You are interested in knowing about the weight distribution of the Green Bay Packers football team. The 2017 team roster gives the weights of each of the players. You can find all the weights (in lbs) below:

188, 188, 189, 192, 195, 195, 196, 197, 200, 200, 201, 202, 207, 208, 209, 211, 211, 212, 215, 216, 216, 217, 218 219, 220, 225, 226, 227, 232, 234, 237, 240, 245, 245, 245, 246, 252, 254, 255, 255, 255, 257, 261, 265, 285. 287 296, 298, 300, 301, 301, 305, 305, 310, 310, 313, 314, 314, 318, 318, 322, 324

a. Create a boxplot of the data above.
b. Looking at the boxplot (and not calculating), would you think the average would be to the right, left, or at the same point as the median? Why?
c. Which measure of center would you choose in this case to best represent the data?

a) Figure attached.

b) For this case as we can see the median seems to be less than the mean and we corroborate this with the summary values that we found on part a, we have that Median = 238.5 <246.8 = Mean, so then the median is to the left of the mean.

c) Since we don't have potentital outliers for this case the  data can be represented with the median or the mean, but the median is a more robust measure in order to analyze central tendency.

Step-by-step explanation:

For this case we have the following data:

188, 188, 189, 192, 195, 195, 196, 197, 200, 200, 201, 202, 207, 208, 209, 211, 211, 212, 215, 216, 216, 217, 218 219, 220, 225, 226, 227, 232, 234, 237, 240, 245, 245, 245, 246, 252, 254, 255, 255, 255, 257, 261, 265, 285. 287 296, 298, 300, 301, 301, 305, 305, 310, 310, 313, 314, 314, 318, 318, 322, 324

Part a

In order to create the boxplot we can use R with the following code:

> data<-c(188, 188, 189, 192, 195, 195, 196, 197, 200, 200, 201, 202, 207, 208, 209, 211, 211, 212, 215, 216, 216, 217, 218, 219, 220, 225, 226, 227, 232, 234, 237, 240, 245, 245, 245, 246, 252, 254, 255, 255, 255, 257,261, 265, 285, 287, 296, 298, 300, 301, 301, 305, 305, 310, 310, 313, 314, 314, 318, 318, 322, 324)

With this following line we have the summary for the data

> summary(data)

  Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  

 188.0   211.0   238.5   246.8   293.8   324.0

And as we can see the Minimum = 188, the Q1 = 211, the median Q2= 238.5, the Q3 - 293.8 and the maximum value is 324

With the following line we create the boxplot:

> boxplot(data, main="Boxplot")

And the result is on the figure attached.

Part b

For this case as we can see the median seems to be less than the mean and we corroborate this with the summary values that we found on part a, we have that Median = 238.5 <246.8 = Mean, so then the median is to the left of the mean/

Part d

Since we don't have potentital outliers for this case the  data can be represented with the median or the mean, but the median is a more robust measure in order to analyze central tendency.

your answer would be 10x - 1

step-by-step explanation: