The normal distribution is a theoretical function commonly used in inferential statistics as an approximation to sampling distributions. In general, the normal distribution provides a good model for a random variable, when:
As an underlying mechanism that produces the normal distribution, we can think of an infinite number of independent random (binomial) events that bring about the values of a particular variable. For example, there are probably a nearly infinite number of factors that determine a person's height (thousands of genes, nutrition, diseases, etc.). Thus, height can be expected to be normally distributed in the population.
Since Gauss used this curve to describe the theory of accidental errors of measurements involved in the calculation of orbits of heavenly bodies, it is also called as Gaussian curve.
The Conditions of Normality
In order that the distribution of a random variable X is normal, the factors affecting its observations must satisfy the following conditions:
Random variables observed in many phenomena related to economics, business and other social as well as physical sciences are often found to be distributed normally. For example, observations relating to the life of an electrical component, weight of packages, height of persons, income of the inhabitants of certain area, diameter of wire, etc., are affected by a large number of factors and hence, tend to follow a pattern that is very similar to the normal curve.
In addition to this, when the number of observations become large, a number of probability distributions like Binomial, Poisson, etc., can also be approximated by this distribution.
Probability Density Function
If X is a continuous random variable, distributed normally with mean m and standard deviation s , then its p.d.f. is given by
Here p and s are absolute constants with values 3.14159.... and 2.71828.... respectively. It may be noted here that this distribution is completely known if the values of mean m and standard deviation s are known. Thus, the distribution has two parameters, viz. mean and standard deviation.
Shape of Normal Probability Curve
For given values of the parameters, m and s, the shape of the curve corresponding to normal probability density function p(X) is as shown in Figure.
It should be noted here that although we seldom encounter variables that have a range from  ∞ to ∞, as shown by the normal curve, nevertheless the curves generated by the relative frequency histograms of various variables closely resembles the shape of normal curve.
Properties of Normal Probability Curve
A normal probability curve or normal curve has the following properties:
Β_{2}= µ_{4}/µ_{2}^{2}= 3 σ^{4}/σ^{4}= 3
Note that the above expression makes use of property .
Probability of Normal Variate in an Interval
Let X be a normal variate distributed with mean m and standard deviation s, also written in abbreviated form as X ~ N(m, s) The probability of X lying in the interval (X_{1}, X_{2}) is given by
In terms of figure, this probability is equal to the area under the normal curve between the ordinates at X = X_{1} and X = X_{2} respectively.
Note: It may be recalled that the probability that a continuous random variable takes a particular value is defined to be zero even though the event is not impossible.
It is obvious from the above that, to find P(X_{1}≤X ≤ X_{2}), we have to evaluate an integral which might be cumbersome and time consuming task. Fortunately, an alternative procedure is available for performing this task. To devise this procedure, we define a new variable z= Xµ/ σ.
Further, from the reproductive property, it follows that the distribution of z is also normal.
Thus, we conclude that if X is a normal variate with mean m and standard deviation s, then z= Xµ/σ is a normal variate with mean zero and standard deviation unity. Since the parameters of the distribution of z are fixed, it is a known distribution and is termed as standard normal distribution (s.n.d.). Further, z is termed as a standard normal variate (s.n.v.).
It is obvious from the above that the distribution of any normal variate X can always be transformed into the distribution of standard normal variate z. This fact can be utilised to evaluate the integral given above.
In terms of figure, this probability is equal to the area under the standard normal curve between the ordinates at z = z_{1} and z = z_{2}.
Since the distribution of z is fixed, the probabilities of z lying in various intervals are tabulated. These tables can be used to write down the desired probability.
Example: Using the table of areas under the standard normal curve, find the following probabilities :
(i) P(0 ≤ z ≤ 1.3) (ii) P(–1 ≤ z ≤ 0) (iii) P(–1 ≤ z ≤ 12)
(iv) P( z ≥ 1.54) (v) P(z > 2) (vi) P(z < 2)
Solution: The required probability, in each question, is indicated by the shaded are of the corresponding figure.
Example : Determine the value or values of z in each of the following situations:
(a) Area between 0 and z is 0.4495.
(b) Area between – ∞ to z is 0.1401.
