 Hello friends, I am Mr. Sanjeev B. Naik working in the mechanical engineering department of the Walshchand Institute of Technology, Sulapur. In this video, I am explaining about the solution to a problem based on the tolerances used in engineering practices. So at the end of the session, learners will be able to use statistical techniques to solve the problem on tolerances. So there is an example where tensile strength of 1000 volts which are produced on certain manufacturing setup is found to be distributed with the mean of 620 Newton per mm square and the standard deviation of about 70 Newton per mm square. So that means whatever the process capability of that particular process where these volts are produced will have the variation of its strength with its mean value as 620 Newton per mm square as a characteristics of that normal distributed curve having standard deviation of 70 Newton per mm square. With this data, what is needed to know that how many number of volts will have the strength less than 540 Newton per mm square. So out of 1000, how many volts will have the strength less than 540 Newton per mm square and also it is expected to know the number of volts that will have strength between particular range 540 Newton to 820 Newton per mm square. So that this analysis will help to use those volts in particular applications by the designer or by the user. That is what expected from this analysis. So we just see that the distribution of manufacturing volts is between having mean of 620 Newton per mm square and standard deviation of 70 Newton per mm square. So pause the video for a while and just recall what is normal distribution curve. So we have seen it and you know it. So just recall it what is normal distribution curve. Now we refer it over here for this problem where it is characterized by plus 3 sigma standard division and minus 3 sigma standard division and which is used for application as manufactured volts. So this standard division is converted to normal deviate value of Z. So we have got two characteristics of this curve as a mu as 620 Newton per mm square as a mean value and standard deviation is 70 Newton per mm square. So we can establish this mu value where Z is 0. So Z is considered to be 0 at mean value 620 and then it increases from 0 to plus 3 and 0 to minus 3 like that. So what is required now is to calculate a number of volts having strength less than 540 Newton per mm square. So now means what we have to want to establish the value of Z with respect to this 540 Newton per mm square as the limit of strength. So first we have to locate this limit on this curve as 540 Newton per mm square for which we want to convert this into value of Z normal deviate. So from 0 how much is the deviation means what we want the variation in terms of deviation Z from 620 to 540 what is the deviation that is been obtained by getting Z at X 540 Newton per mm square for which we refer the equation Z is equal to X minus mu by sigma and that gives this line for which the value of Z obtained is minus 1.14. So this equation is used where X is 540 mu is 620 and sigma is point sorry 70. So if you put on this the value of Z is minus 1.14 that is been obtained. So from 0 to minus 1.14 this much is the deviation or the area of the curve under the curve. So anything below this means this area. So whatever number of observations or number of diameters of the that power number of cells the strength of that bolt varies from 540 to this side will be number of bolts less than 540. So that we can identify. So this is the area where we find the number of bolts having strength less than 540 Newton per mm square. So we want to calculate this area. So to calculate this area we consider the total area of this side like this. So area under the curve between what Z0 at the center to minus 1.14. So this area is obtained from standard table and that is obtained as 0.3729 from the standards that we can obtain from the standard table given by central limit theorem and that is available as a calculation of standard area by using standard normal distribution curve. So this we obtain 0.3729 as the value. So if we get this the shaded area under the curve we are going to obtain how Z is equal to minus 3 to minus 1.14. This we want to get the area Z is between minus 3 to minus 1.14. This area we want to calculate. So that is been considered to be 0.5 because this is symmetrical curve from Z0 to minus 3 it is 50 percent area and Z0 to plus 3 it is 50. So this is 0.5 area and out of that this area which is not shaded is 0.3729. So shaded area will be 0.1270 0.1271 that is 12.71 percent and that is why 12.71 percent volts here will have strength less than 540 Newton per mm square. So total number of volts having strength less than 540 Newton per mm square will be 12 percent 12.71 percent. So we know total population is 1000 volts produced multiplied by percentage. So that results into what is known as 127 volts. So answer to the first bet is that how many volts are less than 540 Newton per mm square strength that is 127 volts. So what is a second part of that solution required is how many volts are having strength between 540 Newton per mm square to 820 Newton per mm square. So 540 Newton to 820 Newton per mm square strength how many volts are there which are going to be manufactured. So for that we have to establish Z for this already we have established minus 1.14. Similarly Z value for this is calculated X minus mu by sigma but here X is upper limit that is 820 Newton per mm square minus that. And now just we have to use the same curve for the analysis in which we have got two limits one is 540 limit 540 Newton per mm square for which Z is minus 1.14 and another is Z at 286 that we have to plot. So first area under the curve which we get from 0 to this minus 1.14 minus 1.14 to 0 means area between this line and this line is from standard table once again 3729 that is 0.3729. Now similarly we establish second limit that is 820 Newton per mm square for which Z obtained is 2.86. So now we get area similar from 0 to this from standard table it is obtained 0.4979 and that is why total area between 540 to 820 that we have to obtain and that total area between minus 1.14 to 286 is this is this area where the strength will lie between 540 to 820. So if you know this area we know the percentage of bolts having this strength between 540 to 820 that we can calculate. So total area is 87.08 percent. So total area at which the bolts are having strength between 540 to 820 Newton per mm square as a strength is 87.08 mm as the total area and that is why the number of bolts having strength between 540 to 820 Newton per mm square is the total manufactured bolts 1000 multiplied by percent is accepted in that range 87 percent and that gives 870.8 bolts so about 871 bolts. So we conclude by this analysis by using the statistical curves normal distribution curves and referring the limits between 540 and 820 we can establish how many bolts are going to be there which are having this strength 871 bolts. So out of 1000 bolts 870 bolts are possible having this strength between 540 to 820. So this is the second part of that problem which you can analyze. So by simple statistical method of using this curve normal curve converting to standard normal curve is converting to applied normal curve with deviation z that is a normal deviation and then getting area from standard table we can analyze how many bolts are there between the range 871 bolts that completes the problem. This is my reference. Thank you.