1. STATISTICAL MODELS

2. Binomial Models
Occurs in experiment which have only two possible outcomes one which is termed “success” and the other “failure.”
Conditions
There must be a fixed number (n) of trials
The trials must be independent
The trials are have only two outcomes: success or failure.
The probability of success (p) is constant for each trial.

3. formula
If the probability that an experiment results in a successful outcome is p and the probability that the outcome is a failure is q, where q=1 –p.
P(X=r) = nCrprqn-r where r = 0, 1, 2, …………, n
And we say X ~ Bin (n, p)

4. Worked Example Question:
A coin is biased so that the probability of obtaining a head in 1/6.
Find the probability of that in 7 tosses of the coin exactly 3 heads are obtained.
Find the probability that of obtaining more than 3 heads

5. Solution:
(a) n = 7, p =1/6, q= 1-p=5/6 X = Bin(n, p) = Bin(7, 1/6) = nCrprqn-r) = 7C3(1/3)3(1/3)4 = 0,078

6. Solution (b) P(X>3) = 1 – P(<=2)
= 1 –(P(X=0) + P(X=1) + P(X=2) )
n = 7, p =1/6, q= 1-p=5/6
= 7C0(1/6)0(5/6)7+7C1(1/6)1(5/6)6+7C2(1/6)2(5/6)56 =
= 0,096

7. Expectation and Variance
If the random variable X is such that X ~ Bin(n, p)
Then E(X) =np
Var(X) = npq
Question
The probability that it is a rainy is 0.3. Find the expected number of rainy days in 2 weeks and the standard deviation.

8. Solutions Let the rainy day be the success .
Then n = 14, p=0.3 and q = 0.7
X ~ Bin(14, 0.3) Var(X) = npq
E(X) = np = 14*0.3*0.7
=14* =
= 4.2

9. Geometric Model
A Geometric distribution arises when we have a sequence of independent trials, each with a definite probability (p) of success and the probability. (q) of failure, where q = 1 –p.
Let X be the r.v the number of trials upto and including the first success.
Now
P(X=1) = P(success on the first attempt) = p
P(X=2) = P(success on the second attempt) = qp
P(X=3) = P(success on the third attempt) = q2p
P(X=4) = P(success on the fourth attempt) = q3p
In general P(X=r) =qr-1p

10. Question
A coin is biased that the probability of obtaining a head is 2/3. If X is r.v the number of tosses upto including the first head.
(a) Find P(X<=2)
(b) P(X>4)

11. Solution (a) X ~Geo(2/3) p = 2/3 and q = 1/3
(a) P(X<=2) = 1 – p(X>2)
= 1 –(1/3) 2
= 8/9
(b) X ~Geo(2/3)
P(X> 4) = (1 /2) 4
= 1/81

12. Expectation and Variance
If X ~ Geo(p), the E(X) =1/p and Var(X) =q/p2
where q = 1 – p
Example : If X ~ Geo(1/5), Find E(X) and Standard deviation of X
Solution:
Here p 1/5 and q = 4/5
E(X) = 1/p
= 1/(1/5)
= 5
Var(X) = q/p2
= (4/5)/(1/5) 2
= 20

13. POISSON DISTRIBUTION -was named after Simeon Dennis Poison(1781-1840)
Poison distribution is a discrete probability distribution for the count of events that occur randomly in a given interval or time.
Poison distribution might be appropriate model in the following cases.

14. Conditions under which a Poisson distribution is likely to arise are:
(a) - the N events occur independently of each other.
(b) - the events occurs singly in continuous space of time.
(c) - the events occur at a constant rate, in the sense that the mean number interval is proportional to the length of the interval.
(d) - the mean and the variance are equal.

15. Applications
The poison distribution has a single parameter (ʎ), which may be estimated from the observed data using
Applications:
The number of radioactive particles emitted by a certain source during 5-minute period.
The number of suicide cases in a city.
The number of accidents occurring
The number of bacteria per 5ml of liquid
The number of air crushes per year.

16. Formulae for Poison Distribution
A discrete random variable X had probability function of the form
Theta, lamda
P (X = r) = e‾ʎʎr r!
For r = 0,1,2,3,……..infinity
e ~ is an mathematical constant which is approximately to ~
where ʎ(lambda) can take positive values

17. Practical example of Poison Distribution
Let X be the average birth of 3 per hour Lambda =3
Now we can calculate the probability of having 4 birth per hour
Then X ~Po(3)
Now p(X=r) = e‾ʎʎr r!
P(X=0) = e‾330 0!
= 0.050

18. Now probability of having a one child P(X=1) = e‾3 311!
= 0.149
Now probability of having two babies P(X=2) = e‾3 32 2!
= 0.224
Now probability of having three babies P(X=3) = e‾3 33 3! =
Now probability of having greater than two babies
P(X>2) = 1 – P(X<=2)
=1 –((P(X=0)+P(X=1)+P(X=2))
= 1-( )
= 0.557

19. Expectation and Variance
If X ~ Geo(p), the E(X) =1/p and Var(X) =q/p2
where q = 1 – p
Example : If X ~ Geo(1/5), Find E(X) and Standard deviation of X
Solution:
Here p 1/5 and q = 4/5
E(X) = 1/p
= 1/(1/5)
= 5
Var(X) = q/p2
= (4/5)/(1/5) 2
= 20

20. CONTINUOUS DISTRIBUTIONS
Continuous random variables can be used to describe random phenomena in which the variable can take on any value in some interval.

21. Exponential Distribution
Usually, exponential distribution is used to describe the time or distance until some event happens. It is in the form of: where x ≥ 0 and μ>0. μ is the mean or expected value. Memoryless property For all s and t greater or equal to 0: P(X > s+t | X > s) = P(X > t)

22. Example: A lamp ~ exp(λ = 1/3 per hour), hence, on average, 1 failure per 3 hours. The probability that the lamp lasts longer than its mean life is: P(X > 3) = 1-(1-e-3/3) = e-1= The probability that the lamp lasts between 2 to 3 hours is: P(2 <= X <= 3) = F(3) – F(2) = 0.145