IBA JU BBA -2Course Instructor: Dr Swapan Kumar Dhar

Probability Distribution (continued) Continuous Probability Distribution: The probability distribution of continuous random variable is continuous probability distribution. If a random variable assumes any numerical value in an interval or collection of intervals, then it is called a continuous random variable.The manner how a continuous variable is defined does not allow us to develop a continuous probability distribution in the way we did in the case of a discrete probability distribution. For instance, while developing the binomial distribution, the basic question raised was: What is the probability that the random variable X takes the value 0,1, 2... or n. A similar question makes no sense when asked in relation to a continuous variable.The reason being that the probability of a continuous variable taking a particular value, as represented by a given point on a line segment, is so small that, for all practical purposes, it can be treated as negligible. This may appear a little hard to accept. However, even a routine example will make the point clear.Consider the weight characteristic of human body as a continuous variable. Given the precise scale of measurement, there are a large number of weights between, say, 55.5 and 56.5 kg. For example, 55.99 kg is one of the weights in this range. The number of weight measurements close to 55.99 kg is so large that it is normally of no avail to distinguish this particular weight from other weights in its close proximity.In fact, as weight measurements can be recorded with maximum possible accuracy, persons with exactly 55.99 kg weight are extremely uncommon to find. The probability of finding such a person so small is that it can safely be treated as zero and ignored. Obviously, the probability distribution of a continuous variable, such is the one being considered, cannot be expressed in a tabular form as was possible in the case of a discrete probability distribution.Importantly, however, a continuous random variable has a probability distribution of its own, which may be called continuous probability distribution. Thus, when X is a continuous variable, its density function, designated as f(x), may take any shape depicted by the continuous curves shown in Fig 1. These are so drawn that the area under each curve above the relevant range on the X-axis is 1.For reasons explained above, no continuous probability function can be used to obtain the probability of X taking an exact value represented by a particular point on the X-axis. Instead, it gives the probability that X takes a value spread over a specified range bound by any two points, say and where < . Denoted as P ( < X < ), it is the

range probability represented by the shaded area between the two ordinates drawn at

and , as shown in Fig. 2..An exact calculation of any such shaded area requires the use of integral calculus. To avoid it for being cumbersome, ready-made area tables have been developed for a typical bell-shaped curve as in Fig. 2. It represents a probability distribution, which we frequently come across in real life, and which best describes the behavior of a large majority of business and economic variables.In essence, it follows that the probability distribution of a continuous random variable X is defined by a continuous curve having the following two properties:In essence, it follows that the probability distribution of a continuous random variable X is defined by a continuous curve having the following two properties:* That the area under the curve above the relevant range on the X-axis is 1.* That the ordinates drawn at any two points on the X-axis and (such that < ) bind an area, which represents the probability of X assuming a value between these points.

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Fig. 1:

Various Forms of Continuous Probability Function To obtain the range probabilities in the manner stated above and below, it is necessary to identify the particular shape a continuous distribution takes. This can be done by obtaining a frequency distribution of the given data on a continuous variable and converting the class frequencies into relative frequencies so that these add to 1.

Fig. 2: Range Probability P ( < X < )The relative frequencies for each class may then be plotted on the y-axis against the corresponding class mid-points measured on the X-axis, and the various points joined by a free hand line. The shape of the resultant relative frequency polygon gives a fairly good idea as to what possible shape among those identified in Figure 1 a continuous distribution takes.Probability Density Function (pdf): To assign the probability of a continuous probability distribution called pdf is used. If a random variable assumes a

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f(X)

X

f(X)

X

d f e

continuous set of values in the range , then its pdf must satisfy the following conditions:

(i) for all values of x, (ii) for all values of x.

Condition (i) indicates that a random variable cannot assume negative value and condition (ii) shows that the probability of the entire range space must equal 1.In general, the probability of the continuous random variable is determined by finding the area under the pdf between the values a and b. Mathematically, the area under pdf between a and b is given by

We can express in terms of a distribution, provided it is differentiable. That is,

.

