Welcome to class 33 in topics in power electronics and distributed generation. We have been looking at the dynamic thermal models of IGBT module and we have seen that it is important to evaluate the junction temperature of the semiconductor chips, transistors, diodes, etcetera. And to do that we looked at concepts of thermal resistance, thermal capacitance and how to model it as a equivalent electrical circuit. And we saw that the electrical circuit can be mapped to a network of to look at the steady state behavior of such a structure of the thermal management system. And to look at the dynamic thermal characteristics, one can then look at a force a core network which looks like the physical R C networks that would correspond to the physical thermal characteristics of the multiple components that go into the IGBT module. The points of interest would essentially be your junction temperature, may be your case temperature, sink temperature or the ambient. And often in a power module the case temperature the data sheets provide transient thermal impedance characteristics for a fixed case temperature. And you are looking at the transfer function between your junction temperature to case temperature and the same behavior can be obtained from equivalent foster network from again from the junction temperature point of view. So, the intermediate nodes of the foster network may not make much physical sense, but from a terminal point of view the behavior of the foster network is similar to that of the power network. The number of R C elements can be typically it is between 2 and 6, the number of elements can be higher to get a better approximation to the thermal response characteristics of your measured measurements with your model of the thermal system. The data sheets provide typically the foster network parameters or some data sheets may just provide the transient thermal impedance curve. And we saw that connecting the module together with the heat sink one has to make transformations between the foster and the power networks. So, if you want to analyze the foster and power network then one has to look at how the impedance can be evaluated for each of this. So, we will look at a foster network as a recursive unit and if you look at the impedance of the nth recursive unit of the foster network Z F T H of a nth unit is 1 by R N plus S C N plus Z of the foster network of the previous stage. So, if you are looking at it this is the impedance of the foster network subsequent to the nth unit and you are adding the nth component to it to get the foster impedance at the nth stage. If you look at so, you can write this as 1 by C N divided by S plus 1 by R N C N plus Z F T H at n minus 1. So, in a similar manner we have Z F T H of n minus 1 can be written 1 by C N minus 1 by S plus by R N minus 1 C N minus 1 plus Z F T H of N minus 2. So, you could write a general expression for the foster network to be of the form with n sections. So, if you look at the form of this expression this is essentially a partial fraction expansion of the transfer function. If you write Z F T H to be a ratio of polynomials P N of S by Q N of S then essentially it is the partial fraction expansion of such a polynomial. You could also then look at what is the recursive unit for core network. So, essentially you are looking at Z C T H at the nth stage in terms of the R N C N parameters and Z Z T H of the n minus 1 stage. So, you have a capacitance in parallel with the resistance plus your subsequent network impedance. So, it is 1 by S C N plus R N plus Z C T H of n minus 1 and you can write Z Z T H of n minus 1. So, if you look at the structure of the impedance the general expression for a core network with n stages you can write Z Z T H of n is 1 by S C N plus 1 by R N plus 1 by S C N minus 1 plus 1 by R N minus 1 plus essentially this would continue along. So, this structure of this is of the form of a continuing fraction. So, if you have impedance that is provided to you as a foster network impedance and you would like to transform it to a core network we had let the foster network that is provided to you Z F T H of n be the ratio of polynomials P of n of S by Q n of S and for thermal systems you would have P n of S to be a polynomial of degree 1 less than the denominator polynomial Q n of S ok. So, we know that the foster to core transformation the cover impedance would also be equal to the same ratio. So, this is also equal to Z Z T H of n. So, you could then evaluate your cover network 1 look at 1 by Z F T H of n which is 1 by Z C T H of n which is Q n of S by P n of S and because Q n is 1 degree higher than P n you can write it as some S C n you can take the S outside plus 1 by R n plus Z T H. So, essentially we have written this S C n plus 1 by R n plus Z C T H from our structure of essentially the cover network which we know is of this particular form ok. So, you could also now we know that Q n of n is 1 degree higher than P n. So, you could apply Euclid's long division for polynomials to get Q n of S by P n of S equal to S C n prime plus K n plus essentially the reminder of the division by P n of S. Now, you the reminder of the division would be 1 degree lower than the degree of P n of S. So, if you compare this expression with that we know that now C n is equal to C n prime. So, essentially the coefficient of S in the numerator of in our quotient gives the capacitance and we will have to look at what is how K n can be related to our resistance ok. So, if you then again compare this particular form with the cover recursive unit which is essentially if you look at this particular term over here when you are looking at 1 by Z T H you would have S C n plus 1 by this particular term in the denominator. So, we have R n plus Z T H of n minus 1 is 1 by K n plus reminder of S divided by P n of S. So, you could then rewrite this particular expression in the form this would also be equal to 1 by K n minus reminder of S reminder at stage n. So, essentially if you simplify this particular expression you will get it to be the same as this expression ok. So, again if you now compare this with our the expression of impedance of our cover recursive unit you can see that the 1 by K term now would correspond to R n ok in this particular if you look at compare these two expressions you will see that 1 by K n corresponds to R n and your Z C T H of n minus 1 is this second term over here which is minus remainder n of S by K n divided by K n P n of S plus reminder n of S. So, this is essentially your P n minus 1 of S by Q n minus 1 of S. So, you apply the same procedure at the next level. So, you will get the terms for C n minus 1 and R n minus 1 till you reach the first C 1 and R 1. So, you would essentially be able to get the R and C parameters of your cover network. So, the other question is how to if you are given a network which is in the cover form how to transform it to the foster form. So, you are given say Z C T H of n is P n of S by Q n of S. So, to actually do such a transformation what you would do is first determine roots of Q n of S. So, this would be your time constants 1 by R i C i and then you do a partial fraction expansion of P n of S by Q n of S. So, you find the residues of the partial fraction expansion P n of S by Q n of S. So, tau i is. So, your 1 by 1 by C i would correspond to your. So, this 1 by R C is your omega i. So, you from your roots of your expression you can find what your R C terms are and from your K i terms which are the residues of your partial fraction expansion you know what your C i's are which are essentially 1 by K i. So, once you have the ability to go from one form the foster form to the cover form and back you can actually build the overall thermal impedance network of your power module your interface layer and the heat sink. And get the overall thermal equivalent model for which can be used for dynamic thermal evaluation. So, essentially what you would have is dynamic thermal model for your junction temperature to your sink. So, this would be your T j of your switch this would be T of your case and this point would be T of your sink and this point is the ambient temperature. Similarly, you could determine the T j of your diode and similarly for the remaining switches that are located within your power module. If you have multiple power modules now located on the same heat sink you could add more modules to the same heat sink your thermal interface material is essentially a very thin layer of the order of tens of microns. So, it has considerable thermal resistance, but not much thermal capacitance. So, you might be able to make use of a simplified model for that your models for your the power module would have some thermal heat capacity and some dynamics associated with it. And your typical power modules can have a thermal time constants which can be of the range of even up to 100s of milliseconds. You look at the thermal dynamics of larger heat sinks it can be of the order of seconds to minutes or very large heat sinks can even take hours to reach thermal equilibrium depending on the cooling that is being used the form the cooling that is being used. So, you one would be able to form the overall thermal model of your overall network. One thing to look at to understand in such a model is that if you are looking at the power loss variation over a fundamental cycle. So, you are talking about 50 milliseconds would be quite small in a typical application. So, your Tj variation during 50 hertz variation might be