 As Salaamu Alaikum, welcome to lecture number 30 of the course on statistics and probability. You will recall that in the last lecture, I discussed with you the Poisson distribution and we did both cases. The case when the Poisson distribution is a limiting approximation to the binomial and the case when we have a Poisson process. Also towards the end of the lecture, I presented to you a brief discussion of the continuous uniform distribution. In today's lecture, students, we will be discussing the most important distribution in statistical theory and that is the normal distribution. I will be presenting to you some of its important and main properties and also its application in various situations. But let us begin first of all with its formal definition. As you now see on the screen, a continuous random variable is said to be normally distributed with mean mu and standard deviation sigma. If its probability density function is given by f of x is equal to 1 over sigma under root 2 pi multiplied by e raised to minus half into x minus mu over sigma whole square and this function is valid for all x values ranging from minus infinity to plus infinity. Students, this is a very complicated expression but it is very interesting to find that when we substitute various values of x in this equation, we get values of f of x which when we plot yield a beautiful bell shaped distribution which is called the normal distribution. The normal distribution was discovered in 1733. It has two parameters mu and sigma and students, it can be mathematically proved that if our function is of the form that I just presented, then mu is the mean of this distribution and sigma is the standard deviation. This brings us to the first property of the normal distribution and as you now see on the screen, the property number one is that for the normal distribution represented by n mu sigma square, mu represents the mean and sigma represents the standard deviation of the distribution. This is the expression I just presented to you. Capital N or bracket is mu comma sigma square students, this is the generally accepted way of presenting a normal distribution. In the bracket, we write the value of the mean before the comma and then the variance. So, for example, if we write capital N 3 comma 4, what does it mean? We are talking about a normal distribution whose mean is equal to 3 and whose variance is equal to 4. In other words, whose standard deviation is equal to 2. Now, students, since mu represents the mean of the distribution, therefore it is a measure of location. As I indicated in an earlier lecture, the mean is also called a measure of location. Therefore, as we change the value of mu keeping sigma constant, the normal distribution shifts its position on the x axis as you now see on the screen. If mu 1 is less than mu 2 and mu 2 is less than mu 3, but the standard deviation sigma remains constant, then we have three identical normal distributions as far as their spread is concerned, but they are located at three different positions on the x axis. The one with the smallest mean value being toward the left side and the one with the largest mean value toward the right side of the x axis. Now, what happens to the normal distribution? If mu is kept constant, but sigma changes students as you now see on the screen, if we have three normal distributions with identical mean mu, but with different standard deviations sigma 1, sigma 2 and sigma 3, such that sigma 1 is less than sigma 2 and sigma 2 is less than sigma 3, then the one which has the smallest standard deviation is the tallest and narrowest and the one having the largest standard deviation is the one which is the most wide and the least peaked one. The second property of the normal distribution is that the normal curve is asymptotic to the x axis as x tends to plus infinity or minus infinity. Students, you must be already aware that it is a pure mathematical concept and we say that a curve is asymptotic to a line if the curve approaches the line, but it does not actually touch the line until you reach infinity. In this distribution, it is a bell shaped distribution and they are approaching the x axis towards minus infinity and plus infinity, but you never actually get to see them touching the x axis because we actually never reach minus infinity or plus infinity. So, we should always whenever we draw this graph, we should always be careful enough not to join the tails of the normal distribution with the x axis and the third property of the normal distribution is as you now see on the screen. Because of the fact that the normal curve is absolutely symmetric around the mean, therefore, 50 percent of the area under the normal curve is to the left of the mean and 50 percent to the right. Since the total area under the curve is equal to 1, therefore, this means that the area from minus infinity up to mu is equal to 0.5 and the area from mu to plus infinity is also 0.5. The next property of the normal distribution is that the density function attains its maximum value at x is equal to mu and this is why the mean median and the mode of the normal distribution they are all equal to mu. Students, we discussed earlier that if our hump shaped distribution absolutely symmetric though mean, median and mode they are all equal to that same central value that we have. The next property of the normal distribution is as you now see on the screen that since the normal distribution is absolutely symmetrical, therefore, mu 3 the third moment about the mean is 0. Property number 6 states that for the normal distribution mu 