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• Sampling Distribution of X

• Mean and Standard Deviation of the Sampling Distribution of X

• Central Limit Theorem

What is Sampling Distribution?

The probability distribution of any statistic (such as the mean, the standard deviation, the proportion of successes in a sample, etc.) is known as its sampling distribution. In this regard, the first point to be noted is that there are two ways of sampling --- sampling with replacement, and sampling without replacement. In case of a finite population containing N elements, the total number of possible samples of size n that can be drawn from this population with replacement is Nn. 
In case of a finite population containing N elements, the total number of possible samples of size n that can be drawn from this population without replacement.


The theorem states that:

“If a variable X from a population has mean µ and finite variance σ2, then the sampling distribution of the
sample mean ⎯X approaches a normal distribution with mean µ and variance σ2/n as the sample size n approaches infinity.
” As n → ∞, the sampling distribution of⎯X approaches normality.

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Replies to This Discussion

 What is different between of the E(X) AND X? and also what is different between EY nad Y?

X is a random variable (A quantity which is determined by the result of the experiment) it may be discrete or continuous while E(X) is the mean of the random variable. Where E(X) = sum{Y*P(X)} for discrete case, E(X) = int{XP(X)}dx where -infinity<x<+infinity for continuous variable.

Y is a random variable (A quantity which is determined by the result of the experiment) it may be discrete or continuous while E(Y) is the mean of the random variable. Where E(Y) = sum{Y*P(Y)} for discrete case, E(Y) = int{YP(Y)}dY where -infinity<Y<+infinity for continuous variable.

For example:

If we toss a coin at 2 times and we are interested that what is the probability for getting the number of heads?

X (represents the number of heads)                P(X)                                    XP(X)
0                                                                    1/4                                         0
1                                                                    2/4                                        2/4  
2                                                                    1/4                                        2/4
E(X) = sum{X*P(X)} = 0+2/4+2/4 = 4/4 = 1.

 Similarly if we are interested that what is the probability for getting the number of tails?
Y (represents the number of tails)                  P(X)                                      XP(X) 
0                                                                    1/4                                          0 
1                                                                    2/4                                         2/4    
2                                                                    1/4                                         2/4
E(Y) = sum{Y*P(Y)} = 0+2/4+2/4 = 4/4 = 1.   

 In this lecture poision process's example how can caluculate the 0.91975, and why minus 0.91975 from 1?

Telephone calls are being placed through a certain exchange at random times on the average of four per minute. Assuming a Poisson Process, determine the probability that in a 15-second interval, there are 3 or more calls.

Here lamda=4 calls recieved per minute, t = 15 second interval. while we are interested duration of recieved calls in minutes. So t = 15/60 = 1/4.

Where x represents the number of occurences of the outcome in t units of time. In this example we want to calculate the probability that 3 or more calls are recieved or P(X>=3). 

It can be written in this form:

P(X>=3) = 1-P(X<3) = 1-{P(X=0)+P(X=1)+P(X=2)}                      

(In poisson distribution the range of x lies between 0 to infinity, so we cannot find the all probabilities and we know that sum of all probabilities is equal to 1. That's why we minus the those probabilities from 1 to require the probability that 3 or more calls recieved).     

where P(X) = {e^-lamdat(lamda t)^x}/x!                     where lamda t = 4*1/4 = 1.
P(X>=3) = 1-{e^-1(1)^x}/x!
               = 1-[{0.3679(1^0)}/0!+{0.3679(1^1)}/1!+ {0.3679(1^2)}/2!]
               = 1-[ 0.3679(1)/1+0.3679(1)/1+0.3679(1)/2]
               = 1-[0.3679+0.3679+0.18395]
               = 1-[0.91975]

What is central limit theorem?

 It is a theorem which states, if the sample size is large then the mean of the sample will approach to the normal distribution in the other words, you can apply the rules of normal distribution on the mean of a large sample.


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