of the order of a few degrees centigrade whereas, the time constants involved in the overall response might have considerable effect when you are considering transients that are of the order of 100s of milliseconds to seconds. So, in applications where say you are starting an induction machine your motor might take many seconds or 10 seconds to start up then your you might have a over current or higher current levels during such duration. You might have say wind turbines where you might have a wind gust which can provide more power. So, you might have transient durations where you are having higher power flowing through your system. You might have say solar applications where you might have periods of shadows clouds going across shadows moving across the system. You can also have faults in your grid and we saw that for many faults your circuit breakers might take 100s of milliseconds to operate. So, if you want your power converter to ride through a fault you might end up having higher current levels for a duration which might be of the order of 100s of milliseconds. Also in standalone applications your loads might start up and your inrush currents of your loads might have a higher power level and so, you are so evaluating the dynamic thermal characteristics of your semiconductors becomes an important issue. We also will see that the evaluating the dynamic thermal characteristics as implications on the reliability of the semiconductor. So, for that we will have to look at a closer take a closer look at the structure of a power semiconductor module. But before we do that we will look at some of the issues of reliability and failure in power semiconductors. So, if you have a power semiconductor and you ensure that under all conditions you are operating the device at a voltage less than its rated voltage when ensuring that the spikes are below your rated voltage you are ensuring that you are operating always at a current level less than the rated current. And you are also ensuring that on a steady state and also on a dynamic basis your junction temperature is held below the maximum junction temperature then one could possibly operate the semiconductor for a very long duration without causing damage ok. But practically people have seen that electronic equipment in the field fail and people have ascribed random failure characteristics to it and done statistical analysis to look at attributes such as what is the mean time between failure. So, for a repairable system mean time between failure. So, the MTBF is the time duration when you send out a component get it once you get a component back from repair and the next point when you are actually sending it out for repair. So, the duration between between is the MTBF. So, these are typically for repairable components for many components you might once it is failed it may not be repairable. So, you might have to throw it out or the duration under which it is repair would be of no use. So, people also talk about mean time to failure in a DG application when your power converter is down the overall DG system is not generating energy. So, MTBF might actually be a better attribute to look at what is the failure characteristics of a power converter. Another way of looking at failure is failure in time and failure in time is actually a quantity which is one failure. So, having one FIT implies that you have one failure in 10 to the power of 9 hours. 10 to the power of 9 hours is actually a very very long time if you convert it into centuries this corresponds to about 1, 1, 4, 2 centuries. So, it is a really long time. If you look at the failure rate associated with IGBT chip the random failure rate associated with the IGBT chip. So, it is of the order of 10 your the FIT rate for the IGBT gate drive is of the order of 500. So, if you now calculate that into years this corresponds to about 228 years. If you look at the FIT rate then for the gate drives of a 3 phase power converter system. So, you would have 6 gate drives. So, you are talking about now 38 years would be the expected say random failure rate, but many times the random failure does not give a true picture of when the device might actually fail because if you look at the random failures they correspond to the point where your device operates under conditions of low failure rates. So, we know that typical devices have this bathtub curve. So, you are looking at your failure rate initially your failure rate might be high because of infant mortality, but once the component enters into service you would have a low failure rate and if you are using this particular component to determine your random failure rates, but you know that you might be underestimating the time because at some point of time your failure rates would start going up again because of end of life issues. So, it is quite important to actually look at what is the end of life, what is the service life of the equipment due to the end of life effects to get a better picture for how long we could expect a component to actually last. So, if you look at typical power semiconductor module you would have ageing effects because of variety of reasons. You might have just a regular material degradation because of the temperature effects. So, one can one knows that higher temperature can actually degrade material at a faster rate. So, there might be material degradation important component of ageing is actually thermal cycling and essentially thermal cycling is because you have a structure of a power module with different materials and you have coefficients of thermal expansion for the different materials that consist that go into the power module. And if your device heats up to some Tj max the maximum junction temperature the expansion of the different materials would be different which means that you might end up with structures that want to buckle or change its shape. So, one is because of the coefficient of thermal expansion mismatch also another factor to keep in mind is that even if you have closely related thermal coefficient of expansions your temperature is not uniform adjacent layers might be sitting at different temperatures. So, your junction temperature might be 125 your case temperature might be 70 degrees. So, the actual temperature is different across the different layers. So, for example, you might have silicon which might be sitting on top of ceramic layer and the ceramic layer say you might have your chip sitting on top of a ceramic layer and at 25 degree centigrade room temperature you might have the silicon chip dimension to be maybe 1 centimeter and that might match your ceramic layer. But at 125 degree centigrade your silicon might expand from say 10 mm to 11 mm, but your ceramic might expand from 10 mm to 12 mm which means that essentially the two structures are now having a mismatch and it is trying to buckle over or develop cracks at the corners. So, you could have a variety of different concerns when you are operating in such a power module you would have stress build up across layers because of your thermal characteristics and you might have cracks in your dielectric. So, your different points of the dielectric might expand at different rates and from cracks wire bonds that are used to attach the chip to your external terminals that is how typically silver or aluminium might have a different expansion coefficient it might actually break or it might lift off you have solder joints within the module. So, those points would develop fatigue your chip itself might get delaminated from your bottom your your insulation itself might get delaminated from your base plate and you might end up with lifted dies etcetera. So, after these problems like wire bond lift off delaminated corners of the chips you end up with non uniform current density within your chip which can actually cause even further heating which can actually further accelerate the temperature related degradation within the power module. If you look at the overall structure of the power module you have essentially the heat sink at the bottom typically commonly used material for the heat sink might be aluminium you might also use copper if you are looking at higher heat storage capacity you have your thermal interface material or essentially the thermal grease between your base plate and the heat sink which is a thin layer you have a base plate and the base plate is essentially the bottom of your module which you tighten with screws to your heat sink commonly used material for the base plate is either copper or people also look at aluminium silicon carbide which has a better match of with the thermal expansion coefficient of silicon above the base plate is actually the ceramic layer. So, the ceramic layer to the base plate is has a direct copper bonding which is essentially a way of bonding a ceramic to copper structure and on the upper side of the ceramic is also direct bonded copper which is used to ensure that you can do a solder joint between your silicon chip to your upper side of your ceramic. So, you have then on top of the DBC layer direct bonded copper you have a solder layer which is used to connect the chips to your bottom of the module. So, essentially if you look at the chip if it is a IGBT chip the bottom would correspond to the collector if it is a MOSFET typically you are looking at the drain at the bottom if you are looking at a diode it is essentially the cathode at the bottom of essentially the module and at the top. So, after that you have the silicon of