Illustration: Consider the function,

For to be a pdf, the condition, must be satisfied, which is true

if

.

Since a>0, the equation, Thus f(x) satisfies both the conditions for a pdf.Expected value of a continuous random variable: For the continuous

random variable, the expected value is given by .

Example 1: A doctor recommends a patient to take a particular diet for two weeks and there is equal chance for the patient to lose weight between 2 kgs and 4 kgs. What is the average amount the patient is expected to lose on this diet?Solution: If x is the random variable, then the pdf is defined as

.

Hence, the amount the patient is expected to lose on the diet is:

kgs.

Example 2: Under an employment promotion programme, it is proposed to allow sale of newspapers inside buses during off – peak hours. The vendor can purchase newspapers at a special concessional rate of Taka 7.50 per copy against the selling price of Taka 8.00. Any unsold copies are, however, a dead loss. A vendor has estimated the following probability distribution for the number of copies demanded.

Number of Copies 15 16 17 18 19 20

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Probability 0.04 0.19 0.33 0.26 0.11 0.07How many copies should be ordered so that his expected profit will be maximum?Solution: Profit per copy = Selling price – Purchasing price = 8.00 – 7.50= 0.50 Taka.                  Expected profit = Number of copies Probability Profit per copy.The calculations of expected profit are shown in the following table:

Calculation of Expected Profit

Number of copies demanded

Probability

Profit per copy (in Taka)

Expected profit (in Taka)

15 0.04 0.50 3016 0.19 0.50 15217 0.33 0.50 280.5018 0.26 0.50 23419 0.11 0.50 104.5020 0.07 0.50 70

The maximum profit of Taka 280.50 is obtained when he stocks 17 copies of the newspaper.Example 3: A bakery has the following schedule of daily demand for cakes. Find the expected number of cakes demanded per day.No of cakes demanded

0 1 2 3 4 5 6 7 8 9

Probability 0.02

0.07

0.09

0.12

0.20

0.20

0.18

0.10

0.01

0.01

Solution : Calculating Table for finding expected number of cakesX = No of cakes Probability = P(x) XP(x)

0 0.02 01 0.07 0.072 0.09 0.183 0.12 0.364 0.20 0.805 0.20 1.006 0.18 1.087 0.10 0.708 0.01 0.089 0.01 0.09

=4.36. Therefore, the expected number of cakes = 436.

Variance of continuous random variable: The expected value measures the central tendency of a probability distribution, while variance determines the dispersion or variability to which the possible random values differ among them.

The variance, denoted by Var (x) or of a random variable X is the squared

deviation of the individual values from their expected value or mean. That is,

,where

, for continuous X.

, for discrete X.

Properties of Expected value and Variance:

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The following are the important properties of an expected value of a random variable:(1)The expected value of a constant c is constant. That is, E(c) = c, for every constant c.(2) E (cx) = c E (x).(3) The expected value of a linear function of a random variable is same as the linear function of its expectation. That is, E(a + bx) = a + b E(x).(4)The expected value of the product of two independent random variables is equal to the product of their individual expected values. That is, E (xy) = E (x) E (y).(5) The expected value of the sum of the two independent random variables is equal to individual expected values. That is, E (x + y) = E (x) + E (y).(6) The variance of the product of a constant and a random variable X is equal to the constant squared times the variance of the random variable X. That is, Var (c x) = Var (x).(7)The variance of the sum (or difference) of two independent random variables equals the sum of their individual variances. That is, Var (x ±y) = Var (x) ± Var (y).