4 the fourth moment about the mean is equal to 3 times sigma raise to 4. Property number 7 is that the moment ratios of the normal distribution come out to be 0 and 3 respectively. As you will recall the first moment ratio beta 1 is defined as mu 3 square over mu 2 cubed and because of the fact that mu 3 is equal to 0 for the normal distribution therefore, beta 1 also comes out to be 0. Also beta 2 is defined as mu 4 over mu 2 square and as I just mentioned mu 4 is equal to 3 times sigma raise to 4 and mu 2 square is equal to sigma square whole square which is also sigma raise to 4 and therefore, dividing the numerator by the denominator. The second moment ratio beta 2 comes out to be equal to 3. Now, once you will remember that when we were discussing the frequency distribution, we will be computing B 2 which is the counterpart of beta 2 and if B 2 comes out to be equal to 3, we will say that our distribution is meso-kertic that is like the normal distribution. So, now this property I have presented to you indicates that this is the reason why that number 3 was taken as the criterion. Because, in the case of normal distribution, it is mathematically proved that beta 2 is equal to 3. Therefore, we take this number as a standard and we say that in any frequency distribution, if the second moment ratio is equal to 3, then our distribution is like normal as far as ketosis or peakedness is concerned. The next property of the normal distribution is concerning the areas under the normal curve. Students as you now see on the slide, no matter what the values of mu and sigma are, the interval mu plus minus sigma will always contain 68.26 percent of the total area. The interval mu plus minus 2 sigma will always contain 95.44 percent of the total area and the interval mu plus minus 3 sigma will always contain 99.73 percent of the total area. Students at this point, I would like to remind you about the empirical rule that I discussed with you in the first part of this course. Do you remember, we talked that if our distribution is approximately normal, then approximately 68 percent of the area lies between x bar minus s and x bar plus s, 95 percent lies between x bar minus 2 s and x bar plus 2 s and approximately 100 percent lies between x bar minus 3 s and x bar plus 3 s. I hope that you will now realize where those values came from. You have seen that if our distribution is absolutely normal, then the areas are 68.26 percent, 95.44 percent and 99.73 percent. And since the real life data can be approximately normal, but it will not be absolutely normal. So, we can have approximations to these exact values. The next property of the normal distribution relates to the points of inflection. As you now see on the screen, the normal curve contains points of inflection where the direction of the concavity changes and these points are equidistant from the mean. They are co-ordinate on the x y plane r mu minus sigma comma 1 over sigma under root of 2 pi e and mu plus sigma comma 1 over sigma under root 2 pi e. Let us try to understand this property. The normal curve is like a bell. It is a beautiful bell shaped distribution. And you note that if these two points, which we call points of inflection, are not there, then it could have been something like an inverted curve. But since these two points are where the direction of the concavity changes, the curve is concave downward. But in points, the curve is concave upward. This is why it is like a beautiful bell. Or in this property, we have said that these two points are equidistant from the mean. One is against the x value mu minus sigma and the other against the x value mu plus sigma. The coordinates are at both points 1 over sigma square root of 2 pi into e. Having discussed the main properties of the normal distribution students, I would now like to draw your attention to the normal distribution, which is called the standard normal distribution. As you now see on the slide, a normal distribution whose mean is 0 and whose standard deviation is 1 is called the standard normal distribution. As you have seen, this is a very simple concept. I had told you at the beginning that mu and sigma are the two parameters of the normal distribution. So, if we keep the value 0 and sigma value 1, then the distribution we get is called the standard normal distribution. So, what is the significance of this? Students, it is actually very significant. I will explain to you that if we have to compute any area under the normal curve, then our ordinary procedure that we should find a certain integral, a definite integral. But the point is that the equation of the normal curve is so complicated as you saw in the beginning when I presented that equation that it is not easy to find the integral or the areas under the curve of the normal distribution by ordinary methods of integration. Rather, we have to resort to the more advanced method that of numerical integration. Therefore, we are going to utilize this particular distribution in order to compute any area or in other words, any probability that we are interested in. The point is that fissure and yates have already constructed for us the table of areas under the standard normal curve and it is a table of the type that you now see on the screen. In the very first column of the table, as well as on the very top of the table in the top row, we have numbers which represent various values of z, where z is the standard normal variable and in the body of the table, we have the areas, any area from z equal to 0 up to any particular z value that we might be interested in. For