the chip on top of that you might have oxides, gate oxides you might have aluminium metallization for contact to external terminals you would have wire bonds that are then used to connect to external terminals. You might have passivation layers you would have potting material and then on top would be the casing typically a plastic casing to ensure that all these structures are held within the power module. So, you can see that it is a sandwich arrangement often you try to select materials for ceramic your base plate for example, the ceramic may be aluminium oxide aluminium nitride aluminium nitride has a better match of thermal expansion coefficient with your silicon. So, it would be able to withstand more thermal cycling. So, to look at the failure when you have such cycling and structure one model that is commonly used when you are looking at fatigue failure in a mechanical structure is the Coffin-Manson model which relates the number of cycles to failure to the strain that happens within the structure. So, essentially the original model was proposed for ductile materials and later on it was seen that you could actually fit the model for brittle structures like silicon chips or ceramic layers etcetera. Essentially you have a exponential relationship between the number of cycles to failure to the strain that is applied within the object under test and a naughter and b 1 are essentially coefficients associated with this model. So, again what we are looking at is a situation where you are looking at strain being the result of a stress being applied on the material and in this particular case the stress that is being applied is your your temperature of your particular junction and because your temperature has increased to some particular point your different layers now apply different levels of force at the interface layers and deform the structure. So, typically when you apply a stress on an object you would initially have something called the elastic range where you take the stress away and it comes back to the normal point, but beyond beyond the yield point you would essentially have plastic strain range where you apply a stress and you then you take away the stress you may not actually return back to the original point. So, what is looked at for the cycles to failure is the range for plastic strain. Also we know that the this particular strain range is proportional to the temperature difference of the cycles through which you heat up and cool down the temperature the temperature of the junction. You might have some temperature range corresponding to the elastic stress and then the balance would correspond to the temperature range where you might actually have plastic deformations. So, overall if you look at it the the delta T naught which might correspond to the temperature variation in the elastic range might be quite small compared to the overall temperature variation. So, you could simplify it to be of the form that the number of cycles to failure depends on your variation of your junction temperatures to some exponent a naught and q are essentially constants. So, some of this effects of temperature cycling people started investigating in the mid 90s and some of the early tests were conducted on traction module in traction applications. So, if you have a train or a tram or a electric vehicle you have a situation where when you accelerate at start you are applying a lot of current through your motors. So, you have higher power loss. So, once you reach steady speed your power level comes down and then when you decelerate you might be regenerating. So, again the power level can go up and then when you are at stop your power level is close to 0. So, traction applications and transportation applications see a lot of thermal cycling. So, in it there are initial concerns when IGBT inverters were failing in the field quite rapidly when initially applied in traction systems and some of the initial studies indicated that the number of cycles to failure can be related to the junction temperature and also the mean temperature through some activation energy type of expressions and one could then make use of expressions such as this to evaluate how many thermal cycles can given module be used before it actually starts failing. More recent work related to thermal cycle failures in IGBT modules have also indicated that there can be dependency on the durations of during which your power loading is being done. The amplitude of the current that is being passed through the connections into the chip. Your voltage rating of your semiconductor we saw that the drift region of the semiconductor depends on the voltage rating. So, a thicker chip can have a different thermal cycle compared to a thinner chip. You would also have dependency on the diameter of the wire bond. So, a thicker wire bond might experience different failure rate compared to a thinner wire bond. So, without we will look at this expression for the cycles to failure and look at an example to look at how we could use this in a design where you are looking at what should be the maximum junction temperature that one could consider when you are looking at a design application. So, in this example we are looking at a load which turns on and off once an hour, 24 hours a day around a year and we are assuming that the thermal steady state is reached quickly. So, when it is off it quickly reaches the ambient temperature of 55 degrees centigrade and the question is when the load turns on what should be the system design should we choose at Tj max of 125 degree centigrade or 120 or 110. So, as to ensure that your life of this particular equipment meets your target. So, the parameters for this thermal cycling model used are the exponent being equal to 5, Q the activation energy related term is 7.8 into 10 power 4 joules, R is the gas constant and A naught is 640. So, if you have your Tj min which is essentially your ambient temperature when it is off and Tj max to be 125 your delta Tj would then be 70 degrees and your mean temperature is essentially 125 plus 55 by 2. In Kelvin this is 363 degrees centigrade and you can plug this in to the expression over here to evaluate the cycles to failure. And you can get an expression for then the life of the equipment because we know there is one cycle in R. So, you take this divided by the number of hours number of days a year and you get 7.2 years. So, you can expect life of the order of 7.2 years if you are designing with a junction temperature of 125. If you reduce your junction temperature maximum junction temperature to 120 then your delta T reduces your Tj mean reduces by some extent. Your number of cycles to failure now increases and you get about 12.5 years. If you reduce further reduce your Tj max you can end up now with a lifetime of the order of 41 years. Again these are design numbers it does not mean that your actual inverter might last for 41 years you might have other factors which can cause different types of failure, but these are indicative numbers to see whether this particular issue might be a concern in your particular application. So, one way of addressing say you have an equipment which is intended to last for 10 years then definitely this may not be a good way to actually do the design you might have to select a different temperature. And one way of ensuring that you can actually do that is by going for a higher rated current rating of the device which would have typically lower on state resistance with lower participation. Also your higher current rating device would have lower thermal impedance resulting in a lower Tj max you can get a improvement in your cycle lifetime. One thing to keep in mind is typical data sheets do not provide this level of detail of IGBT data. So, to get detailed data from on the modules to do evaluate cycle life etcetera one might actually have to contact design engineers or application engineers from the company to actually get the detailed information. And the thing to keep in mind is that in typical applications it may not be just two temperatures that one is operating it may not be just operating between say 55 degree centigrade and 120 degree centigrade you might have number of temperatures over which the equipment would operate. So, you might have so you might operate at stress sigma 1 for some duration followed by sigma 2 up to say sigma n and you might have different cycles to failure rate for the different temperature stretches stresses that are being applied on the component. So, if corresponding to sigma 1 you might have n 1 cycles to failure and corresponding to sigma 2 you might have n 2 cycles to failure and n n cycles to failure. So, using the expression we could actually calculate the cycles to failure for the actual for under these varying conditions of stress, but your actual applications may not just operate at n 1 or n 2 cycles you might actually operate for some small n n i small n 1 cycles under stress condition sigma 1 small n 2 under sigma 2 and small n n for sigma n. So, the question is how to actually consider conditions when you are operating under conditions of varying stress rather than just switching back and forth between a minimum and maximum. So, this can be done by one way to look at this is what is called the palm green minor rule for damage accumulation. So, let n i be the expected cycles to failure under stress condition sigma i and small n i be the actual number of cycles operated under this particular stress level sigma i. So, essentially