(c) Area between – ∞ to z is 0.6103.
(d) Area between – 1.65 and z is 0.0173.
(e) Area between – 0.5 and z is 0.5376.
Solution:
(a) On locating the value of z corresponding to an entry of area 0.4495 in the table of areas under the normal curve, we have z = 1.64. We note that the same situation may correspond to a negative value of z. Thus, z can be 1.64 or  1.64.
(b) Since the area between –∞ to z < 0.5, z will be negative. Further, the area between z and 0 = 0.5000 – 0.1401 = 0.3599. On locating the value of z corresponding to this entry in the table, we get z = –1.08.
(c) Since the area between –∞ to z > 0.5000, z will be positive. Further, the area between 0 to z = 0.6103  0.5000 = 0.1103. On locating the value of z corresponding to this entry in the table, we get z = 0.28.
(d) Since the area between –1.65 and z < the area between –1.65 and 0 (which, from table, is 0.4505), z is negative. Further z can be to the right or to the left of the value –1.65. Thus, when z lies to the right of –1.65, its value, corresponds to an area (0.4505 – 0.0173) = 0.4332, is given by z = –1.5 (from table). Further, when z lies to the left of  1.65, its value, corresponds to an area (0.4505 + 0.0173) = 0.4678, is given by z = –1.85 (from table).
(e) Since the area between –0.5 to z > area between –0.5 to 0 (which, from table, is 0.1915), z is positive. The value of z, located corresponding to an area (0.5376 – 0.1915) = 0.3461, is given by 1.02.
Example : If X is a random variate which is distributed normally with mean 60 and standard deviation 5, find the probabilities of the following events:
(i) 60 ≤ X ≤ 70, (ii) 50 ≤ X ≤ 65, (iii) X > 45, (iv) X ≤ 50.
Solution: It is given that m = 60 and s = 5
(i) Given X_{1} = 60 and X_{2} = 70, we can write
z_{1} = X_{1}µ/σ = 6060/5=0 and z_{2}= X_{2}µ/σ= 70605 = 2.
∴ P(60 ≤ X ≤ 70) = P(0 ≤ z ≤ 2) = 0.4772 (from table).
(ii) Here X_{1} = 50 and X_{2} = 65, therefore, we can write
z_{1}= 5060/5 = 2 and z_{2}=6560/5 = 1
Hence P(50 ≤ X ≤ 65) = P(–2 ≤ z ≤ 1) = P(0 ≤ z ≤ 2) + P(0 ≤ z ≤ 1)
= 0.4772 + 0.3413 = 0.8185
= P(3≤ z≤0)+ P(0≤z≤∞)= P(0≤z≤3)+P(0≤z≤∞)
=0.4987+0.5000=0.9987
Example : The average monthly sales of 5,000 firms are normally distributed with mean Rs 36,000 and standard deviation Rs 10,000. Find :
(i) The number of firms with sales of over Rs 40,000.
(ii) The percentage of firms with sales between Rs 38,500 and Rs 41,000.
(iii) The number of firms with sales between Rs 30,000 and Rs 40,000.
Solution: Let X be the normal variate which represents the monthly sales of a firm. Thus X ~ N(36,000, 10,000).
Example : In a large institution, 2.28% of employees have income below Rs 4,500 and 15.87% of employees have income above Rs. 7,500 per month. Assuming the distribution of income to be normal, find its mean and standard deviation.
Solution: Let the mean and standard deviation of the given distribution be mand srespectively.
Example : Marks in an examination are approximately normally distributed with mean 75 and standard deviation 5. If the top 5% of the students get grade A and the bottom 25% get grade F, what mark is the lowest A and what mark is the highest F?
Solution: Let A be the lowest mark in grade A and F be the highest mark in grade F. From the given information, we can write
Example : The mean inside diameter of a sample of 200 washers produced by a machine is 5.02 mm and the standard deviation is 0.05 mm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 4.96 to 5.08 mm, otherwise the washers are considered as defective. Determine the percentage of defective washers produced by the machine on the assumption that diameters are normally distributed.