Normal Distribution: It has been observed that most business and economic variables generate continuous data whose behavior is often best described by a bell-shaped continuous curve. Since this is what we normally come across in the case of most populations on these variables, a bell-shaped curve has come to be universally known as a normal curve. Accordingly, the probability distribution described by a normal curve is called the normal (probability) distribution.The concept of normal distribution was initially developed by Abraham De Moivre (1667-1754). His work was further refined by Pierre S. Laplace (1749-1827). But the latter contribution remained unnoticed for long till it was given concrete shape by Karl Gauss (1777-1855). That is why the normal distribution is sometimes also referred to as the Gaussian distribution.The normal distribution has since come to acquire a wide range of applications in many areas of human knowledge. It is being used in almost all data - based research in the field of agriculture, trade, business, and industry. As mentioned, the probability of a particular value of random variable cannot be calculated and is always zero. Therefore the probability that a random variable value lies in a given interval can be calculated by finding the area under the probability density function and within the boundaries of that interval. The probability density function of a normal distribution is given by

,

where, = constant 3.1416, =constant 2.7183, = mean of the normal distribution, =

standard deviation of normal distribution.In symbols, if a random variable X follows normal distribution with mean and variance

, then it is also expressed as: .

Characteristics of Normal Distribution:A unique normal distribution may be defined by assigning specific values to the mean and standard deviation in the normal density function given in (1). Figure 3 shows three normal distributions with different values of the| mean and a fixed standard deviation , while in Figure 4 normal distributions are shown with different values of the standard deviation and a fixed mean .

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Fig 3: Normal Distributions with different mean values but fixed standard deviation

Fig 4: Normal Distribution with fixed mean and variable standard deviation From Figure 3 and 4 the following characteristics of a normal distribution may be derived:

(1) For every pair of values of and , the curve of normal probability density is bell shaped and symmetric.(2) The normal curve is symmetrical around a vertical line erected at the mean

with respect to the area under it, that is, fifty per cent of the area of the curve lies on both sides of the mean and reflect the mirror image of the shape of the curve on both sides of the mean . This implies that the probability of any individual outcome above or below the mean will be same. Thus, for any normal random variable x,

(3) Since the normal curve is symmetric, the mean, median, and mode for the normal distribution are equal because the highest value of the probability density function occurs when x = .(4) The two tails of the normal curve extend to infinity in both directions and theoretically never touch the horizontal axis.(5) The mean of the normal distribution may be negative, zero, or positive.(6) The mean determines the central location of the normal distribution, while standard deviation determines its spread. The larger the value of the standard deviation , the wider and flatter the normal curve, thus showing more variability in the data, as shown in Figure 4. Thus deviation determines the range of values that any random variable is likely to assume.(7) The area under the normal curve represents probabilities for the normal random variable, and therefore, the total area under the curve for the normal probability distribution is 1.Properties of the Normal distribution:(1) If several independent random variables are normally distributed, then their sum will also be normally distributed. The mean of the sum will be the sum of the all the individual means, and by virtue of the independence, the variance of the sum will be the sum of all the individual variances. We can write this a algebraic form as

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If are independent random variables that are normally distributed,

then their sum S will also be normally distributed with

and

.Example 4:Let and be independent random variables that are normally distributed with means and variance as follows:

Mean Variance10 120 230 3

Find the distribution of the sum . Report the mean, variance

and the standard deviation of S..Solution: The sum S will be normally distributed with mean 10 + 20 + 30 = 60 and variance 1 + 2 + 3 = 6. The standard deviation of .Example 5:The weight of a module used in a spacecraft is to be closely controlled. Since the module used a bolt-nut-washer assembly in numerous places, a study was conducted to find the distribution of weights of these parts. It was found that the three weights, in grams, are normally distributed with the following means and variance

Mean VarianceBolt 312.8 2.67Nut 53.2 0.85

Washer 17.5 0.21

Find the distribution of the weight of the assembly. Report the mean, variance, and standard deviation of the weight.Solution: The weight of the assembly is the sum of the weights of the three component parts, which are three normal random variables. Furthermore, the individual weights are independent since the weight of any one-component part does not influence the weight of the other two. Therefore, the weight of the assembly will be normally distributed. The mean weight of the assembly will be the sum of the mean weights of the individual parts: 312.8 + 53.2 +17.5 = 383.5 grams. The variance will be the sum of the individual variances: 2.67 + 0.85 + 0.21 = 3.73 gram The standard deviation grams.(2) Another interesting property of the normal distribution is that if X is normally distributed, then a X + b will also be normally distributed with mean a E(X) + b

and variance .

