16 Continuous distribution – Normal distribution : Standard Normal probability curve.

Dr. Nidhi Handa

0. Introduction :Normal Distribution

1. Characteristic of Normal Probability Distribution

2. Empirical Rule

3. Table

4. Standard Normal Distribution

5. Related Problems

6. Summary

THE NORMAL DISTRIBUTION

A very important continuous probability distribution is the normal distribution. It is applicable to many situations in which it is necessary to make inferences by taking samples. The normal distribution comes close to fitting in actual observed frequency distributions of many phenomena, including human characteristics (weights, heights, and IQs), output from physical process (dimensions and yields) and other measures of interest to managers in both the public and private sectors.

Characteristics

Characteristics of the Normal probability Distribution

The diagram suggests important features of normal probability distribution :

1. The curve has a single peak thus it is uni modal. It has the bell shaped curve

2. The mean of a normally distributed population lies at the centre of its normal curve.

3. Because of symmetry of the normal probability curve the mean, median and mode lie on the same point.

4. The two tails of the normal probability curve extend indefinitely and never touch the horizontal axis.

5. To define a particular normal probability distributions, we need only two parameters, the mean () and

the standard deviation().

EMPIRICAL RULE

Further, in the discussion on the use of standard deviation, we also discussed the empirical rule for a bell shaped curve. That empirical rule is based on the standard normal distribution table. By using the normal distribution table, we can now verify the empirical rule as follows.

The total area within one standard deviation of the mean is 68.26%. This area is given by the sum of the areas between z = -1.0 and z =0 and between z=0 and z = 1.0. As shown in Figure 1, each of these two areas is .3413 of 34.13%. Consequently, the total area between z = -1.0 is 68.26%

Figure 1 Area within one standard deviation of the

2. The total area within two standard deviations of the mean is 95.44%. This area is given by the sumof the

area between z = -2.0 and z =0 and between z = 0 and z =2.0. As shown in Figure 2 each of these two areas is

.4772 or 47.72%. Hence, the total area between z = – 2.0 and z = 2.0 is 95.44%.

Figure 2 Area within two standard deviations of the mean

3. The total are with in three standard deviations of the mean is 99.74%. This area is given by the sum of the

areas between z = -3.0 and z =0 and between z = 0 and z = 3.0. As shown in figure 3, each of these two areas

is .4987 or 49.87%. Therefore, the total area between z = -3.0 and z = 3.0 is 99.7%.

Figure 3 Area within three standard deviation of the mean

Again, note that only a specific bell-shaped curve represents the normal distribution. Now we can state that a bell-shaped curve that contains (about) 68.26% of the total area within one standard deviations of the mean, and (about) 99.74% of the total area within three standard deviations of the mean represents a normal distribution curve.

The standard normal distribution table, Table A only goes up to z = 3.09. In other words, that table can be read only for z = 0 to z = 3.09 (or to z = –3.09). Consequently, if we need to find the area between z = 0 and a z value greater than 3.09 (or between a z value less than – 3.09 and z =0) under the standard normal curve, we cannot obtain it from the normal distribution table because it does not contain a z value greater than 3.09. In such cases, the area under the normal distribution curve between z = 0 and any z value greater than 3.09 (or less than – 3.09) is approximated by .5. From the normal distribution table, the area between z= 0 and z = 3.09 is .4990. Hence, the area between z = 0 and any value of z greater than 3.09 is larger than .4990 and can be approximated by .5

Table : A STANDARD NORMAL DISTRIBUTION

The entries in the table give the areas under the standard normal curve from 0 to …

Z	.00	.01	.02	.03	.04	.05	.06	.07	.08	.09
0.0	.0000	.0040	.0080	.0120	.0610	0.199	.0239	.0279	.0319	.0359
0.1	.0398	.0438	.0478	.0517	.0557	.596	.0636	.0675	.0714	.4753
0.2	.0798	.0832	.0871	.0910	.0948	.0987	.1026	.1064	.1103	.1141
0.3	.1179	.1217	.1255	.1293	.1331	.1368	.1406	.1443	.1480	.1517
0.4	.1554	.1591	.1628	.164	.1700	.1736	.1772	.1808	.1844	.1879
0.5	.1915	.1950	.1985	.2019	.2054	.2088	.2123	.2157	.2190	.2224
0.6	.2257	.2291	.2324	.2357	.2389	.2422	.2454	.2486	.2517	.2549
0.7	.2580	.2611	.2642	.2673	.2704	.2734	.2764	.2794	.2823	.2852
0.8	.2881	.2910	.2939	.2967	.2995	.3023	.3051	.3078	.3106	.3133
0.9	.3159	.3186	.3212	.3238	.3234	.3289	.3315	.3340	.3365	.3389
1.0	.3413	.3438	.3461	.3485	.3508	.3531	3554	.357	.3599	.3621
1.1	.3643	.3665	.3686	.3708	.3729	.3749	.3770	.3790	.3810	.3830
1.2	.3849	.3869	.3888	.3907	.3925	.3944	.3962	.3980	.3997	.4015
1.3	.4032	.4049	.4066	.4082	.4099	.4115	.4131	.4147	.4162	.4177
1.4	.4192	.4207	.4222	.4263	.4251	.4265	.4279	.4292	.4306	.4319

