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# Normal Distribution

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In the binomial distribution, there are a number of probabilities, like:

• the trial number is indeterminately large, e., n -> âˆž
• porq is minor, then the controlling form of the binomial distribution is known as ‘normal distribution’. It is thus a continuous distribution and the probability density function is given by,

f(x) = 1/ Ïƒ âˆš2Ï€. e-(x-Âµ)2/2Ïƒ2,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  -âˆž<x<âˆž, -âˆž <Âµ<âˆž

= 1/ Ïƒ âˆš2Ï€. exp [ – (x-Âµ)2/2Ïƒ2]

It is thus represented by N (11, cr2) where 1.1. = Mean and cr2 = Variance.

Thisarrangement was established by Carl Friedrich Gauss, which is also known as ‘Gaussian Distribution’.

Properties

(i) The normal curve is bell shaped and symmetrical about the .line x = Âµ.

(ii) Median: Let M be the median, then

MÊƒ-âˆžf(x) dx = 1/ 2

= 1/ Ïƒ âˆš2Ï€ MÊƒ-âˆžexp. [-(x-Âµ)2/2Ïƒ2]dx = 1/ 2

= 1/ Ïƒ âˆš2Ï€ MÊƒ-âˆžexp. [-(x-Âµ)2/2Ïƒ2]dx + 1/ Ïƒ âˆš2Ï€ MÊƒ-âˆžexp. [-(x-Âµ)2/2Ïƒ2]dx = Â½

In the first integral we consider z= x-Âµ/Ïƒ, then what we obtain is

1/âˆš2Ï€ âˆžÊƒ0 exp. [-z2/2] dz = Â½

So therefore, 1/ 2 + 1/ Ïƒ âˆš2Ï€ MÊƒÂµ exp. [-(x-Âµ)2/2Ïƒ2]dx = 1/ 2

1/ Ïƒ âˆš2Ï€ MÊƒÂµ exp. [-(x-Âµ)2/2Ïƒ2]dx =0

Âµ = M

(iii) Mode: It is the value of x for which f(x) is maximum i.e., f’ (x) = 0 and f” (x) < 0.

Therefore, we obtain, mode = Âµ

Note. Mean, Median and Mode might match.

(iv) Mean-deviation (M.D.)

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