theoretical distribution statistics)

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5 DECEMBER 09 www.afterschool.tk 1

THEORETICAL DISTRIBUTION(STATISTICS)

by :

DR. T.K. JAIN

AFTERSCHOOL

centre for social entrepreneurship

sivakamu veterinary hospital road

bikaner 334001 rajasthan, indiaFOR PGPSE PARTICIPANTS

mobile : 91+9414430763


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WHAT IS IT?

When we can allocate probability for differentfrequency points/ class intervals, out of the

total frequency, it is called theoreticaldistribution. It exists only in theory. The

observed (actual) frequency distribution maybe similar to it. This helps us in prediction,

projections, and in decision making. Forexample: we may say that out of 1000 toss ofcoins, we will get 500 heads (actually, it may

not be 500).


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Uses of theoretical distribution

It is used in following purpose :

statistical analysisprediction of future -

prediction of the data on the basis of the the

distributiondata analysis


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Characteristics

Theoretical distribution contains all thecharacteristics of population as it is based on

population data. So it has all the characteristics

of population parameters


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Types of distribution

There are two types of distributions :

discreet individual values for example :1,2,3,4

continuous continuous values for example : 1to 2, 2 to 3, 3 to 4 etc.


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Discrete distributions

Important ones are :

binomial

poisson


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Continuous distributions

Important ones are :

1. normal

2. chi-square

3. student (or T distribution)

4. F distribution


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Binomial distribution

When only two outcomes are possible,binomial distribution is prepared example

throw of coins (head or tail)trials are independent

P = probability of success

Q= probability of failurefinite number of trials with exclusive and

exhaustive outcomes


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Properties of binomial distribution

Mean = N*P

example : throw coin 100 timesN=100, P=.5, thus mean is 100*.5 = 50

variance =N*p*q

for coins : 100*.5*.5 = 25standard distribution = sqrt(25) = 5


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Ex: a coin is tossed 8 times, whatis the probability of getting 4

heads?

Formula of probability mass function ofbionmial distribution : Ncr*p^r*q^(n-r)

N=8, P=.5, Q=.5, r=4

8C4*(.5)^4*(.5)^4

=105/512


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Ex: a coin is tossed 8 times, whatis the probability of getting at

least 4 heads?

Formula of probability mass function of

bionmial distribution : Ncr*p^r*q^(n-r)N=8, P=.5, Q=.5, r=4,5,6,7,8

when r is 4, =8C4*(.5)^4*(.5)^4

when r is 5, =8C4*(.5)^5*(.5)^3so answer : = 837/1024


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What is the probability of getting2 sundays out of 15 days selected

at random?

N = 15, R = 2, P = 1/7, because only 1 Sundayin a week Q = 6/7 (all other days )

=15C2*(1/7)^2*(6/7)^13


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Poisson distribution

We use it when the possibility of success is in avery small time interval is given. There is very

little possibilty of getting 2 successes. It isgenerally used for situations like queue etc.

Formula : e^kt*(kt)^x / x!

Where e=2.71828 k = chance of success(between 0 to 1)

t = time and x is the value that we want to test.


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Find the mean and standard deviation of Xwhere X is poisson variate satisfying the

conditions P(x=2) = P (x=3) Options : mean=3,4,5 standard deviations : sqrt (2), sqrt (3), 2

e^kt*(kt)^x / x!

2.71828^kt*(kt)^3 / 3! = 2.71828^kt*(kt)^2/2!

Kt^3/6 = kt^2/2

kt^3 = kt^2*3

thus kt = mean = 3

standard deviation = square root of mean (in poisson

distribution) so standard deviation is sqrt(3)


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Between 5 to 6 PM the average number ofphone clls per minute is 4. what is the

probability that there is no phone callduring a minute?

e^kt*(kt)^x / x!

E = 2.71828, t = 1 minute, k = 4, x = 0 (nophone call)

solving we get : = .018 answer


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2% Cars are defective. What is theprobability that out of 150 cars,

there is exactly one defective car?

E =2.71828, KT = 2% of 150 = 3

x = 1.

solving the formula we get : .15 answer


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Normal distribution

Developed by Karl Gauss

perfectly symmetrical, mean=mode=medianvalues = equally divided from mean

+ or one sigma = .68

+ or 2 sigma = .9546+ or 3 sigma = .9973 _


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Out of 500 workers mean and standarddeviation of wages are 500 and 48

respectively. What is the number ofworkers with more than 600 as wages.

Here 500 is mid point, so it represents .5

we are interested in 600, which is more than 500

600 500 / 48 = 100/48 = 2.083

now find the value 2.083 in the table of normal distribution,which is .4812 add .5 to it : .9812,

so answer : 1 - .9812 = .0188 or 1.8% or 9 workers. Answer.


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What is the standard deviation ofthe number of recoveries among

48 patients, the probability ofrecovering is .75? Options :

36,81,9,3

Formula : N*P*Q

= 48*.75*.25 = 9 answer


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The incidence of occupational disease in anindustry is such that the workman have a 10%

chance of suffering from it. What is the

probability that out of 5 workmen 3 willcontract the disease?

Formula : NCR*P^r*Q^(n-r)5C3*(.1)^3*(.9)^2

10*.001*.81

= .0081 answer


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What is the probability of success of thebinomial distribution satisfying the

following condition: 4P(X=4) = P(X=2)and having other parameter as 6? options :

.25, .33, .4, .5

Here N = 6 (as this is the only parameter missing).

4* 6C4(p)^4*(q)^2 = 6C2*(p)^2*(q)^4

try with options, .33 gives answer so P = .33


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