1.5	.4332	.4345	.4357	.4370	.4382	.4394	.4406	.4418	.4429	.4441
1.6	.4452	.4463	.4474	.4884	.4495	.4505	.4515	.4525	.4535	.4545
1.7	.4554	.4564	.4573	.4582	.4591	.4599	.4608	.4616	.4625	34633
1.8	.4641	.4649	.4656	.4664	.4671	.4678	.4686	.4693	.4699	.4706
1.9	.4713	.4719	.4726	.4732	.4738	.4744	.4750	.4756	.4762	.4767
2.0	.4772	.4778	.4783	.4788	.4793	.4798	.4803	.4808	.4812	.4817
2.1	.4821	.4826	.4830	.4834	.4838	.4842	.4846	.4850	.4854	.4857
2.2	.4861	.4864	.4868	.4871	.4875	.4878	.4881	.4884	.4887	.4890
2.3	.4893	.4896	.4898	.4901	.4904	.4906	.4809	.4911	.4913	.4916
2.4	.49.18	.4920	.4922	.4925	.4927	.4929	.4931	.4932	.4934	.4936
2.5	.4938	.4940	.4941	.4943	.4945	.4946	.4948	.4949	.4951	.4952
2.6	.4953	.4955	.4953	.4957	.4959	.4960	.4961	.4962	.4963	.4964
2.7	.4965	.4966	.4967	.4968	.4969	.4970	.4971	.4972	.4973	.4974
2.8	.4974	.4975	.4976	.4977	.4977	.4978	.4979	.4979	.4980	.4981
2.9	.4981	.4982	.4982	.4983	.4984	.4984	.4985	4985	.4986	.4986
3.0	.4987	.4987	.4987	.4988	.4988	.4989	.4989	.4989	.4990	.4990

.3 THE STANDARD NORMAL DISTRIBUTION

The standard normal distribution is a special case of the normal distribution. For the standard normal distribution, the value of the mean is equal to zero and the value of the standard deviation is equal to 1.

STANDARD NORMAL DISTRIBUTION

The normal distribution with = 0 and = 1 is called the standard normal distribution.

Figure 4 displays the standard normal distribution curve. The random variable that possesses the standard normal distribution curve are denoted by z and are called the z values ro z scores.

Fig. 4 the standard normal distribution curve 1The equation of the normal distribution is

f(x) =5

where e = 2.71825 and = 3.14159 approximately; f(x), called the probability density function, gives the vertical distance between the horizontal axis and the curve at point x. For the information of those who are familiar with integral calculus, the definite integral of this equation from a to b gives the probability that x assumes a value between a and b.

Z VALUES OF Z SCORES

The units marked on the horizontal axis of a standard normal curve are denoted by z and are called be z value or z scores. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.

In Figure 4, the horizontal axis is labelled z. The z values on the right side of the mean are positive and those on the left side are negative. The z value for a point on the horizontal axis gives the distance between the mean and that point in terms of the standard deviation. For example, a point with a value of z = 2 is two standard deviations to the right of the mean. Similarly, a point with a value of z = –2 is two standard deviations to the left of the mean.

The standard normal distribution table, Table A lists the areas under the standard normal curve between z = 0 and the values of z from 0.00 to 3.09. To read the standard normal distribution table, we always start at z = o, which represents the mean of the standard normal distribution. We learned earlier that the total area under a normal distribution curve is 1.0. We also learned that, because of symmetry, the area on either side of the mean is 0.5. This is shown in Figure 5.

Fig.5 Area under the standard normal curve.

NOTE: Although the values of z on the left side of the mean are negative, the area under the curve is always positive.

The area under the standard normal curve between any two points can be interpreted as the probability the z assumes a value within that interval. Following Examples in describe how to read Table A to find areas under the standard normal curve.

Example 1 : Find area under the standard normal curve between z = 0 and z = 1.95.