for a given component let d f be the total damage that you can actually subject the object before it becomes non-functional and let d i be the number of the damage caused by operation under condition i. So, if you are operating under stress condition i and d i is the damage that is caused under stress condition i then essentially what you are doing is operating under varying stress conditions you are saying that you could sum over all the stress conditions and once you reach d f then the component is considered failed or in the way to look at it you can look at the damage on a normalized basis. So, you take the damage under your ith condition divided by your damage to failure and once that sums up to 1 essentially here what you are done is we have divided both sides by d f essentially when your normalized damage reaches 1 then essentially you have reached your end of life point. So, for the cycles to failure condition you can then express this as the summation over the number of actual cycle of operations divided by the cycles to failure numbers when that accumulates to 1 then essentially you would end up having thermal cycling failure of this particular component. One thing to consider when you are applying such a concept is that essentially what you are doing is you are assuming that the damage accumulates and essentially you are assuming that there is linearity in the way the damage accumulates. For example, you might have a situation where you apply stress 1 which is larger and stress 2 which is smaller and your total damage at the end point of after application of stress 2 might be something and you could reverse the sequence you could apply stress 2 first and then stress 1. So, the accumulation of damage assumes that you get the same level of stress irrespective of whether you apply the larger stress first or the larger stress second. So, you are neglecting sequence effect because if you have already damaged applying the same effect might cause further damage you are neglecting sequence effects when you are actually applying this particular rule. So, this is an approximation, but it gives you a way of looking at how to accumulate damage when you are operating under conditions of varying stress. So, we will look at an example now of a situation where you are operating under conditions of varying stress we before that example we will look at what could be the procedure that we could use to actually sum up the stress when you are operating under such varying stress conditions. So, the contributions to stress can be not just the loading can be just the ambient itself varying over time. And many in many power applications you would also have some repetition rate of loading or ambient situations. For example, in a factory you might have the production duration might repeat on a daily basis. So, the equipment in the factory might get loaded under some set of conditions on a daily basis or in a transportation application you might have a typical drive cycle which is repeated on a daily basis. There might be some variations on weekends, but you might have some then weekly patterns. For DG applications for example, in a wind turbine you might have patterns of wind which might seem random, but over longer durations your loading levels might have statistical distributions from which might lead to similar level of loading repeating over multiple months. So, depending on your application you can have repetition durations of loading over which you are looking at how the temperature junction temperature is varying over time. So, the first thing to consider is your t-rap. So, the second step is to identify cycles to failure or lifetimes over shorter intervals and then you calculate what is the normalized damage and then you calculate the number of repetition periods over which the normalized damage would reach one would give you an indication of what your expected cycle life service life of your equipment is for an actual operating condition. So, we will look at an example for doing this particular exercise in the next class, but we can see that evaluating the junction temperature the temperatures of various components actually a very critical part of power electronic design. Thank you.


As Salaamu Alaikum, welcome to lecture number 14 of the course on statistics and probability. Students you will recall that in the last lecture I discussed with you in detail the box and whisker plot and towards the end of the lecture I discussed the Pearson's coefficient of skewness. As you will remember the Pearson's coefficient of skewness is defined as mean minus mode over the standard deviation and in those situations where the mode is not very easily found we can apply the empirical relation between the mean, median and the mode and in that situation the formula for the Pearson's coefficient becomes 3 times mean minus median over the standard deviation. . . . . . . . . . . . The answer that you will get will be in the same units as the units of the data itself, but when we divide it by the standard deviation we get a pure number and now we are in a a position to compare the skewness of any one distribution with the skewness of another distribution. A pure number can be compared with a pure number, but for example, if there is a distribution of a pound or a second distribution of inches, then the pounds cannot be compared with inches in a very proper way. So as you noticed, the Pearson's coefficient of skewness requires the computation of the mean as well as the standard deviation. Now if you want to measure the skewness of a distribution without having to compute the mean and the standard deviation, there is another formula and that is the one given by Baale. This formula students is based on the median and the quartiles and these are of course measures of partition and these also enable us to measure the skewness in a very interesting way. For an absolutely symmetric distribution, the distance between the median and the first quartile is exactly the same as the distance between the median and the third quartile. But in case of a positively skewed distribution, the distance between the median and the third quartile is longer than the distance between the median and the first quartile as you now see on the screen. So if I subtract x tilde minus q 1 from q 3 minus x tilde students, I will obtain a positive answer as you now see on the screen. q 3 minus the median minus median minus q 1 is the same thing as q 1 plus q 3 minus 2 times the median and as I just explained, in case of a positively skewed distribution, this quantity will be greater than 0. Only the opposite situation prevails in case of a negatively skewed distribution. In this case, the distance between q 1 and x tilde is more than the distance between x tilde and q 3 and if you subtract x tilde minus q 1 from q 3 minus x tilde, you will obtain a negative answer. q 3 minus the median minus median minus q 1 is again the same thing as q 1 plus q 3 minus 2 times the median and because of the fact that in this case, median minus q 1 is greater than q 3 minus median, hence the quantity q 1 plus q 3 minus 2 times the median comes out to be less than 0. So, this is the way you have seen that quartiles and the median also provide a way of determining the skewness of your distribution q 1 plus q 3 minus 2 times x tilde compute. If it is a positive quantity, it indicates that the distribution is positively skewed and vice versa, but students, the Baoules coefficient is given by as you now see on the screen q 1 plus q 3 minus 2 x tilde divided by q 3 minus q 1. The reason is that when we divide the numerator by q 3 minus q 1, once again we obtain a number which is unit less a pure number and hence we are able to compare the skewness of one distribution with another. Another important point is that it has been mathematically shown that this coefficient that I have just presented to you lies between minus 1 and plus 1. So, when you compute the closer it is to minus 1, the greater will be the skewness in the negative direction and the closer it is to plus 1, the greater will be the skewness in the positive direction or agar aapka answer 0 k bahut nisdeek, positive ya negative ho to is ka matlab ye hai, k your distribution is approximately symmetric. Let us apply this concept to the example of the ages of the children of the manual and non-manual workers that we discussed in the last lecture. As you will recall, we had data regarding the age of onset of nervous asthma in children and we had this information for the children of manual workers and the children of non-manual workers. Also, we had computed various sample statistics pertaining to the ages of these children and we had the figures that you now see on the screen. First, let us concentrate on the distribution of the ages of the children of the manual workers. Now, because q 1 is equal to 6.0, x tilde is equal to 8.5 and q 3 is equal to 11.0, you will realize that the distance between q 1 and x tilde is exactly the same as the distance between x tilde and q 3 and this leads to the symmetric picture that you now see on the screen. On the other hand, the situation of the ages of the children of the non-manual workers is quite different. q 1 is equal to 5.5, x tilde is equal to 9.2 and q 3 is equal to 10.8 and this implies that the distance between q 1 and x tilde is much greater than the distance between x tilde and q 3 and this leads to a negatively skewed picture that you now see on the screen. So, the distribution