example, if you look in the first column after the column of z and you look at the third value students, that number is 0.0793 and this number represents the area under the standard normal distribution from z equal to 0 up to z equal to 0.20. The reason why I am saying this is that if you look at the first column, you have z equal to 0.2 against the number 0.0793 and if you look directly above the number 0.0793, then you find that in the very very top row of this table, the number is 0.00. Actually, this number in the top row indicates the second decimal and we combine the first column's z number and the top row's z number and we know the z value of the top row. In this case, as I said earlier, 0.0793 is the area under the standard normal distribution curve from z equal to 0 up to z equal to 0.20. If you look at the third value in the column which is headed by the number 0.01, you find that this number is 0.0832 and this means that the area under the standard normal curve between z equal to 0 and z equal to 0.21 is 0.0832. Students, I have presented to you a detailed explanation of how to use the a table of areas under the standard normal curve and the reason is that you will be referring to this table again and again and again in solving various problems and the procedure will be that in any problem involving the normal distribution, first we will convert our normal distribution into the standard normal distribution by the process of standardization and once this conversion has happened, then we will apply this table that I just presented to you. Now, what do I mean by the process of standardization? As you now see on the screen, the standardization formula is z is equal to x minus mu over sigma and if we apply this formula, then we find that if the random variable x is normally distributed with mean mu and standard deviation sigma, then z will be normally distributed with mean 0 and standard deviation will be standard deviation 1. In other words, this formula z is equal to x minus mu over sigma is the standardization formula, jis ke zar ye koi bhi normal distribution, standard normal distribution me convert ki jaasakti. Let me now present to you a very interesting example, students which will not only convey to you how to use the table of areas, but also it will hopefully convey to you the importance of the normal distribution in real life situations. As you now see on the screen, suppose that the length of life for an automatic dishwasher is approximately normally distributed with a mean life of 3.5 years and a standard deviation of 1.0 year. If this type of dishwasher is guaranteed for 12 months, what fraction of the sales will require replacement students, i a is kozara step by step analysis and analysis. Pali baat ye hai ke hamara variable of interest jo hai that is the life length of the dishwasher. Kisi ek factory me dishwashers produce ho rein aur pichle 5-10 saal ka jo un ka record hai usse unne maloom hai ke on the average, their dishwasher lasts 3.5 years. Therefore, we say that the mean life is 3.5 years. Lekin zahir hai ke puch dishwashers jyada der chalte hain, puch jaldi kharaab ho jayenge. So, we have variation and the measure of variation is the standard deviation and that is equal to 1 year. We have the most important point that based on the past record of all the dishwashers sold and consumed by people in the past, they found that if they draw the histogram of this particular frequency distribution in which x represents the time, the life of the dishwasher and f of course represents the number of dishwashers falling in various time classes, they find that this histogram is approximately normally distributed. Yani, bahut kam life wale dishwashers ki tadad bhi kam hai, bahut zyada life wale dishwashers bhi kam hai tadad mein. Aur 3.5 years ya 3 ya 4, yani bahut jo average life hain, bahut zyada dishwashers hain which are over there. Therefore, our histogram rises and falls and if we draw a curve on top of the histogram it is like a normal distribution. Ye sari assumption jo hai students, this is not unrealistic, jyasa ke mein aap se pehle gaya chuki hoon, bohot se phenomena hain which are like a hump which is approximately normal. So having understood this basic point, now what is the problem in this example? The problem is ke boh jo dishwashers banane wali factory hain, unke jo managers hain ya unke jo owner hain, unne garanti jo dete hain ke agar ish mudhat se pehle dishwashers kharaab hojai, to then we will replace it for free. That guarantee has been placed as 12 months and the question is that we want to find what fraction or what proportion of the dishwashers will have to be replaced for free. In other words, students we can say that we would like to find the probability that a dishwasher fails before 12 months. Now, since 12 months is equal to 1 year, therefore what we have to do is to find the area under our normal curve in this problem and the area from minus infinity up to 1 as you now see on the screen. As I stated earlier, the theoretical normal distribution starts from minus infinity and goes up to infinity and although of course all of us will agree that the minimum life length of any dishwasher can be 0 and not minus infinity, but if we try to model this particular real life situation by the normal distribution, then we will be finding the area under the normal curve from minus infinity up to x equal to 1. As I have already told you, it is not easy to find this area directly. So, students we will first use the process of standardization and we will convert our x to z to be equal by the formula