Example : The average number of units produced by a manufacturing concern per day is 355 with a standard deviation of 50. It makes a profit of Rs 1.50 per unit. Determine the percentage of days when its total profit per day is (i) between Rs 457.50 and Rs 645.00, (ii) greater than Rs 682.50 (assume the distribution to be normal). The area between z = 0 to z = 1 is 0.34134, the area between z = 0 to z = 1.5 is 0.43319 and the area between z = 0 to z = 2 is 0.47725, where z is a standard normal variate.
Solution: Let X denote the profit per day. The mean of X is 355*1.50 = Rs 532.50 and its S.D. is 50*1.50 = Rs 75. Thus, X ~ N (532.50, 75).
Example : The distribution of 1,000 examinees according to marks percentage is given below:
Assuming the marks percentage to follow a normal distribution, calculate the mean and standard deviation of marks. If not more than 300 examinees are to fail, what should be the passing marks?
Solution: Let X denote the percentage of marks and its mean and S.D. be mand s respectively. From the given table, we can write
Example : In a certain book, the frequency distribution of the number of words per page may be taken as approximately normal with mean 800 and standard deviation 50. If three pages are chosen at random, what is the probability that none of them has between 830 and 845 words each?
Solution: Let X be a normal variate which denotes the number of words per page. It is given that X ~ N(800, 50).
The probability that a page, select at random, does not have number of words between 830 and 845, is given by
Thus, the probability that none of the three pages, selected at random, have number of words lying between 830 and 845 = (0.91)3 = 0.7536.
Example : At a petrol station, the mean quantity of petrol sold to a vehicle is 20 litres per day with a standard deviation of 10 litres. If on a particular day, 100 vehicles took 25 or more litres of petrol, estimate the total number of vehicles who took petrol from the station on that day. Assume that the quantity of petrol taken from the station by a vehicle is a normal variate.
Solution: Let X denote the quantity of petrol taken by a vehicle. It is given that X ~ N(20, 10).
Normal Approximation to Binomial Distribution
Normal distribution can be used as an approximation to binomial distribution when n is large and neither p nor q is very small. If X denotes the number of successes with probability p of a success in each of the n trials, then X will be distributed approximately normally with mean np and standard deviation √ npq.
ICorrection for Continuity
Since the number of successes is a discrete variable, to use normal approximation, we have make corrections for continuity.
Example : An unbiased die is tossed 600 times. Use normal approximation to binomial to find the probability obtaining
(i) more than 125 aces,
(ii) number of aces between 80 and 110,
(iii) exactly 150 aces.
Solution: Let X denote the number of successes, i.e., the number of aces.
Normal/ Approximation to Poisson Distribution
Normal distribution can also be used to approximate a Poisson distribution when its parameter m ≥10. If X is a Poisson variate with mean m, then, for m ≥ 10, the distribution of X can be taken as approximately normal with mean m and standard deviation √m so that z=xm/√m is a standard normal variate.
The fittings of Normal distribution problems are given below
Example : A random variable X follows Poisson distribution with parameter 25. Use normal approximation to Poisson distribution to find the probability that X is greater than or equal to 30.
Solution:
Fitting a Normal Curve
A normal curve is fitted to the observed data with the following objectives:
1. To provide a visual device to judge whether it is a good fit or not.
2. Use to estimate the characteristics of the population.
The fitting of a normal curve can be done by
(a) The Method of Ordinates or
(b) The Method of Areas.
(a) Method of Ordinates: In this method, the ordinate f(X) of the normal curve, for various values of the random variate X are obtained by using the table of ordinates for a standard normal variate.
Example : Fit a normal curve to the following data :
Note: If the class intervals are not continuous, they should first be made so.
∴ µ= 4510*(10/100)=44
(b) Method of Areas: Under this method, the probabilities or the areas of the random variable lying in various intervals are determined. These probabilities are then multiplied by N to get the expected frequencies. This procedure is explained below for the data of the above example.
Exercise with Hints
At such an examination, it is found that 10% of the candidates have failed, 5% have obtained distinction. Calculate the percentage of students who were placed in the second division. Assume that the distribution of marks is normal. The areas under the standard normal curve from 0 to z are
z: 1.28 1.64 0.41 0.47
Area : 0.4000 0.4500 0.1591 0.1808


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