Problem 5 Easy Difficulty

The sum and sum of squares corresponding to length $x$ (in $\mathrm{cm}$ ) and weight $y$ (in $\mathrm{gm}$ ) of 50 plant products are given below:
$$
\sum_{i=1}^{50} x_{i}=212, \sum_{i=1}^{50} x_{i}^{2}=902.8, \sum_{i=1}^{50} y_{i}=261, \sum_{i=1}^{50} y_{i}^{2}=1457.6
$$
Which is more varying, the length or weight?

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Video Transcript

hell in this problem? The sum and sum of square corresponding to length X in centimeter and wait by Ingraham or 50 plant products are given below that is sigma X I Is equals to 212. Well, I goes from 1 to 50. Sigma X i square is equal to 902.8. And we are asked to find which is more wearing the length or weight. First of all, we will find mean variance and standard deviation of the length of a plant. So we have sigma excise equals to 212 and sigma X i square as equals to 902.8. From here we get mean is equal to 200 to upon 50, which is equal to 4.24 variants of length is equals two. When I've gone 50 into 902.8 minus 212 upon 50 whole square. By simplifying it, we get 18.056 17.978 It would be equals two, 0.078. Now, standard deviation of length is equal to underwrote 0.78 which is equals to 0.28 Now we have given Sigma YI is equal to 261. Well, mhm I goes from 1 to 50 and sigma phi I square is equal to 1457.6 where I goes from 1 to 50. Now, we will find the mean variance and standard deviation of the videos plant. So we have mean is equal to 261 upon 50, which is equals to 5.2. Now, variants of weight is equal to Run up on 50 into 1457.6 261 upon 50 whole square. By simplifying it, we get 29.152 My News 27.248 It would be equal to 1.904. Now, a standard deviation of weight would be equals two. Underwrote 1.904 which is equal to 1.38 Therefore, Stevie of length as equals two 0.28 upon 4.24 in 200, Which is equals to 6.6 and CVS will read is equals two, 1.385.22 in 200, Which is equals to 26.4. Since CV of weight is greater than Cv of length, does the weight has more variability than length?

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