z is equal to x minus mu over sigma. So, as you now see on the slide in this problem z is equal to x minus 3.5, the whole thing divided by 1.0 and since our x value to the left of which we wish to compute the area is 1.0. Therefore, substituting this value in the expression for z, z comes out to be minus 2.5 over 1 and that is equal to minus 2.5. So, this means that we need to find the area under the standard normal curve from minus infinity up to minus 2.5, but students, you note that the area table I have just presented in front of you that all the values of z were positive and the area table usually is constructed in this manner only that you do not find negative values of z. So, how do we tackle the problem? It is quite simple. The point to be understood is that the curve is absolutely symmetrical about the mean. So, you will have exactly the same area on the right side of the curve between 2.5 and plus infinity. So, what we need to concentrate on first is the area under the standard normal curve from 2.5 to plus infinity. As you now see on the screen, this area is on the extreme right of the normal distribution, but according to the table of areas that I just presented to you students, you will be able to find first the area from z is equal to 0 to z is equal to 2.5. In the first column of the table, we are not saying 2.51 or 2.53. So, the first column of the table or we are not saying 2.51 or 2.53. So, under the value 0.00 against the value 2.5, we find that the number is 0.4938 and this is exactly the area that lies between 2.5 between z equal to 0 and z equal to 2.5. Zero to 2.5, how do we find that? Because the distribution is absolutely symmetrical, therefore the area from 0 to infinity is exactly 0.5 and the area from minus infinity to 0 is of course, is also exactly 0.5. This is the total area 0.5 here between z equal to 0 and infinity or we have to find here that is 0.4938. So, the tail area we want that is obtained by subtracting this area internal that we just found from 0.5 and as you now see on the screen, this area comes out to be 0.5 minus 0.4938 is equal to 0.0062. The area between 2.5 and plus infinity is exactly equal to the area between minus infinity and minus 2.5 and hence we can say that the required area, the one that we wanted to find in this particular problem that is equal to 0.0062. This key interpretation kya hai? Deke 0.0062 johena that is even less than 0.01 or is ka matlab yeh hua that the probability that the life length of a dishwasher is less than 1 year, this probability is extremely small and it is even less than 1 percent. Is ka matlab yeh hua kya even less than 1 percent of the dish washers that are sold will require replacement. Yani joh log baapis aayenge yeh kahte hua kya ek saal bhi nahi chala, bara maine bhi nahi chala aur aapki guarantee ke matabik toh yeh usse pehle fail ho gya lehaza aap replace ke jhe, aisa hone ki probability joh hai that is 0.0062 even less than 1 percent. Deke yeh students iss example se aap yeh understand ke jhe ki iss khasam ki probabilistic calculations kis qadar helpful saabith ho sakti hai in making important decisions. Agar iss tara se past record ke zare yeh iss tara ki calculation ki jaye toh management joh hai they can decide whether it is appropriate to give a guarantee of 1 year or maybe it is better that they should give a guarantee of 2 years. Agar bo aise hi computation 2 years ke liye kar leh and if they find ke even when we put x equal to 2 that area is still quite small then they can decide that they can afford that agar fars ki jhe 3 percent nikal aata hai wo area. It means only 3 percent of the people will come back and if they think that they can afford to replace for free even 3 percent of the dishwashers then why should they not write 2 years on the guarantee rather than 1 year. Saab zahere ke agar bo 2 years likhenge to zyada customers attract honge as compared with if they write 1 year. Hence the example that we just did was of the direct use of the area table jab kabhi bhi aap kisi z value ke against area find karna chahti hain this is called the direct use of the area table and the procedure is as I just explained. But we can also use the area table in the other order and that is called the inverse use of the area table. Ke amatlab, ke humare pass ab ek area available hain and we would like to find the corresponding z value and also the corresponding x value. Pele z value maloom thi or area nikala ab area available hain or z value nikalingi. Let me explain this with the help of the example that you now see on the screen. The heights of applicants to the police force in a certain country are normally distributed with mean 170 centimeter and standard deviation 3.8 centimeter. If 1000 persons apply for being inducted into the police force and it has been decided that not more than 70 percent of these applicants will be accepted and the shortest 30 percent are to be rejected. Then what is the minimum acceptable height for the police force? Students, aapne dekhah ke this is also quite an interesting problem. Aur ab iss baat pe aghor ki jhe ke jo mai aat thodideh pehle aapse kairi thi. In some situations an area is available and we want to find the z value ye baat yaha pe kishtara apply hoti hai. Iss baat ko understand karne ke liye students. First of all let us visualize the normal distribution that we have at our disposal. So as you now see on the slide our normal distribution has a mean of 170 and the standard deviation is equal to 3.8. As the value 170 is in the exact middle of the distribution therefore 50 percent of the area under