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Normal Distribution

  1. In the binomial distribution if (i) the number of trials is indefinitely large, i.e., n ->∞ and (ii) neither p nor q is very small, then the limiting form of the binomial distribution is called 'normal distribution'. It is a continuous distribution and probability density function is given by, -∞<x<∞, -∞ <µ<∞

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

Since this distribution was developed by Carl Friedrich Gauss, sometimes it is also referred as 'Gaussian Distribution'.

 

II. Properties

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

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

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

Here we obtain, mode = µ

Note. Mean, Median and Mode coincides.

(iv) Mean deviation (M.D.)

M.D.

taking Z2/2=t

(v)        Q.D : M.D : S.D. = 2/3 : 4/5 : 1

10: 12 : 15

(vi) Points of inflexion of the normal curve is given at x = µ±

 

(vii) Central moments   :µ2n+1 = 0 (odd-order)

And                                µ2n= 1,3,5,…, (2n-1)  2n. (even order)

 

 

(i)                 Skewness :              β1= 0 ( µ3=0 ), ᵧ1=0

 

(ii)                Kurtosis :                           β2=3

(ix) Kurtosis :                          β2 = 3, γ2 = β2 - 3 = 0.

(x) Distribution function         F(x) =  dx.

(xi) It is a pdf   ,                        dx = 1.

(xii) Standard Normal Distribution

If X  N(µ, ), then z =  L is called standard normal variate with E[z] = 0 and

var [z] = 1 and zN(0,1).

(z) = the pdf =

(xiii)     P[ x1<X<x2] 

 

 

 

 

 

These values are obtained from the standard normal table.

 

(xiv) Area under the normal curve:

 

III. Probable Error

Any manufactured items or measurement of any physical quantity shows slight error. All the errors in manufacturing or measurement are random in nature and follow a normal distribution. We define the probable error is such that the probability of an error falling within the limits µ- and µ+. is exactly equal to the chance of an error falling outside these limits which implies that the chance of an error lying within µ- and µ+ is .

IV. Normal Approximation To Binomial Distribution

If the number of trials is sufficiently large (i.e., n ≥ 30) the binomial distribution B(n. p) is approximated by the normal distribution N(µ, ) with µ = np and = npq. But a continuity correction is required. The discrete integer x in B(n, p) becomes the interval [x - 0.5, x + 0.5] in the N(µ, ). Thus

Example 1: Given a random variable having the normal distribution with µ = 18.2 and σ = 1.25, find the probabilities that it will take on a value

 

(a)   less than 16.5,

(b)   greater than 18.8,

(c)    between 16.5 and 18.8,

(d)   between 19.2 and 20.1.

 

 

P (z < -1.36) = 0.5 - (l.36)

= 0.5 - 0.4131

= 0.0869.

 

 

 

P (z < 0.48) = 0.5 - (0.48)

= 0.5 – 0.1844

= 0.3156.

(c) (Fig. 8.4) P(-1.36 < z < 0.48) = >(l.3

= 0.4131 + 0.1844

= 0.5975.

 

 

Example 2. In a large institution 2.28% of employees receive income below $ 4500 and 15.87% of employees receive income above $ 7500 p.m... Assuming the income follows normal distribution. Find the mean and S.D. of the distribution.

Solution. Let µ and σ be the mean and S.D. of the normal distribution.

Given 2.28% of employees receive income below $ 4500.

Again, 15.87% of employees receive income above $ 7500.

 

We obtain two equations as follows:

Then the solution gives σ = 1000 and µ= 6500.

 

Example 3: Samples of 40 are taken from a lot, which is on the average 20 percent defective.

(a)    What is the probability that a sample of 40 will contain exactly 11 defectives? (b) What is the probability that it will contain 11 or more defectives?

 

Solution: This problem can be solved using normal distribution as an approximation to the binomial.

Here n = 40, p = 0.2, q = 0.8.

 

(a)    Since the normal distribution is continuous, the probability of exactly 11 defectives should be interpreted as meaning the probability of defectives from 10.5 to 11.5.

The required probability is

 

P(z1 < z < z2) = P(0.99 < z < 1.38)

= 1.38) - 0.99)

= 0.4162 - 0.3389

= 0.0773.

 

(a)    The probability of 11 or more defectives should be interpreted to mean the probability of 10.5 or more defectives.

 

Therefore,                    P(z > 0.99) = 0.5 - 0.99)

= 0.5 - 0.3389

= 0.1611.

Example 4:  Fit a normal curve to the following distribution: