 Hello and welcome to the session. In this session we discussed the following question which says, a discrete random variable capital X takes values X only from the set n equal to 2579. The probability that capital X takes the value X is given by probability of the random variable capital X equal to X is equal to lambda into X plus to the whole, where this lambda is some constant. In the first part, determine the value of lambda. In the second part, find the mean and variance of capital X. Consider a random variable capital X takes the values X1, X2 and so on up to Xn and they have the probabilities P1, P2 and so on up to Pn. These are the corresponding probabilities of the random variable X taking the values X1, X2 and so on up to Xn. Then we have the mean denoted by X bar is equal to summation Pi Xi where this i goes from 1 to n and we have each of these probabilities is greater than equal to 0 and also summation Pi where i goes from 1 to n is equal to 1. Then next we have the variance of the random variable capital X denoted by sigma square is equal to summation Pi into Xi minus the mean which is X bar the whole square and i takes the values from 1 to n. Now we have sigma square is equal to summation Pi into Xi minus X bar the whole square and i takes the values from 1 to n. This could be further written as summation Pi into Xi square minus 2 Xi into X bar plus X bar square the whole and i takes the values from 1 to n. Further we can write this as summation Pi into Xi square minus 2 X bar into summation Pi into Xi i takes the value from 1 to n and here also i takes the values from 1 to n plus X bar square into summation Pi where i takes the values from 1 to n. So now we have that this is equal to summation Pi into Xi square i takes the values from 1 to n minus 2 X bar into now summation Pi into Xi where i takes the values from 1 to n is X bar that is mean itself plus X bar square into now summation Pi that is sum of the probabilities is 1. So this is equal to summation Pi into Xi square where i takes the values from 1 to n minus X bar square. This is another form of writing the variance. So this could also be written as summation Pi into Xi square where i takes the values from 1 to n minus X bar whole square. This is also the variance. This is the key idea that we use in this question. Let's now move on to the solution. In the question we have that the random variables capital X takes the values X from the set n of numbers 2579 and probability of the random variable X equal to X is equal to lambda into X plus 2 the whole where this lambda is some constant and the first part we have to find the value of lambda in the second part we have to find the mean and variance of the random variable X. So first of all we will form a table from the given data as we are given set n as the set containing the elements 2579 and X takes the values from the set only so X would have values 2579 and it's also given to us that probability of the random variable X equal to X is equal to lambda into X plus 2 the whole. So when X is equal to 2 probability would be lambda into 2 plus 2 that is 4 lambda when X is 5 so probability would be lambda into 5 plus 2 that is 7 lambda when X is 7 probability would be 9 lambda when X is 9 probability would be 11 lambda. Now let's find the value of lambda. Now we have obtained a table so now let's find the value of lambda. Now we know the summation of the probabilities of the random variable X taking different values is 1. So here we have summation probability of the random variable X equal to X is equal to 1. So this means probability of the random variable X equal to 2 plus probability of the random variable X equal to 5 plus probability of the random variable X equal to 7 plus probability of the random variable X equal to 9 is equal to 1. So, from here we have 4 lambda plus 7 lambda plus 9 lambda plus 11 lambda is equal to 1 that is 31 lambda is equal to 1 and from here we get the value of lambda as 1 upon 31. So, this is the answer for the first part of the question. First we have to find the mean, the mean is equal to summation p i into x i where i takes the values from 1 to n. Now in the next part we have the mean deloted by x bar or by mu is equal to summation x into probability of the random variable capital X equal to x and here x takes the values 2, 5, 7, 9. So, this is equal to when x is 2 we have 2 into probability of the random variable x at 2 which is 4 lambda. So, here we have 2 into 4 lambda plus 5 into probability of the random variable x at 5 is 7 lambda. So, here we have 5 into 7 lambda plus x is 7. So, 7 into probability of the random variable x at 7 which is 9 lambda. So, here we have 7 into 9 lambda plus x is 9. So, 9 into probability of the random variable x at 9 is 11 lambda. So, here we have 9 into 11 lambda. So, this is equal to 8 lambda plus 35 lambda plus 63 lambda plus 99 lambda and this is equal to 205 lambda. Now, the value of lambda is 1 upon 31. So, putting the value of lambda here we have 205 into 1 upon 31. So, this is equal to 205 upon 31 which is equal to 6.6. So, we have x bar that is the mean as 6.6. Now, you can find out the variance which is given by sigma square is equal to summation p i x i square where i goes from 1 to n minus x bar square. So, next we have variance denoted by sigma square is equal to summation x square into probability of the random variable x equal to x minus x bar square and here takes the values 257 and 9. Now, we will prepare this table in which we have the values of x taken as 279 and the corresponding probabilities are given as 4 lambda is the probability where x is 2 then we have 7 lambda is the probability where x is 5 then 9 lambda is the probability where x is 7 and 11 lambda is the probability where x is 9. Let us now find out x i square that is 2 square is 4 5 square is 25 7 square is 49 9 square is 81. Let's we find out p i into x i square. So, that becomes 16 lambda and here we have 175 lambda, 441 lambda and 891 lambda. Now, summation p i x i square is 1523 lambda. So, putting these values in this formula we have that sigma square which is the variance is equal to 1523 lambda minus x bar square that is 6.6 square. Now, as the value of lambda is 1 upon 31. So, here we have 1523 into 1 upon 31 minus 43.56 that is 6.6 whole square is 43.56 and now here we have sigma square is equal to 1523 upon 31 is 49.13 minus 43.56 and this gives us sigma square is equal to 5.57 that is we have the variance given by sigma square is 5.57 that we have got to mean as 6.6 and the variance as 5.57. So, this comes the easy session. Hope you have understood the solution of this question.