the normal curve lies to its left and 50 percent to the right. But students we are interested in that particular value of X to the left of which the area is 30 percent and to the right it is 70 percent. Aisa kyu isliye ke jo cutoff jaha pe hunahe na that is that point. Hame ne kaha tha na ke short test 30 percent johen they have to be rejected. So we have to determine that particular cutoff point and before we can find it in terms of X we will first find it in terms of Z. As you now see on the screen the standardization formula Z is equal to X minus mu over sigma can be rewritten as X is equal to mu plus sigma Z. So this is and substituting the values of mu and sigma in this problem we obtain X is equal to 170 plus 3.8 Z. Now we will first find the value of Z using the area table and substituting that Z value in this particular equation. We will be able to find the required X value that is that height which is the cutoff point. Hame jis Z value ki baat kare hain that is on the left hand side of Z equal to 0 which is the mean. Aur ye wo Z value hain jis ke left side per area hain 30 percent aur jis ke right side per area hain 70 percent. Lekin jo area Z equal to 0 aur iss particular Z value ke dhar maan hain that is equal to 20 percent. Ye sari baat jo maini kahin this pertains to that Z value which is to the left side of Z equal to 0. Iska matlab ye hua ki agar hain iss value ko determine karle students it will be a negative number. Kyuke Zero ke left side pe jo number hain obviously that is going to be negative. Lekin jaisa ke maini thori der pehle aapse kaha tha jo humari area table hain usme we only find positive values of Z. Toh we can do something similar to what we did in the last problem. We should realize that whatever happens to the left of Z equal to 0 happens in a very similar manner to the right of Z equal to 0. As you now see on the screen we can think of that particular Z value which is to the right of Z equal to 0 and the area from 0 to that particular Z value is 20 percent whereas the area from that particular Z value up to plus infinity is 30 percent. Now we are in a position to consult the area table of the standard normal distribution. Ye jo area hume nazar aara hain between 0 and our Z value this is 20 percent which is the same as 0.200 and this is exactly the value that we would like to find. In the body of the area table but students jab aap area table ko study karenge toh aap dekhenge ke uske body ke andar 0.20000 aapko nahi milleega. Then how do we tackle this problem? As you now see on the screen the value that we have against Z is equal to 0.52 is 0.1985 and the value that we have against Z equal to 0.53 is 0.2019. Now 0.1985 is closer to 0.20000 than 0.2019 hence the value that we consider is going to be 0.52. Ye jo kuch main abhi explain kia iska mafum ye hain ke humne ye determinant kar liya ke Z equal to 0 or Z equal to 0.52 ke dar mian the area is approximately 0.20000. Jaisa ke aap ne dekhaha ke agar exact wo value nahi mille aapko toh jo uske closest value aapko us area table ke body ke andar se milti hai that is the value that you consider. Now having found this Z value 0.52 students we realize that this is the one which is on the right side of Z equal to 0. Lekin aapko yaad hai na humto left side main interested hai. Now because of the absolute symmetry of the normal distribution I hope that you realize that whatever is happening on this side for Z is equal to 0.52 exactly the same kind of a thing is happening on the left hand side for Z equal to minus 0.52 and hence minus 0.52 is that Z value that I will substitute in the equation that I presented to you earlier. As you now see on the slide our equation was X is equal to 170 plus 3.8 Z and putting Z is equal to minus 0.52 I obtain X is equal to 168.024 which is approximately equal to 168. Hence we have arrived at the conclusion that the minimum acceptable height is 168 centimeter for induction into the police force according to the criterion that they had set. So, aapne dekhaha ke hum area table ko directly use karne ke elawa inversely bhi istimal kar sakte hain and we can arrive at some very important and useful results regarding our problem. Just as I mentioned to you that the binomial distribution and the Poisson distribution and other distributions can be fitted to real data similarly we can fit a normal distribution to real life data provided that we are reasonably confident by looking at the histogram or by looking at the proportions of areas within the intervals X bar plus minus S X bar plus minus 2 S and so on that our distribution is approximately normal. If we feel that our distribution is normal then we have a whole procedure through which we can fit a normal distribution to our data and then applying the chi square test of goodness of fit we can also determine whether or not our fit is good. What I will discuss with you now is the normal approximation to the binomial distribution. Yeh bara interesting sa concept hain aapko maloom hain ke binomial distribution to discrete distribution hain aur jab hum uska graph banaate hain to that is a line chart in which we have separate lines or normal distribution zahir hain ke because it is a continuous distribution is liye the curve is a continuous curve to fir yeh kaise hoga ke hum yeh kahin ke we are using the normal distribution which is a continuous distribution to approximate the binomial distribution which is a discrete distribution students. I will explain this to you with the help of an example and you will realize that there is something called the continuity correction which