For example, if X is normally distributed with mean 10 and variance 3, then 4X + 5 will be normally distributed with mean 4 10 + 5 = 45 and variance

.(3) We can combine the above two properties and make the following statement:

If are independent random variables that are normally

distributed, then the random variable Q defined as

will also be normally distributed with

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And .

Example 6:The four independent normal random variables and have the following means and variances:

X Mean Variance12 4-5 28 5

10 1

Find the mean and variance of . Find also the standard deviation of Q.Solution:

Example 7: A cost accountant needs to forecast the unit cost of a product for next year. He notes that each unit of the product requires 12 hours of labor and 5.8 pounds of raw material. In addition, each unit of the product is assigned an overhead cost of $ 184.50. He estimates that the cost of an hour of labor next year will be normally distributed with an expected value of $45.75 and standard deviation of $1.80; the cost of the raw material will be normally distributed with an expected value of $62.35 and a standard deviation of $2.52. Find the distribution of the unit cost of the product. Report its expected value, variance and standard deviation. Solution: Let L be the cost of labor and M be the cost of the raw material. Denote the unit cost of the product by Q. Then Q = 12L + 5.8M + 184.50. Since the cost of labor L may not influence the cost of raw material M, we can assume that the two are independent. This makes the unit cost of the product Q a normal random variable. Then

Standard Normal Distribution: The computation of requires

that we either make use of integral calculus or construct separate probability tables for each normal curve. But an efficient method of obtaining range probabilities of the type stated above requires that the whole family of normal curves is transformed into a standard normal curve. This facilitates construction of a standard normal distribution, whatever be the values of and for different normal distributions.So deal with problems more simply, it is necessary to compute a value of Z by applying the following formula corresponding to a particular value of the random variable X. The variable Z is also called standardized normal random variable.

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Thus any normal distribution with any set of and values can be translated into a particular normal curve called standard normal distribution with mean equal to zero and standard deviation equal to one.

Here .

In the figure 5 it is given that = -3

Similarly other x and z values are calculated.

Figure 5: Standard Normal DistributionThe standardized variate Z represents the number of standard deviations that an x-value lies away from the mean of any normal distribution. For example z = 2 implies that the value x is 2 standard deviations above or below the mean. Figure 5 shows the graph of the probability density function of z with mean zero and standard deviation one. This curve is symmetric, bell- shaped and is centered around the mean equal to zero has most of the area contained within the range 3.

Table1: Area under the Normal Curve

Area under normal curve between x and

1.0 0.341342.0 0.477253.0 0.498754.0 0.49997

Since the normal distribution is symmetrical, Table 1 and figure 5 indicate that about 68.26 per cent of the normal distribution lies within the range - to + . The other relationships derived from Table 1 are shown and in Figure 5.

Table 2: Percentage of the area of the normal distribution lying within the given range

Number of standard deviations from Approximate percentage of area under

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mean normal curve68.2695.45

99.75

Area under the Normal Curve: Since the range of normal distribution is infinite in both the directions away from , the pdf function f (x) is never equal

to zero. As X moves away from , f (x) approaches x – axis but never actually touches it.The Normal Table: The following table is a table of cumulative areas under the standard normal curve. This table is also called cumulative normal table or Z table. The areas under the normal curve are given in the body of the table, accurate to four decimal places. Table: Cumulative Areas under the Standard Normal Curve z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003-3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005-3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007-3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010-2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014-2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019-2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026-2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036-2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064

-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084

-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110

-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 -2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 -1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233

-1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294-1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367-1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455-1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559-1.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681-1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823-1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985-1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170-1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379-0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .4660 .1635 .1611-0.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867-0.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148-0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451-0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776-0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121-0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483-0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859-0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247-0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

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1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

° z 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .994 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .996 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .997 .9973 .9974 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .997 .9980 .9981 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .998 .9986 .9986 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .998 .9990 .9990 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .999 .9993 .9993 3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .999 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .999 .9996 .9997 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .999 .9997 .9998

