Below are sur mary statistics for the sizes lin acres) of some local vinayans. U

Question

Below are sur mary statistics for the sizes lin acres) of some local vinayans. Using the summary statisti s complete pats a through ¢ belo Var able N Mean SIDen Mnimu Mecian Q3 Maun Acres 36 48.00 46.34 18.12 32.50 57.31 250 a) From tha summary statistis, would you dascnbe this cistrbution as symmetric or skawad? Explain. Choosa tha correct answar helow. OA. This distrbulion is symmetric, because the meen and mecian are reasonsbly close to one another and the eatrene values are approximately the same distance from the median. ?B. This distrbutiar symmatric necausa the mean and mncian are rnasonably claseto one another and tha quarties ana appronmately the sama distance from ???" maan O C. This distrbution is skewed, because the mesn and mecian are not ressonably close to one another b) From the summary statistics, are there any outliers? Explain, Choose the correct answer below. OA Yes, becsuse the maximum value is far greater than the upper fence. B. Yas, bacause the minimum value in the disribtion is lss than the lwer fanca. ° C. No, because all the values in the distrbution are between the fences. ? D- Yas, because the minimum value in the distribution is lessthan the lower fenca and tha maximum value is greater than the uppar force. Create a baxplat of these data. What additional information would you need to comp ce the booxp ot? Using the ghan values, draw as much of the boxplot as possible. Choasa tha cornact plot balaw. O C. O D. Whal additional infomation wouid you need to complete the bexplor? Select all that apply [ A The smallast valua inside th? fena is needad in crder to plat harw tar the low whiskar should go. There is no additional information needed to complete the boxplet. B. C. Tha largast value ns da tha fanca is neodad in arder to piat haw far the high whiskar shauld ga D. The value of any outiers that are above the top fence ard below the maximum value are needed 80 lhat they can be added 1* the plot.

Explanation / Answer

a) C. This distribution is skewed because the mean and median are not reasonably close to one another.

b) IQR = Q3 - Q1 = 57.31 - 18.12 = 39.19

Hence,

Lower fence = Q1 - 1.5IQR = 18.12 - 1.5*39.19 = -40.665

Upper fence = Q3 + 1.5IQR = 57.31 + 1.5*39.19 = 116.095

Whatever data which are not between these two numbers are outliers. " A. yes, because the maximum value is far greater than the upper fence.

C) option B is correct as there is one dot for 250(outlier)

D) the value of any outliers that are above the top fence and blow the maximum value are needed so that they can be added to the plot.

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