enables us to do exactly what I just said to approximate a discrete distribution by the continuous normal distribution. As you now see on the slide suppose that the past records indicate that in a particular province of an underdeveloped country the death rate from malaria is 20 percent. Find the probability that in a particular village of that particular province the number of deaths is between 70 and 80 inclusive out of a total of 500 patients of malaria. Students, sab se pehle yeh dekhye ke kya hum ek binomial experiment ke sath deal kar rahe hain? Pehle baat yeh ke what is success and what is failure? Now we are talking about malaria patients. Either a patient will survive or a patient will die from malaria. And if we regard dying as success then we can deal with this problem quite conveniently because as you just noticed our query was that out of a total of 500 patients what is the probability that the number of patients who will die that number lies between 70 and 80 including 70 and 80. I had already told you that success does not mean that it is something very good. It is a technical term and it represents that outcome that we are interested in in any particular problem. So, yeh toh ho gayi pehle baat, dosri baat ke the various trials are independent. To yeh hain par agar hum yeh assume karein ke all these patients are independent of each other. To phir yeh jo past record se hum yeh baalum hua ke probability of dying from this disease in that province is 20%. We can say that this probability is constant from patient to patient and thus the third property of the binomial experiment is also satisfied. P is equal to 20% that is 0.2. Aur aahri baat kya thi that the number of trials is fixed in advance. Tu ke hum karein hain ke we have a total of 500 patients therefore the number of trials N is 500 and it is a fixed number. Now that we are confident that we are indeed dealing with a binomial experiment students the next point is when can we apply the normal approximation? The answer is that if neither p nor q is close to 0 and if n is large then we can apply the normal approximation to the binomial. The rule of thumb in this regard is that if both n p and n q are greater than or equal to 5 then we can apply this approximation. In this problem n p is equal to 500 into 0.2 and that is equal to 100 and n q is equal to 500 into 0.8 and that is equal to 400. Now both these quantities are much much larger than 5 and hence we can happily apply the normal approximation to the binomial in this particular problem. The next very important concept is continuity correction. I had told you that binomial distribution is a discrete distribution and we have a line chart as you know we have separate bars and the normal distribution is a continuous curve. So, in order to find the probability that my x variable lies between 70 and 80 because x represents the number of deaths from malaria students. If I do this by binomial then you can understand that the lines or bars that we have made are the lengths which present the probabilities of those x values. We will have to consider those lengths and adding them all up we find the probability. But if we want to do this through the normal distribution we will be super imposing a normal curve on top of those bars as you now see on the screen. And in order to justify this process what we need to do is to replace every x value which is an isolated point on the x axis by an interval. The x value 70 is replaced by the interval 69.5 to 70.5 the x value 71 is replaced by the interval 70.5 to 71.5 and so on. And in this way our region from x equal to 70 to x equal to 80 is replaced by the region x equal to 69.5 to x equal to 80.5. Yani because of the continuity correction our region is expanded in this particular problem slightly. Next we have to find the area under the normal distribution between x equal to 69.5 and 80.5. And for that of course we use the process of standardization that I have already explained to you. As you now see on the slide z is equal to x minus mu over sigma. And in this particular problem since we are basically dealing with a binomial experiment and a binomial distribution. Therefore mu is equal to n p and sigma square is equal to n p q. Hence our formula becomes z is equal to x minus 100 over 8.94 because mu is n p and n p is equal to 100. And sigma square is n p q giving us a variance equal to 80 whose square root 8.94 is the standard deviation that we require in our formula. Students I now leave it to you to compute the area under the normal curve between x equal to 69.5 and 80.5. You will be transforming x to z and then you will consult the area table of the standard normal distribution in order to find the required area. And it may be a few steps before you arrive at the final answer which is 0.0145 indicating that the probability that the number of deaths from malaria in this particular village of this particular province out of a total of 500 patients. The number of deaths lies between 70 and 80 including these two values. This probability is 0.0145 or in other words one and a half percent. All right students this brings us to the end of the second segment of this particular course and the segment on probability theory. Next time inshallah we will begin the third and last segment of this course a very interesting portion and that is inferential statistics. In the meantime I would like to encourage you to attempt a lot many questions on the normal distribution so that you feel at home and comfortable with this distribution. My best wishes to you and until next time Allah Hafiz.