 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says the sum and sum of squares corresponding to length x in centimeter and weight y in gram of 50 plant products are given below. Submission x i i going from 1 to 50 is 212 and so on we have others which is more varying the length or weight. So let us see the solution to this question. Now to tell which is more varying the length or weight we have to find out the coefficient or variation for each one of them. So first of all we calculate for length. Now for length we see that submission x i i goes from 1 to 50 is 212. Submission x i square i goes from 1 to 50 is 900 and 2.8. Therefore we say that mean that is x bar is equal to submission x i divided by n that is 212 divided by 50 that is equal to 4.24 centimeter. The deviation is equal to sigma that is equal to 1 by n into square root of n into submission x i square submission x i the whole square. We simply put in the values here and we get 1 by 50 into square root of 50 into 902.8 minus 212 the whole square. On calculating this we get sigma is equal to square root of 196 divided by 50 divided by 50 that is equal to 0.28. So standard deviation for length is 0.28. Now we see that will be equal to sigma divided by mean multiplied by 100 that is standard deviation divided by mean into 100 that is 0.28 divided by mean that is 4.24 into 100 and that comes out to be 6.6. Now we have to find out coefficient of variation for weight. Now submission y i where i goes from 1 to 15 is equal to where i goes from 1 to 15 is 61. Submission y i square where i goes from 1 to 15 is 7.6. Now these two things they are given to us in the question for weight we see that mean let this be y bar be equal to submission y i divided by n that is 261 divided by 50 that is equal to 5.22 gram. We find out SD that is standard deviation for weight that is sigma that is equal to 1 by n into square root of n into submission y i square minus submission y i the whole square that is equal to 1 by 50 into square root of 5.7.6 minus 261 the whole square. On calculating this we get square root of 4759 divided by 50 and that is approximately equal to 1.38. Therefore CV that is equal to 6.6. Now coefficient of variation will be equal to standard deviation divided by mean into 100 divided by 5.22 multiplied by 100 and that comes out to be 26.4. Therefore we see that coefficient of variation of weight is more than that of length that was 6.6. Therefore weight is more varying than length. So to this question is we shall enjoy the session. Have a good day.