of the children of the manual workers is symmetric and the distribution of the ages of the children of the non-manual workers is negatively skewed. If we, for both data sets, compute the coefficient of skewness, as you see on the screen, the coefficient comes out to be 0 for the children of the manual workers and for the children of the non-manual workers it comes out to be minus 0.37. And the negative answer indicates that the distribution of the children of the non-manual workers is indeed negatively skewed. Students, the next concept that I am going to discuss with you is the concept of kurtosis. This term was introduced by Karl Pearson and it literally means the amount of hump in your distribution. In other words, this is the concept of the peakedness or the flatness of your distribution. Students, when the values in your data set are closely bunched around the mode, we say that the distribution is leptocurtic and it is a relatively high peaked distribution, as you now see on the screen. On the other hand, if the curve is flat topped, we say that the curve is platicurtic. Students, the normal curve is a curve which is neither very peaked nor very flat and this is the one which is taken as a basis for comparison. The normal curve itself is called a mesocurtic distribution and it is like a bell shaped curve as you now see on the screen. Students, the concept of the normal distribution is of the utmost importance in the theory of statistics and I will discuss this concept with you in a lot of detail, when we discuss probability distributions. It is of a moderate amount of hump and a beautiful bell shaped curve that is the kind that is called the normal distribution. Superimposing the three curves on the same graph, we get a picture as the one that you now see. The leptocurtic curve is the tallest one, the mesocurtic is the intermediate one and the platicurtic is the flat topped distribution. Students, peakedness or ketosis concept, I have discussed with you. Now, the next question is how we measure the degree of peakedness? As you now see on the screen, there is a formula which involves the quartiles and the percentiles and it is known as the percentile coefficient of ketosis. It is defined as quartile deviation divided by p90 minus p10. Students, this measure of ketosis is denoted by k and for the mesocurtic distribution, which is neither very peaked nor flat, its value comes out to be 0.263. Also, this measure, its value can range from 0 to 0.5 and for a leptocurtic distribution, it is less than 0.263, lies somewhere between 0 and 0.263 and for the platicurtic distribution, it is greater than 0.263. In other words, lies somewhere between 0.263 and 0.5. I am sure that you will understand that the closer your answer is to 0.263, the closer your curve will be to the normal curve. The next concept that I am going to discuss with you students is a very important concept and I would like to encourage you to listen to it and to study it very carefully. This is the concept of moments. What do I mean by moments? A moment designates the power to which deviations are raised before averaging them. I know. Let me explain this to you bit by bit. I have just talked to you. There are two or three terms involved. Let us first consider the term deviation. When we say that we are taking the deviation of the x values from a certain value, it is simply a morphoom that we are considering the difference between our x values and that particular value. So, if we say deviations from the mean, that we are talking about the column that we might have of x minus x bar, that is, if we have 10 x values and we have computed their mean, then we can measure 10 other deviations and they will be denoted by x minus x bar. For example, if the x values are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10, then what will be their mean? 5.5. And when we are talking about deviations, then what are we saying? 1 minus 5.5, 2 minus 5.5, 3 minus 5.5 and so on. This is the first thing.  The power, a certain power to which the deviations are raised before averaging them. If we, I have just talked about these 10 deviations, if we take their first power and then sum them all up or divide them by the number of deviations, then this will be the first moment about the mean. As you now see on the screen, the first moment about the mean is given by sigma x minus x bar raised to 1 divided by n. We can say that the first moment about the mean is equal to sigma x minus x bar over n. In an earlier lecture, I said to you that the sum of the deviations of the observations from the mean is always equal to 0. And in this example which I have just discussed with you, you can see that this will be 1 minus 5.5 is minus 4.5 and the last value which is 10, 10 minus 5.5 is 4.5. Similarly, 2 minus 5.5 is minus 3.5, but 9 minus 5.5 is 3.5. So, your upper deviations are negative and your lower deviations are positive. And when you are adding, then your sum is 0. If sigma x minus x bar is equal to 0, then it is obvious that the first moment about the mean will also be 0. Students, in this first moment we denote it as m 1 and this is a small m. And this is of course, the situation when we are dealing with sample data. If we collect data from the entire population or compute these things, then our notation will be mu 1 for the first moment, mu 2 for the second and so on. Abhi me ne second moment ki baat ki students, second moment se meri kya muradhe? According to the definition that I gave earlier, we are talking about the power to which the deviations have to be raised before averaging them. So, if I talk about m 2, the second moment about the mean, it is given by sigma x minus x bar whole square over n. Students, ye formula jo abhi screen pe dekhaha ye to koi neya formula hi nahin ye. Do you not remember the variance when you squared the deviations of the x values from the mean, found their sum and divided by n? Yes indeed the second moment about the mean is exactly the same thing as the variance. And if you take the positive square root of this quantity, then of course, you obtain the standard deviation. Now that I have discussed the first and second moments, students, we are in a position to formally define the arath moment about the mean. The arath moment about the mean is the arithmetic mean of the arath power of the deviations of the observations from the mean. Symbolically, m r is equal to sigma x minus x bar whole raise to r over n. Ye sare moments jo main abhi tak aap ke sath discuss ke ye, these are moments about the mean and students, they are also called central moments. But then we can define moments in another way as well. Ye saroori to nahi hai ke aap mean he subtract kare. It can be any arbitrary value or it can be the number 0. So, let me define for you the moments about an arbitrary value alpha. As you now see on the screen, the arath moment about an arbitrary origin alpha is given by m dash r is equal to sigma x minus alpha whole raise to r divided by n. Students, aap ne note kia ke ab jo moment main define kia uske liye main notation is temal ki hai m dash r. Ye hi basic farkh hai between the moments about the mean and the moments about any other arbitrary value. We will denote the moments about the mean by m r and the moments about arbitrary value by m dash r. Aap se kushter pehle main aap ke sath discuss kia that the first moment about the mean is equal to 0. Now, let us consider what the first moment about alpha will be equal to. As you now see on the screen, the first moment about alpha is equal to sigma x minus alpha over n and if we open the bracket, we obtain sigma x over n minus alpha which is equal to x bar minus alpha. So, this is a simple relationship between the first moment about alpha and the arithmetic mean x bar. Isitara, we can develop a number of relationships and a short while later, I will be discussing with you formally certain relationships that exist between moments about the mean and moments about a and relationships which are very useful in computing the moments that we require. Lekin iss se pehle main aap ke sath moments about arbitrary origin ka ek special case discuss karna chaatin. Students, if you put alpha equal to 0, you obtain what we call moments about 0 or moments about the origin and as you now see on the screen, the alpha rth moment about 0 will be equal to sigma x raise to r over n. Obviously, if we put alpha equal to 0 in our basic expression for m dash r, the quantity x minus alpha will become simply x and the formula will become sigma x raise to r over n. So, in this, you note a point that when I was talking about moments about alpha, I said moments about an arbitrary origin and when I put alpha equal to 0, I said moments about the origin. Now, you know that if you visualize a graph paper, then the x equal to 0 will represent the origin. So, moments about 0 are also called moments about the origin. Let us apply the concept of moments to a simple example. As you now see on the screen, suppose we have data regarding the examination marks of just 9 students and the marks of the marks are 45, 32, 37 and so on. If we desire to compute the first 4 moments about the mean for this data, the first step of course, is to find the arithmetic mean and once we have done that, we will construct the columns of x minus x bar, x minus x bar whole square, x minus x bar whole cubed and x minus x bar whole raise to 4. Dividing the sums of these columns by the total number of observations, we will obtain m 1, m 2, m 3 and