insurance at a very nominal
charge. Records show that the probability that
a motorist will have an accident during
weekend and file a claim is 0.0005. Suppose
Alden wrote 400 policies for the forthcoming
weekend. What is the probability that exactly
two claims will be filed?
Example 9: It is thought that, on an average,
0.01% of the workforce in a particular
industry will suffer from an acute form of
industrial disease. Hot Flush Inc., employs a
workforce of 2,000 men, and the senior
medical adviser of the company has stated
that if three people in the workforce suffer
from the disease, the cases should be
treated as a sign that there are unacceptable
health hazards.
Calculate the probability that three or more
men in the workforce will catch the disease.
Examples
10. Suppose the proportion of defective items in a
production process is 0.01. A random sample
of 100 items is selected. What is the probability
that there are
(a) no defective items
(b) one defective item
(c) 2 defective items
(d) 3 or more defective items
Use Poisson Probabilities table.
Examples
11. An advertising executive receives an average
of 6 telephone calls each afternoon between 2
and 4 P.M. The calls occur randomly and
independently of one another.
•
•
•
Find the probability that the executive will
receive 8 calls between 2 and 4 P.M. on a
particular afternoon.
Find the probability that the executive will
receive four calls between 2 and 3 P.M. on a
particular afternoon.
Find the probability that the executive will
receive at least five calls between 2 and 4 P.M.
on a particular afternoon.
PROBABILITY DISTRIBUTIONS
The Poisson distribution can be a reasonable
approximation of the binomial under the
following conditions:
1. n is large, and
2. p is small.
The rule most often used is when n is
greater than or equal to 20 and p is
less or equal to 0.05.
Normal Distribution
• This is the most important continuous
distribution.
– Many distributions can be approximated by a
normal distribution.
– The normal distribution is the cornerstone
distribution of statistical inference.
Normal Distribution
• A random variable X with mean  and
variance 2 is normally distributed if its
probability density function is given by
 x 
 (1 / 2 ) 

  
2
1
f ( x) 
e
  x  
 2
where   3.14159... and e  2.71828...
Finding Normal Probabilities
• Two facts help calculate normal probabilities:
– The normal distribution is symmetrical.
– Any normal distribution can be transformed into a
specific normal distribution called…
“STANDARD NORMAL DISTRIBUTION”
Example
The amount of time it takes to assemble a computer
is normally distributed, with a mean of 50 minutes
and a standard deviation of 10 minutes. What is
the probability that a computer is assembled in a
time between 45 and 60 minutes?
Finding Normal Probabilities
• Solution
– If X denotes the assembly time of a computer, we
seek the probability P(45<X<60).
– This probability can be calculated by creating a
new normal variable the standard normal
variable.
Every normal variable
with some  and , can
be transformed into this Z.
X  x
Z
x
E(Z) =  = 0
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
V(Z) = 2 = 1
Finding Normal Probabilities
• Example - continued
45 - 50
X 
60 - 50
P(45<X<60) = P(
<
<
)

10
10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution
Finding Normal Probabilities
• Example - continued
45 - 50
X 
60 - 50
P(45<X<60) = P(
<
<
)

10
10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = -.5
z0 = 1
Finding Normal Probabilities
• Example - continued
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328
.3413
.1915
.1915
.1915
-.5
.5 1.0
• Example 12
– The amount of soda pop in each bottle is
normally distributed with a mean of 32.2
ounces and a standard deviation of .3
ounces.
– Find the probability that a bottle bought by
a customer will contain more than 32
ounces.
Finding Values of Z
• Sometimes we need to find the value of Z
for a given probability
• We use the notation zA to express a Z
value for which P(Z > zA) = A
A
zA
Finding Values of Z
• Example 13
– Determine z exceeded by 5% of the population
– Determine z such that 5% of the population is below
• Solution
z.05 is defined as the z value for which the area on its
right under the standard normal curve is .05.
Characteristics of Normal Probability
Distribution
1. The curve is bell shaped, that is, it has the same
shape on either side of the vertical line from
mean.
2. It has a single peak. As such it is unimodal.
3. The mean is located at the centre of the
distribution.
4. The distribution is symmetrical.
5. The two tails of the distribution extend
indefinitely but never touch the horizontal axis
(asymptotic).
The mean, median and mode have the same
value, that is, mean = median = mode.
The percentage distribution of area under
standard normal curve is broadly as follows:
± 1 =68.27%; ± 2 =95.44% and
± 3 =99.73%.
The units for the standard normal distribution
curve are denoted by Z and are called the Z
values or Z scores. They are also called
standard units or standard scores. The Z
score is known as a ‘standardised’ variable
because it has a zero mean and a standard
deviation of one.
Bell-shaped curve
99%
95%
68%
  3   2
 
     2   3
Area under the normal curve,
between  ± k 
k
1
1.96
2
2.58
3
Area in %
68.26
95.00
95.46
99.00
99.73
Normal approximation to the
Binomial
– Normal approximation to the binomial works
best when
• the number of experiments (sample size) is
large, and
• the probability of success, p, is close to 0.5.
– For the approximation to provide good results
two conditions should be met:
np

5; n(1 - p)

5
Example 14
•
•
•
If X is a normal random variable with a
mean of 100 and a standard deviation of
10, find the following probabilities:
P(X ≥ 128)
P(X ≤ 113)
P(87 ≤ X ≤ 98)
Example 15
If Z is a standard normal random
variable, find the value z for which:
•
•
•
•
•
•
the area between 0 and z is 0.3729
the area to the right of z is 0.7123
the area to the left of z is 0.1736
the area to the left of z is 0.7673
the area to the right of z is 0.1841
the area between –z and z is 0.6630
Examples
16. Suppose the owner of a bakery knows that the daily
demand for his wholemeal bread is a random variable
having the mean of 400 loaves and the standard
deviation is 20. What is the probability that the demand
for its bread will exceed 450 loaves?
17. The average monthly sales of 5000 firms are normally
distributed with mean and standard deviation of
Rs.36,000 and Rs.10,000 respectively. Find
a) the number of firms having sales over Rs.40,000
b) the percentage of firms having sales between
Rs.38,500 and Rs.41,000
c) the number of firms having sales between
Rs.30,000 and Rs.40,000