Solution : We divide the number 1.95 into two portions : 1.9 (the digit before the decimal and one digit after the decimal) and .05 (the second digit after the decimal). (Note that 1.9 +.05 = 1.95.) to find the required area under the standard normal curve, we locate 1.9 in the column for z on the left side of Table 1and 0.05 in the row for z at the top of Table A. The entry where the row for 1.9 and the column for .05 intersect gives the area under the standard normal curve between z = 0 and z = 1.95. The relevant portion of Table A. is reproduced below as Table 1. From Table1 the entry where the row for 1.9 and the column for.05 cross is .4744. Consequently, the area under the standard normal curve between = z = 0 and z = 1.95 is .4744. This area is shown in Figure 6.19. (It is always helpful to sketch the curve and mark the area we are determining).

Table :1. Area Under the Standard Normal Curve Between z = 0 and z = 1.95

Z	.00	.01	…	.05		…	.09
0.0	.0000	.0040	…	.0199		…	.0359
0.1	.0398	.0438	…	.0596		…	.0753
0.2	.0793	.0832	…	.0987		…	.1141
…	…	…	…	…		…	…
…	…	…	…	…		…	…
…	…	…	…	…		…	…
1.9	.4713	.4719	…	.4744		…	.4767

..	..	..	…	..		…	…
..	..	..	…	..		…	…
..	..	..	…	..		…	…
3.0	.4987	.4987	…	.4989		…	.4990
					Required area

Fig. 6 Area between z = 0 and z = 1.95

The area between z = 0 and z = 1.95 can be interpreted as the probability that z assumes a value between 0 and 1.95. That is,

Area between 0 and 1.95 = P(0 < z < 1.95) = .4744

the probability that a continuous random variable assumes a single value in zero. Therefore,

P(z = 0) = 0 and P(z = 1.95) = 0

Hence

P(0 < z < 1.95) = P(0 < z < 1.95) = .4744

Finding area between a negative z and z = 0.

Example 2 : Find the area under the standard normal curve from z = –2.17 to z = 0.

Solution : Because the normal distribution is symmetric about the mean, the area from z = –2.17 to z = 0 is the same as the area from z = 0 to z = 2.17, as shown in Figure 7.

Figure : 7

To find the area from z = –2.17 to z = 0, we look for the area from z = 0 to z = 2.17 in the standard normal distribution table (Table A).

To do so, First we locate 2.1 in the column for z in that table. Then we read the number at the intersection of the row for 2.1 and the column for 0.7. The relevant portion of the table A is produced below as Table: 2. As shown in Table : 2 and fig 8 ,this number is .4850

Table : 2 Area Under the Standard Normal Curve Between z = 0 and z = 2.17

Fig 8 Area from Z= -2.17 to Z= 0

Area from Z= -2.17 to Z= 0

The area from z = -2.17 to z = 0 gives the probability that z lies in the interval – 2.17 to 0. That is,

Area from -2.17 to 0 = P(-2.17 ≤ z ≤ 0) = .4850

Example 4 : Find the following areas under the standard normal curve.

(a) Area to the right of z = 2.32

(b) Area to the left of z =-1.54

Solution

(a) As mentioned earlier, to read the normal distribution table we must start with z = 0. To find the area to the right of z = 2.32, first we find the area between z = 0 and z = 2.32. Then we subtract this area from .5, which is the total area to the right of z = 0. From Table A, the area between z = 0 and z = 2.32 is .4898. Consequently the required area is .5 – .4898 = .0102, a shown in figure—-

Figure 9 Area to the right of z =2.32

The area to the right of z= 2.32 gives the probability that z is greater than 2.32 Thus,

Area to the right of 2.32 = P(z > 2.32) =.5 -.4898 =.0102

(b) To find the area under the standard normal curve to the left of z = – 1.54, first we find the area between z = -1.54 and z = 0 and then we subtract this area from .5, which is the total area to the left of z = 0. From Table A, the area between z = – 1.54 and z = 0 is .4382. Hence, the required area is .5 – .4382 = .0618. This area is shown in figure

Fig: 10 The area to the left of z = – 1.54

The area to the left of z = – 1.54 gives the probability that z is less than – 1.54. Thus,

Area to the left of -1.54 = P (z < – 1.54) = .5 – .4382 = .0618

Example 5: Suppose the training- programme director wants to know the probability that a participant chosen at random would need between 550 to 650 hours to complete the required work.Mean is 500 and standard distribution is 100.

Solution : This probability is represented by shaded area in the fig.11.