m 4. The important point to note is that the second moment is expressed in square units. The third moment is expressed in the cubes of the units and the fourth moment in the fourth powers of the units. So, that in this example, since we are dealing with the marks of the students, hence we say that m 2 is equal to 25.78 marks squared. Similarly, m 3 is equal to 20.67 marks cubed and so on. Students, you have noted that I computed the first 4 moments or what about the fifth, sixth and so on. Actually, if we go like this, there will be no end to it because obviously, r can be given any value, but a very interesting and important point to note is that the first 4 moments about the mean contain a lot of useful information regarding our data set. I would like to discuss with you the formula of the moments in the case of grouped data. Of course, they were for the case of the raw data and if we have grouped data, then that is, a frequency distribution students, the formula will be very similar to the case of the raw data with the only difference being that the frequency f is inserted in the formula as you saw in earlier cases. So, as you now see on the screen, the r th moment about the mean is given by sigma f into x minus x bar whole raise to r over n and the r th moment about alpha is given by sigma f into x minus alpha whole raise to r over n. Students, in the case of the frequency distribution of a continuous variable, you will recall that x represents the midpoints of the classes. And as I explained in an earlier lecture, this assumption that all the values lying in a particular class are equal to the midpoint of the class, this introduces a certain amount of error and that is called grouping error. May be you remember that I told you that in the case of the arithmetic mean, this error is not significant and it can be ignored. But students, in the case of the variance and the higher moments, this error becomes quite important and it is advisable to apply some sort of a correction for this error. W F Shepard has introduced certain corrections which are known as Shepard's corrections for grouping error. And as you now see on the screen, the Shepard's corrections are given by the formula a. M 2 corrected is equal to M 2 uncorrected minus h square over 12. M 3 corrected is the same thing as M 3 uncorrected. In other words, the third moment about the mean does not need any correction and M 4 corrected is equal to M 4 uncorrected minus h square over 2 into M 2 uncorrected plus 7 over 240 times h raised to 4. All these corrections are valid in the case of frequency distributions of continuous variables and these distributions should be of the form that they should tail off to 0 at each end. It is to be noted that in all these formulae h will represent the uniform class interval. Shepard's corrections ko ham abhi thori der ke baad ek example pe apply karenge. Usse pehle students, I would like to convey to you the mathematical relationships that exist between the moments about the mean and the moments about alpha. As you now see on the screen, M 1 is equal to 0 as stated earlier. M 2 is equal to M 2 dash minus M 1 dash whole square. M 3 is given by M 3 dash minus 3 times h times M 2 dash into M 1 dash plus twice M 1 dash whole cubed and M 4 is equal to M 4 dash minus M 3 dash into M 1 dash plus 6 times M 2 dash into M 1 dash square minus 3 times M 1 dash raised to 4. I will not be discussing the mathematical derivation of these relationships in this course, but of course, if any of you is interested, you are welcome to study the mathematics behind these formulae and they are very much in your own textbook. What I would like to discuss with you is that there is an easy way of remembering these relationships with which will be useful to you when you would like to compute the moments for real life data sets. As you now see on the screen, the two points to remember are number 1 in each of these relations, the sum of the coefficients of the various terms on the right hand side equals 0 and number 2 each term on the right is of the same dimension as the term on the left. Alright, let us now apply all these concepts to an example. Suppose that we have data regarding the marks obtained by students in a test, the total marks were 20 and the students are obtaining marks between 5 and 15, the number of students obtaining these marks are 1, 2, 5, 10 and so on. Suppose that we are interested in computing the first four moments about the mean for this data. If we wish to do so directly, the first step is to compute x bar and for this data set students x bar comes out to be 10.06. Now the point is that x bar equal to 10.06, this is not a very convenient number to work with because in all those columns that you have to construct x minus x bar, x minus x bar square cubed and force power, they will involve a lot of decimals. So, what we do is that first we compute the moments about alpha, where alpha is a convenient number. To work with and then we utilize those relationships that I can wait to you to compute the moments about the mean. So, in this example we consider alpha equal to 10, which is the x value against the maximum frequency 51 and using this number we construct the column of capital D, which is the same thing as x minus 10.   Dividing the sums of these columns by 131, the total number of students we obtain m 1 dash, m 2 dash, m 3 dash and m 4 dash as you see on the screen. Applying the relationships that exist between moments about the mean and moments about alpha, we obtain m 2 equal to 2.64, m 3 equal to 0.08 and m 4 equal to 28.30. Applying the shepherds corrections for grouping error, m 2 corrected comes out to be 2.56, m 3 corrected is the same as m 3 uncorrected and m 4 corrected comes out to be 27.01. Students we have been talking about moments for the last 10-15 minutes and you must be wondering why are we going through all these lengthy calculations. The reason is that as I said earlier moments give you a lot of information about your distribution and in this connection I will now introduce the concept of moment ratios and once we are through that you will be able to understand clearly what the important role of moments is in describing a frequency distribution. Ye jum nai moment ratios ki baat ki iss se murad wo quantities hain jin mein both the numerator and the denominator consist of moments and the most important moment ratios are b 1 and b 2 and as you now see on the screen, b 1 is given by m 3 square over m 2 cubed and b 2 is equal to m 4 over m 2 square both of these quantities are pure numbers and b 1 is used to measure the skewness of the distribution whereas b 2 is used to measure the kurtosis or the peakedness of the distribution. Aayi pehle b 1 pe concentrate karte hain. Students it has been mathematically shown that for a symmetrical distribution b 1 will be equal to 0 but for skewed distributions b 1 will be positive. So in any data set if your b 1 comes out to be 0 or approximately 0 you can conclude that your distribution is approximately symmetric. This means that the distribution is skewed. Now as far as the direction of skewness is concerned students you should look at the sign of the third moment about the mean that is m 3. If m 3 is a positive number it means that your distribution is positively skewed. If m 3 is negative it implies that the distribution is negatively skewed. b 1 cho hain that is m 3 square over m 2 cubed or squaring ki wajase b 1 to bahar hal positive hi aega. Next, let us talk about b 2. It has been mathematically proved that for the normal distribution which is neither very peaked nor very flat b 2 is equal to 3 and students therefore this number 3 acts as the basis for us to decide whether our distribution is leptocortic, mesocortic or platicortic. If for any data set your b 2 comes out to be equal to 3 you can say that your distribution is like the normal curve, the bell shaped curve which is neither very peaked nor flat. But if your b 2 comes out to be greater than 3 this implies that your distribution is leptocortic that is the one which is more peaked than the normal distribution. And if your b 2 comes out to be less than 3 then your distribution is flat that is platicortic. Students b 1 or b 2 ki discussion se aap pe ye waje ho jaan ha chahiye that the third moment and the fourth moment play a very important role in indicating the skewness and the kurtosis of your distribution. After all m 3 occurs in the numerator of b 1 and m 4 is in the numerator of b 2. What about the dispersion and the center of your distribution? Do you not remember that a short while ago I indicated to you that the second moment about the mean is none other than the variance and the positive square root of the variance is the standard deviation the most important measure of dispersion. What about the center? You will be interested to realize that the first moment about zero is none other than the arithmetic mean. After all how will you define the first moment about zero? Sigma x minus 0 over n which is exactly the same as sigma x over n minus the arithmetic mean. So, in this manner students moments indeed play a very very important role in describing frequency distributions. Next time I will discuss with you quite a different and a very interesting concept the concept of correlation and regression. In the meantime I would like to encourage you to practice all these formulae that you have learnt today and to make yourself at home with the concept of moments. Best of luck and Allah Hafiz.