0 -Z-

For Negative Values of Z

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How to find probabilities:As an example, suppose that we wish to find the area under the standard normal curve at or below a z value of 1.00. This area is illustrated in Figure 6. To find this area, we scan down the far left column of the table (starting at the top) until we find the value 1.0. We now scan across the row in the table corresponding to the z value 1.0 until we find the column corresponding to the heading 0.00. The desired area (which we have shaded blue) is in the row corresponding to the z value 1.0 and in the column headed .00. This area, which equals 0.8413, is the probability that the random variable z is less than or equal to 1.00. That is, we have found that P (z 1.00) = 0.8413. As another example, the area under the standard normal curve at or below the z value 2.53 is found in the row corresponding to 2.5 and in the column corresponding to 0.03. We find that this area is 0.9943—that is,P(z 2.53) =0.9943.We now show how to use the cumulative normal table to find several other kinds of normal curve areas. First, suppose that we wish to find the area under the standard normal curve at or above a z value of 2—that is, we wish to find P(z 2). This area is illustrated in Figure 7 and is called a right-hand tail area. Since the total area under the normal curve equals 1, the area under the curve at or above 2 equals 1 minus the area under the curve at or below 2. That is, we find that P (z 2) = 1 - P(z 2) = 1 - 0.9772 =0 .0228.Next, suppose that we wish to find the area under the standard normal curve at or below a z value of — 1. That is, we wish to find P (z -1). This area is is called a left-hand tail area. The needed area is found in the row of the cumulative normal table corresponding to -1 and in the column headed by .00. We find that P (z — 1) = 0.1587. Notice that the area under the standard normal curve at or below — 1 is equal to the area under this curve at or above 1. This is true because of the symmetry of the normal curve. Therefore, P (z 1) = 0.1587.Finally, suppose that we wish to find the area under the standard normal curve between the z values of — 1 and 1. we can see that this area equals the area under the curve at or below 1 minus the area under the curve at or below -1. Referring to Table 5.3, we see that P(-l z 1) = P(z 1) - P(z -1) = 0.8413 - 0.1587 = 0.6826.

=0.8413

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Figure 6: Finding 0.9772

Figure 7: Finding The table of areas gives the proportion of the area under the curve, which lies between the vertical lines, erected at two points along the x-axis. A portion of the table is shown in Table 1. For example, as shown in Figure 5, if x is away from , that is, the distance between x and is one standard deviation or

=1, then 34.134 per cent of the distribution lies between x and .

Similarly, if x is at 2 away from , that is, = 2, then the area will

include 47.725 per cent of the distribution, and so on, as shown in Table 1.The standard normal distribution is a symmetrical distribution and therefore

for any value a.

, which is also shown in Figure 5.

Illustration: Let us consider our random variable X with mean 50 and S.D. 10 i.e. Suppose we want to find the probability that X is greater than 60, i.e. we want to find P (X>60). We cannot evaluate this probability directly, but we can transform x to Z and then we will be able to find the probability directly from the normal table. As stated the required transformation is

. Hence

Example 8: Suppose that the time it takes the electronic device in the car to respond to the signal from the toll plaza is normally distributed with mean 160 microseconds and S.D. 30 microseconds. What is the probability that device in the car will respond to a given signal within 100 to 180 microseconds?

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Solution: Let X denote the time to take. Then we can write Our

required probability is

Example 9: Fluctuations in the prices of precious metals such as gold have been empirically shown to be well approximated by a normal distribution when observed over short intervals of time. In May 1995, the daily price of gold (1 troy ounce, where one troy ounce = 31.10 grams and 3.75 troy ounces = 10 tolas) was believed to have a mean of $383 and a S.D. of $12. A broker, working under these assumptions, wanted to find the probability that the price of gold the next day would be between $394 and $399 per troy ounce. In this eventuality, the broker had an order from a client to sell the gold in the client’s portfolio. What is the probability that the client’s gold will be sold the next day?

Solution: Let X denote the price of gold. Then Then the

required probability is

Example 10: 1000 light bulbs with a mean life of 120 days are installed in a new factory and their length of life is normally distributed with standard deviation of 20 days.(a) How many bulbs will expire in less than 90 days?(b) If it is decided to replace all the bulbs together, what interval should be allowed between replacements if not more than 10% should expire before replacement.