Step :1

We look up at z = 1.5 in Normal probability distribution table and find the corresponding probability value as 0.4332 i.e. p(z) = 0.4332

Step : 11 :

Now Calculate value of Z when X = 550.

Z = = 0.5

Again we look up at Z = 0.5 is Normal Probability distribution table and find the corresponding probability value as 0.1915 i.e. p(z) = 0.1915.

Step : 111 :

Now to find the final answer that is the chance that random variable will fall between the 650 and 550 hrs. is:

P(random variable will lie between 650 – 500 hrs.) = 0.4332

–P(random variable will lie between 550 – 500 hrs.) = 0.1915

=P(random variable lies in shaded area that is between 550 – 650) = 0.2417

Example 6: An aptitude Test was conducted on 900 employee of the metro tyres limited. In which the mean score was found to be 50 with s.d = 20 on the basis of this information what was the.

(a) No. of employee chose mean score was less than 30.

(b) No. of employees whose mean score exceeds 70.

P (–1 < z < 0) +P (0 < z < 1)

= 3413 + 0.3413

= 6826

> 70 = 900 x 0.1587

143

the no. of employees whose mean score was

< 30

= 900 x 0.1587 143

No. of employees whose mean score b/w 30 & 70.

0.6826 x900 = 614.

Example 7: In a normal distribution 7% of the items are under 35 & 89% are under 63. What are the mean & S.d. of the distribution.

Solution :

P (X < 35) = 0.07————[1]

P(X < 63) = 0.89————[2]

For X = 35 Z = = –Z2 (say)

Using the given probability in [1] & [2] and

P(Z < Z2) = 0.07

P(–Z2 < Z < 0) = 0.5 – 0.47

we can see that P(0 < Z < Z2) = 0.43

& P(O < Z < Z1) = 0.39

(35 – ) x 123 = (63 – ) x (–148)

4925 – 123 = –9324 + 148

4925 + 9325 = (148 + 123)

13619 = 271

= = 50.25

~ 50.3

Put the above value in eqn [3]. We get

= –1.48

Example 8: Of a large group of man, 5% are under 60 inches in height & 40% are between 60 & 65 inches. Assuming a normal distribution, find the mean height & standard Here both point lie to the left of mean so the corresponding value will be negative when

From Table-

P (0 < Z < Z1) = 0.45 = P(0 < Z < 1.645) &

[5] & [6] we get

(60 – ) 0.13 = (65 – ) 1.645

7.8 – 0.13 – 1.645

78 – 106.925 = 0.13 – 1.645

99.125 = 1.515

Example 9: A random variate X is Normally distributed with mean 12 * s.d. = 2 find prob.

P(9.3 < x < 13.8) given that = 0.9, A = 0.3159 & = 1.2, A = 0.3849.

Solution : P (9.6 < X < 13.8)

Example 10: If skulls are classified as A,B,C according as the length- breadth index as under 75, between 75 & 80, or over 80, find approx. (assuming that dist is normal) the mean & Standard .deviation of a series in which A are 58%, B are 38% & C are 4% being given that.

Solution : Let M be the mean & be the S.d. according to the given condition. area between t =0 & t = 0.20 is 0.08 so that the area to the left is 0.5 + 0.08

Which corresponds to X = 75

Example 11: In a normal distribution, 31% of the items are under 45 & 8% over 64. Find the mean & standard deviation of the distribution.

Summary

The standard normal distribution is a special case of normal distribution.For the standard normal distribution the value of mean is equal to zero and the value of standard deviation is equal to 1.The Z values or Z scores which is known as standard units or standard scores. They represent random variable that possesses the standard normal distribution. To see the tabular values for finding area under the curve has been explained here.Different problems are taken into consideration and their solutions are also given.

Learn More:

‘Fundamentals of Mathematical Statistics’, S.C. Gupta & V.K. Kapoor Public, Sultan Chand & sons, New Delhi.
‘Solution and Problems on Mathematical Statistics’, Dr. B.D. Gupta, Publ., CBS Publishers & Distributors.
‘Probability & Statistics for Engineers’, Miller & Freund & Richard A. Johnson, Pearson Edition.
‘Modern Probability theory and its applications,’ E-Porzen, Wiley, New York.
‘Theory of Statistics’ Schervish Mark J., New York, Springer.
‘Theory and problems of Probability and statistics, Murray R.Spiegel, John J. Schiller R. Alu Arinivasan, Tata Mc Graw – Hill Publishing company limited.
“Introductory Statistics” (second edition), Prem S.Mann, Publ. John Wiley & Sons, INC,(1995).