The Interpol Company of Wilmington, Delaware, makers of better faith or better living through chemistry, presents the Cavalcade of America. Tonight's star, Nina Foch, as Captain Johnita Bonham, flight nurse, United States Air Force. Tonight's story, The Nurse Who Forgot Fear.  The sun is where it always was. The world still turns in the same old way. You don't feel any different, but they say you're a hero. You, you're just a girl in Montgomery, Alabama. But that's your picture in that magazine. And when the phone rings, it's for you. Bonham? Yes? Miss Bonham, this is Cy Harris. I'm calling from New York City for Cavalcade of America. We heard about an experience you went through off of it. I wonder if I could ask you a few questions. Well, sure, but I don't know if I could... Well, tell me, you're a nurse, right? A flight nurse. And how does this differ from a regular nurse, Miss Bonham? Well, a flight nurse?  We'd get up every morning at three and then... Yeah, between two and three. Vera and I used to room together. She was the other nurse. We worked together and lived together. We'd set her alarm for two and mine for two-thirty. That's the way we'd be sure and get up. When the second clock went off... Was that the first or the second? Bonnie... I think it was the second. Oh, that's nice. Well, what do you think? Oh, just know how to ask. Come on, Op. How to sleep your life away. That's an interesting idea. Well, you don't have to sleep in the air for it. She can stimulate. Yeah, yeah. The truth is, I wouldn't mind just getting up. It's only the prospect of having to climb to those long flannels that kills me. I think I'm going to get some of those black lace jobs and wear them under my long flannel. Just to remind me, I'm still a woman. Okay, woman. You fill the canteens and I'll call operations for the trucks. First, we had about three cups of black coffee. Then a causal plane took us to Korea. In Korea, things were almost the same. It was... It wasn't. Then there was the rain and the mud and the rain and the mud. You're a good voice today. Thank you. The blues I can't lose when it rains. Bonnie, look, I finished another song. You know, I might get to pay for a class meter if I stay in the service long enough. Who do you know with a foot shape like that? Two heels. You're looking at it upside down. It looks more like a sweater to me. Well, that's an idea. I'll cut two holes in it and make... I'll get it. Hello? Dog pass 47. Go ahead. Dog pass 40. Okay, target 9-1. We'll wait. Double-header again today, huh? Yeah. We'll be lucky if we hit the sack by midnight. I get the blues. Hey, what happened to your voice all of a sudden? You, uh... Henry? Not much. I'll trade you my candy bar for that can of beans. It's a deal. I sure do pity those people back in the state. Imagine having to sit down to a hot meal. Horrible thoughts. Oh, well, how do you say... say, like, in French? That's about the way it was. Yeah, I see. Just working and waiting and mud and rain. Too much of everything in that country.  We just worked along from day to day doing our jobs. It's about all. The office anything like that? No excitement? Well... And it was shooting all around? Is that what you mean? And sometimes on the flight when the weather would get bumpy, the patients would... No, no, no, no, no. Romance. Deep someplace, the Japanese moon is hanging over you like a paper lantern. Do you get the picture? Romance. He's off to the front. Once in a while. Sometimes a fellow I knew would manage to get a flight down from Tokyo where he was stationed. He'd be at the air base waiting for my plane to get in. He sounds busy. I'd land the engine. We'll land it. Just let me out of here. Hey, now, wait a minute, soldier.  You'll pull out those pictures. Take it easy. Need any help, Bonnie? No, they just opened the door. Right. And we'll get you fellas unloaded in a minute. Hey, do you leave us here, nurse? That's right. That's the end of the line for us. Lieutenant, who do I have to see to get you for my private nurse? Well, you're just right to your congressman. All right, here we go, men, and cheer up! You'll be back in the station less than a week.   Oh, well, am I dehydrated? Just lead me to that hot shower. Did you say hot shower? I can drink. Look, there's Tom. Bonnie, what happened? I thought you were due back this afternoon. We had to make two trips, Tom. Hi, boy. Don't I get it alone? I hear a hard bubble head. Well, that's a matter of opinion. Tom, when do you have to leave? Well, let's see. It's 11 now. I've been there. Yeah, if you want me, which I know you won't, I'll be under the first shower.  Oh, don't look now, but somebody's calling you. Oh, well, it was good seeing you, Bonnie. Yeah, Tom. Listen, you look really beat. Just hope I can stay awake long enough to make it to my sack. Yeah. Well, get some rest, Bonnie. Sure, Tom. You too.  Just sitting around waiting. I'll get him a special note from the chaplain. Any hot water? Ice cold. Very wet, though. There's a purpose. I don't think I'll ever get this mud out of my pores. But what I do is rub it in, so it doesn't show. You ever wonder what it's for? What it means? Everything? Us. Here. The fighting? Yeah, everything. Who's got the strength to wonder? Those guys. I haven't seen a woman in months. You just don't want to look at her. Makes you feel kind of special. You're out of clean, Randy. That kid today. He was so bashful. I thought he'd never get around to asking if he could take our picture. I didn't think it was possible in this day and age for men to blush. How does this sack feel good? We could be home right now, sleeping in a real bed. Instead of a miserable car. I know. I was all set to go home when the fighting started. 30 months in this mud hole. When I was discharged in 46, I figured I'd never want to see another plane. He'll tell me a remission. I don't know. Everything is so dull. I missed all this, I guess. I'm a woke character. What can you do? I know so much. Yeah? Now the plane went down today. I'm going to take home. I know I've been thinking about it. Bonnie, I've got a feeling. It's crazy. It's the kind of feeling I know I shouldn't let myself get. But I don't think I'll ever get home, Bonnie. Oh. You don't think I'll ever get home? No, I mean it. I'm just... Just so sure of it. It's going to happen. And it's going to happen quickly. And it's going to happen here. Thank you, Sophia Rose. Yeah, you know what I got the guys for? A chocolate mallet. Oh, boy. I can almost taste it. They used to make them great back home in my hometown. Old Doc Grissom. He used to fill up a full ice cream right to the top. Huh? A chocolate mallet. I don't care when you get it. You just get it. And I want it to be filled with ice cream. Sick.  And I want you to drink a toast to me. You know what I think? You've been reading too many war novels. You know what I think? What? You're right. This is Captain Jolita Bollum. Living in Montgomery, Alabama. Home from air evacuation duty in Korea. And one day the phone rang. It's Cy Harris of the radio program, Cable Cate of America in New York. Do you want to do your story? I need something exciting. I told Mr. Harris about the murder. But Ms. Pod... I went to high school there. I made the commencement speech on a football field. Oh, you were first in your class. No. You could be heard the length of the field. I wish I had something to tell him. All I did was my job. You train, you train for all the emergency. And then if one comes up, they make a big production out of it.  