Solution: (a) For =120, =20, and x = 90, = .

The area under the curve between z = 0 and z = - 1.5 is 0.4332 ( found from the normal table). Therefore area to the left of – 1.5 is 0.5 – 0.4332 = 0.0668. Thus the expected number of bulbs to expire in less than 90 days will be 0.06681000=67 (approx).(b) The value of Z corresponding to an area 0.4 (0.5 – 0.10), under the normal

curve is 1.28. Therefore, =

Hence, the bulbs will have to be replaced after 94 days.Example 11: In a certain examination, the percentage of passes and distinctions are 46 and 9 respectively. (i) Estimate the average marks obtained by the candidates, (ii) the minimum pass and distinction marks being 40 and 75 respectively (assume the distribution of marks to be normal). (iii) Also determine what would have been the minimum qualifying marks for admission to a re-examination of the failed candidates, had it been desired that the best 25 per cent of them should be given another opportunity of being examined.Solution: (i) Let X be the marks scored by the candidates. The area to the right of the ordinate at x = 40 is 0.46 and hence the area between the mean and the ordinate at x = 40 is 0.04.Now from the normal table, corresponding to 0.04, the standard normal variate, z = 0.1. Therefore, we have

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.

Similarly, .

Solving these two equations, we get and Hence, the average marks obtained by the students = 37.(ii) Let us assume that y is the minimum qualifying marks for re-examination of the failed candidates.The area to the right of x = 40 is 46 percent. Thus the percentage of students failing = 54 and this is the area to the left of 40. We want that the best 25 per cent of these failed candidates should be given a chance to reappear. Suppose thus area is equal to the shaded area in the diagram. This area is, 25 per cent of 54 = 13.5 per cent = 0.1350.The area between mean and ordinate at y = - (0.1350 – 0.04) = -0.0950 (negative sign is included because the area lies to the left of the mean ordinates).Corresponding to this area, the standard normal variate z = -0.0378. Thus, we write

Therefore, the minimum qualifying marks = 36.Example 12: The income of a group of 10000 persons was found to be normally distributed with mean Taka 1750 and S.D. Taka 50. Show that of this group 95% had income exceeding Taka 1668 and only 5 per cent had income exceeding Taka 1832. What was the lowest income among the richest 100?Solution: (a) Given that x = 1668, and Therefore, the standard

normal variate corresponding to x = 1668, is

The area to the right of the ordinate at z = - 1.64 ( or x = 1668) is (0.4495 + 0.5000) = 0.9495 ( because z = - 1.64 to its right covers 95 per cent area).The expected number variate corresponding to x = 1832 is

The area to the right of ordinate at Z = 1.64 is : 0.5000 – 0.4495 = 0.0505.The expected number of persons getting above Taka 1832 is:

This is about 5 per cent of the total of 10000 persons.

This probability of getting richest 100 out of 10000 is

The standard normal variate having 0.01 area to its right is, Z = 2.33. Hence,

Taka 1866 (Approx).

This implies that the lowest among the richest 100 is getting Taka 1866 per month.

Uniform Distribution: The uniform distribution is the simplest of continuous distributions. The probability density function is

Where a is the minimum possible value and b is the

maximum possible value of X. The graph of f (x) is shown in the following figure.

f (x)

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0 a b XBecause the curve of f (x) is flat line, the area under it between any two points

and , where , will be rectangle with height and

width ( - ). Thus . If X is uniformly distributed

between a and b, we shall write .

The mean of the distribution is the midpoint between a and b which is and

variance of the distribution is .

A common instance of uniform distribution is waiting time for a facility that goes in cycles. Two good examples are a shuttle bus and an elevator, which move, roughly, in cycles with some cycle time. If a user comes to a stop at a random time and waits till the facility arrives, the waiting time will be uniformly distributed between a minimum of zero and a maximum equal to the cycle time. In other words, if a shuttle bus has a cycle time of 20 minutes, the waiting time would be uniformly distributed between 0 and 20 minutes.

