 case students, let me present to you an application of the Chebyshev's inequality. Let the random variable capital X have the pdf, small f of x is equal to 1 over 2 into square root of 3 where X itself lies between minus square root of 3 and plus square root of 3. Alright, now for this particular distribution, if you do the normal calculations and try to find the mean and the variance, you will find that for this particular one, mu is equal to 0 and sigma square, that is the variance, that is equal to 1. Alright, so we have the conditions by which we can apply the Chebyshev's inequality. Usme yehi kondition hoti hai ke mean or variance, finite quantities ho or yeh naho ke mean does not exist or variance does not exist. Alright, now suppose that we put k equal to 3 by 2, yani jo Chebyshev's inequality ke andar jo k exist karta hai, that can be any positive number, positive real number. So let us choose it as 3 by 2, yani one and a half. So now if I insert the values of mu, sigma and k in the left hand side of the Chebyshev inequality and try to compute that probability so that I get the exact probability for that interval, what will I have? I will have the following. The probability of the modulus, the absolute value in other words of x minus mu being greater than or equal to k sigma, that is the same as the probability of the absolute value of x minus 0, is leke mu is equal to 0, being greater than or equal to 3 by 2 into 1, yani ke ke ke jaga 3 by 2 or sigma ke jaga 1, alright, so what do we have? What is this equal to? It is the probability of the modulus of x being greater than or equal to 3 by 2. Now this is the same as the 1 minus, 1 minus the probability of the modulus of x being less than 3 by 2. Ab jab yeh humne iss form meh hum aadhe hai, to ab hum integrate kar sakte hai in order to find the exact probability. So what do we get? 1 minus ab yeh jo expression hai yeh integral ke form meh kya baneega? The integral from minus 3 by 2 to plus 3 by 2 of small f of x dx, aur small f of x kya tha, bo jo pdf hai f of x wo kya tha? That is 1 over 2 into square root of 3, ache yeh shayad aap confuse ho rehin ke yeh integral meh meh ne minus 3 by 2 se 3 by 2 kaise lilya. My dear students, QK iss ko meh shayad rekhega, yeh agar aap kya rahe na ke the probability of the modulus of x being less than 3 by 2, that actually means the probability of x lying between minus 3 by 2 and plus 3 by 2. You can work it out. It's very simple and it's not difficult to understand. Agar aap thoda sa concentrate kane, absolute value ke concept par. Okay, now that I have got my integral, obviously I will try to solve it. To wo poora expression jo hai wo kya baneega? 1 minus ab wo jo integral ke andar hai, 1 over 2 into square root of 3, wo constant to pehli hum bahar nikal nege. So wo bahar nikal yeh hum ne. Ab humare integral ke andar, we only have 1. So students, what is the integral of 1? You all know that it is x. Aur peh rham yeh jo upper limit or lower limit jo hai humare integral ke. We will apply it on that. So what do we get? We get that entire expression is now as follows. 1 minus 1 over 2 into square root of 3 multiplied by 3 by 2 minus minus 3 by 2. Upper limit minus the lower limit. To abis ko aap solve kalee jhe. And once you solve it, finally, we obtain 0.134 as the probability of modulus of x being greater than or equal to 3 by 2. Ya yeh kalee jhe ke jaha se hum ne start liya tha. OK, yeh jo probability aai hai, this is the probability of my random variable x assuming a value which is beyond one and a half standard deviation on the left side of the mean and one and a half standard deviation on the right side of the mean. Mager yeh jo answer aaya hai, this is the exact probability. Let us try to apply the Shebhi-Shev inequality in order to see ke uske hisaab se kya aana tha. OK, yeh jo jab hum beyond ki baat kartein to phir hum wo 1 minus 1 over k square nahi kare hote. Hum kare rahe hotein 1 over k square that the probability of that expression being greater than or equal to k sigma, this probability is less than or equal to 1 over k square. Ab yaha pe ke kya hai, k is 3 by 2, so 1 over 3 by 2 square, yeh nahi nahi jahi, alright. So what is that? 1 over 9 by 4 and what is that equal to? That is 4 by 9, but what is 4 by 9 in decimal form? It is 0.444, to aapne dekha ke there is no contradiction, yeh ek simple hum nahi jo ke ek simple PDF leh liya, humare liya aasan tha exact probability nikal nahi, aur wo nikal liye 0.134, aur shebe-shebe inequality yeh kare yeh ke wo probability jo bhi whatever it is, but it cannot be greater than 0.444, wo bilkul dikh hai, humaree jo aai hai, it is not greater than 0.444, it is considerably less. So ishi tara agar hum k ke ko 3 by 2 nahi rakhte kuch aur rakh lehte, aapne dekhte ke jo exact probability hai, that is actually lower than the upper bound that you will get for that particular value of k from the shebe-shebs inequality.