That's his job. I remember the day it all happened. It was raining. As usual. And Zira and I were at base operations trying to get a plane out of Korea. Yeah, I got a plane leaving in about 10 minutes. You're going to K-2, right? Isn't that what they said there? K-2, the biggest mud hole in Korea. So who's pilot? Oh, good. He's standing right over there with a door. Why don't you ask him? That's the kind of an answer I like. It's practical. Come on. Let's tear ourselves away from the happy jacket. Come on, Florence. Hey, Bill. Bill, you're going to have company. Good deal. Sure, why not? I like the ride. Good to cargo. Replacements again? Great. We're off on the Sardine Express.  The engine. I've never been up in a plane before. Well, it's just like riding a bus only loaded. Oh, Bill, are you going up forward? Might as well. I think I'll take me a nap. Are you coming? I'll be up in a few minutes. It's only exhaust. You said it. Just relax now. Take it easy.  Everybody make sure your seat belt is fastened. Oh, and I felt the familiar pull of speed as the big plane raced down the runway. And within a matter of seconds we were airborne. Then I went forward. I guess we might be out of over the bay by now. That's right.  Yeah. How long have you been on it? Three months. I figure I've made, well, somewhere between 150 and 200 round trips between Korea and Japan so far. Just to make you a commuter. Good night. Get to sleep. Oh, that's like a night. She needs it. She's been averaging about three hours a night. That's really rugged.  OK. Maybe I will. If we run into anything interesting, I'll let you know. OK. They tell me the crash broke the plane in two. Things were floating, or hands in the water. Some men were swimming. Some just floated. Some just floated. Give me your hand. Pull this man in. You first. Go ahead now. Now you're going. I thought I helped these men. Here's a rope. Grab it. Hey, nurse, come on. Get in the rope. That water's ice cold. If you don't drown, you'll freeze to death. I'm all right. Is anybody seen here? Captain Brown? No, ma'am. At least he's not this bad. Hey, nurse, are you OK? Grab her. She's passing her out. I got her. Split her head. Split wide open. How about that? What a girl. Didn't last long. Anyway, as the night wore on, I began to fear our little party, but it just never end. I've been so sure someone on shore would learn what had happened. But they didn't come. Water poured into our rafts and made the sign to give the first aid tougher than ever. A minute and an hour. I'm going to swim to shore. I'm not going to wait here until the waves turn us over. Yeah, we'll on prime.  You'll catch up. I'm going to swim to shore. Yeah, that's better. You just sit down there. If we keep under control, we'll be OK. It might take a few hours, but they'll find us. Now, here's what we're going to do. You, over there. You keep watching the way you're looking now. You, Slim, you watch on this side. You two keep bailing out the water. Here, you just tell me. Is everybody clear what their job is? Can we try it if we don't? No, it'll be going soon. Just a few more minutes. We're not going to lose hope. I know we'll be risked. Watch out! I see it! Fishing boat, he's coming along to throw us a line. Now, what's wrong with you? Well, I just think my wrist is broken. I think so, too. In about ten places. You've also got a fractured skull, a split cheekbone, a broken shoulder, and six broken ribs. Now, how do you feel? Terrible, Doctor. Where am I? You're in the first aid station, Bonnie. Doctor. Is... Did they find beer? Just nothing, Bonnie. No, no, thanks. I don't think so. Bonnie... You mustn't brood about it. Don't forget, you saved 27 lives. That's a pretty good score. I can't forget the other. 26. I can't. Yes, Doc. There is something you can get for me. Sure, what is it? A chocolate mold. Oh, I think that can be arranged. With a lot of ice cream. Filled with ice cream.  I want it. Propose a toast. That's about all. That's the story. Did you think of something special? What made you do it, Miss Bonham? Well, I don't know. I just did it. There was nothing else I could do. There was nothing else I could do. Those seven words sum up the career and character of Captain Johnita R. Bonham. Plightners, United States Air Force. But there are other words about the story. I don't know. I just did it. There was nothing else I could do. But there are other words about her, too. Words that go with the decorations she received. Distinguished Flying Cross. Citation reads, for meritorious service in medical air evacuation flights and for carrying out her missions willingly and without complaint, Lieutenant Bonham's unselfish devotion to duty has reflected great credit upon herself, her profession and the United States Air Force. One week after the Battle of Fort Sumter, he can only be one thing. A spy. Listen next week when the DuPont Cavalcade presents Fly High, Fly Low, the story of a startling event that was to influence the course of military tactics and strategy from that day to this. Our star, Lee Bowman. The nurse who forgot fear was written by Arnold Shulman and Irving Ellman, and based on an article from every woman's magazine as condensed in the readers' digest. Original music was composed by Arden Cornwell, conducted by Donald Borey. The program was directed by John Zoller. Nina Fosh may soon be seen co-starring with Glenn Ford in the MGM picture, Young Man with Ideas. With Nina Fosh tonight, our Cavalcade cast feature Judith Parrish as Vera John Newland Esville. We wish to thank Captain Raymond Hausman of the United States Air Force for his technical assistance in tonight's Cavalcade. This is Sy Harris speaking. Don't forget next week, our star, Lee Bowman. The DuPont Cavalcade of America comes to you from the Velasco Theater in New York and is sponsored by the DuPont Company of Wilmington, Delaware. Makers of better things for better living through chemistry. Next, it's Barry Craig, confidential investigator on NBC.


As the world's population grows and countries develop, so do the challenges billions of people around the world face on a daily basis. A wide range of nuclear techniques can be used to tackle many of these issues. The IAEA helps countries to use these versatile tools. But as the demand for the agency's services increases, so does the need for more resources. The Peaceful Uses Initiative, or PUI, is an effective way to supplement the IAEA's funding mechanism and generate additional resources when they're urgently needed. Since it was launched in 2010, the initiative has received over 100 million euros in contributions and supported more than 200 projects in over 150 countries. These funds help farmers breed healthier cattle by controlling the pests that make them sick. They finance a network of laboratories in over 40 countries that use nuclear techniques to rapidly diagnose livestock diseases and support research that makes our oceans cleaner and helps countries to find safe and sufficient water supplies. Thanks to the PUI, disused radioactive sources are now stored safely. Patients are receiving timely and appropriate treatment for cancer. The initiative supports IAEA activities that assist countries planning to develop nuclear power programs to meet their energy needs. When crisis hits, the PUI can mobilize funds for fast and emergency action. The initiative is investing in the future of nuclear technologies by contributing to the renewal of the IAEA's nuclear applications laboratories and funds projects that play a key role in achieving many of the sustainable development goals. The IAEA will continue to use the Peaceful Uses Initiative to promote atoms for peace and development. The PUI, taking the initiative for peaceful uses.