The Exponential Distribution: This distribution is closely related with the Poisson distribution. For example, if the Poisson random variable represents the number of arrivals per unit time at a service window, the exponential random variable will represent the time between two successive arrivals. Suppose that the number of times that a particular event occurs over an interval of time or space has a Poisson distribution. Furthermore, consider an arbitrary time or space unit ( for example, minute, week, inch, square or the like) and let x denote the number of time or space units between successive occurrences of the event. Then x is described by an exponential distribution having parameter , where is the mean number of events that occur per time or space unit.When X is exponentially distributed with frequency , we shall write The probability density function f(x) has the form

, where is the frequency with which the event occurs.

The mean and variance of x that has an exponential distribution are and

.

Example 13: Suppose that the number of people who arrive at a hospital emergency room during a given time period has a Poisson distribution. It follows that the time x, between successive arrivals of people to the emergency room has an exponential distribution. Furthermore, historical records indicate that the mean time between successive arrivals of people to the emergency room is

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seven minutes. Therefore =7 which implies that = 0.14286.

Example 14: A particular brand of handheld computers fails following an exponential distribution with a of 54.82 months. The company gives a warranty for 6 months.(i) What percentage of the computers will fail within the warranty period?(ii) If the manufacturer wants only 8% of the computers to fail during the warranty period, what should be the average life?Solution: (i) Here = 0.0182. (i) . Thus, 10.37% of the computers will fail within the warranty period.(ii) Here = 0.0139, for which value is 71.96 months. Therefore, the average life of the computers must be 71.96 months.The density curve corresponding to any normal distribution is bell-shaped and therefore symmetric. There are many practical situations in which the variable of interest to an investigator might have a skewed distribution. One such distribution is exponential distribution. This is also a continuous distribution. This distribution is widely used in engineering and science disciplines.This distribution is closely related with the Poisson distribution. For example, if the Poisson random variable represents the number of arrivals per unit time at a service window, the exponential random variable will represent the time between two successive arrivals. Suppose that the number of times that a particular event occurs over an interval of time or space has a Poisson distribution. Furthermore, consider an arbitrary time or space unit (for example, minute, week, inch, square or the like) and let x denote the number of time or space units between successive occurrences of the event. Then x is described by an exponential distribution having parameter , where is the mean number of events that occur per time or space unit.When X is exponentially distributed with frequency , we shall write The probability density function f(x) has the form

, (1)

Where is the frequency with which the event occurs. Here the parameter is a real, positive constant.Some sources write the exponential distribution in the form

, so that . (2)

The mean and variance of x that has an exponential distribution are and

.

Example 13: Suppose that the number of people who arrive at a hospital emergency room during a given time period has a Poisson distribution. It follows that the time x, between successive arrivals of people to the emergency room has an exponential distribution. Furthermore, a historical records indicate that the mean time between successive arrivals of people to the emergency room is

seven minutes. Therefore =7 which implies that = 0.14286.

Example 14: A particular brand of handheld computers fails following an exponential distribution with a of 54.82 months. The company gives a warranty for 6 months.(i) What percentage of the computers will fail within the warranty period?(ii) If the manufacturer wants only 8% of the computers to fail during the

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warranty period, what should be the average life?Solution: (i) Here = 0.0182. (i) . Thus, 10.37% of the computers will fail within the warranty period.(ii) Here = 0.0139, for which value is 71.96 months. Therefore, the average life of the computers must be 71.96 months.Example: The average time needed to send an e-mail from a cyber café is 5 minutes. Find the probability that an e-mail will be sent (i) after 10 minutes (ii) by 10 minutes (iii) Within 5 to 10 minutes.If a man is in queue for 25 minutes, what is the probability that he will be able to send e-mail by 30 minutes?Solution: let be the time needed to send an e-mail. It is given that

.

Here ,x>0.

(ii)

(iii)

In the last example, the person is in queue for 25 minutes that is not the time to send mail. Therefore, mail will be sent by (30 – 25) = 5 minutes.

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