- 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 Z^{2}/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 :µ_{2n+1 }= 0 (odd-order)

And µ_{2n= 1,3,5,…, (2n-1) }2n. (even order)

(i) Skewness : β_{1}= 0 ( µ_{3}=0 ), ᵧ1=0

(ii) _{ }Kurtosis : β_{2}=3

*(ix) *Kurtosis : β_{2} = 3, γ_{2} = β_{2} - 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 *z*N(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: