 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age x of the selected student is recorded. What is the probability distribution of the random variable x? Find mean, variance and standard deviation of x. Now we know that the mean of x denoted by mu is the number sigma xi pi i varying from 1 to n where x takes possible values x1, x2 so on till xn with probabilities p1, p2 till so on pn. Now the mean of a random variable x is also called the expectation of x and this is denoted by Ex. So Ex is equal to mu which is again equal to sigma xi pi i varying from 1 to n. Now again the variance of x is given by Ex square minus Ex square. Now the non-negative number sigma x which is called the standard deviation of the random variable x is given by under root of variance of x. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. So here in this question we are given the ages of 15 students of a class. Now one student is selected in such a manner that each has the same chance of being chosen and the age x of the selected student is recorded. So the sample space x is equal to so x has 15 elements. Now we have to find the probability distribution of the random variable x. So here we observe that the random variable x can take some values 14, 15, 16, 17, 18, 19, 20 and 21. So the probability distribution x is so when x is 14 px is equal to 2 over 15 because there are two students whose ages are 14 years. Now when x is 15 probability of x is 1 over 15 because there is only one student whose age is 15 years. Again when x is 16 probability of x is 2 over 15 because from here we can see there are only two students whose ages are 16 years. Similarly when x is 17 px is 3 over 15. Again when x is 18 px is 1 over 15 similarly when x is 19 px is 2 over 15. Now when x is 20 px is 3 over 15 similarly when x is 21 px is 1 over 15. So this is the probability distribution of x. Now we have to find mean variance and standard deviation of x. Now according to our key idea mu is equal to ex which is equal to sigma xi pi i varying from 1 to n. So this is equal to 14 into 2 over 15 plus 15 into 1 over 15. Plus 16 into 2 over 15 plus 17 into 3 over 15 plus 18 into 1 over 15 plus 19 into 2 over 15 plus 20 into 3 over 15 plus 21 into 1 over 15 and this is equal to 28 over 15 plus 15 over 15 plus 32 over 15 plus 51 over 15 plus 18 over 15 plus 38 over 15 plus 60 over 15 plus 21 over 15. Now this is equal to 263 over 15 which is again equal to 17.533 So this is the mean of the variable x. Now for the variance we will first find out ex square and this is equal to 14 square into 2 over 15 plus 15 square into 1 over 15 plus 16 square into 2 over 15 plus 17 square into 3 over 15 plus 18 square into 1 over 15 plus 19 square into 2 over 15 plus 20 square into 3 over 15 plus 21 square into 1 over 15 and this is equal to 196 into 2 over 15 because 14 square is 196 plus 225 into 1 over 15 plus 256 into 2 over 15 plus 289 into 3 over 15 plus 324 into 1 over 15 plus 361 into 2 over 15 plus 400 into 3 over 15 plus 441 into 1 over 15 and this is equal to 392 over 15 plus 225 into 1 over 15 plus 512 over 15 plus 867 over 15 plus 324 over 15 plus 722 over 15 plus 441 over 15 and this is equal to 4683 over 15 this is again equal to 312.20 Now we will find ex square which is equal to 17.53333 square and this is again equal to 307.41766 Now according to our key idea we have variance of x is equal to ex square minus and this is equal to 312.20 minus 307.41766 and this is equal to 4.78 again equal to 4.78. Now again according to our key idea the standard deviation of the random variable X that is sigma x is equal to under root of variance of x so this is equal to under root of 4.78 which is equal to 2.19 So this is the answer for the upper question that is this is the probability distribution of X now 17.53 is the mean of X 4.78 is the variance of X and 2.19 is the standard deviation of X So this completes our session I hope the solution is clear to you my have a nice day.