6. Consider the following observations on shear strength of a joint bonded 30.0

Question

6. Consider the following observations on shear strength of a joint bonded 30.0 4.4 33.1 66.7 81. 22.2 4 in a particular manner: 40.4 16.4 73.7 36.6 109.9 a. Determine the value of the sample mean. b. Determine the value of the sample median. Why is it so different fro c. Calculate a trimmed mean by deleting the smallest and largest What is the corresponding trimming percentage? How does the value of this and median? observations. compare This question related to Statistical Methods. Please solve it and make it clear to understand. Thanks.

Explanation / Answer

6.

a. Sample mean is given by:

xm = [30.0+4.4+33.1+66.7+81.5+22.2+40.4+16.4+73.7+36.6+109.9]/11 = 46.809

b. Arranging the data in increasing order :

4.4 , 16.4 ,22.2 , 30.0 ,33.1 , 36.6 , 40.4 , 66.7 , 73.7 , 81.5 , 109.9

median is the middle number , here it is 36.6

mean is just average of the given data while median is the middle number of data arranged in increasing order.

c.

smallest observation :4.4

largest observation: 109.9

removing these observation , the trimmed mean can be given as :

[16.4+22.2+30.0+33.1+36.6+40.4+66.7+73.7+81.5]/9 = 44.511

we have removed one date each side , so

trimming percentage can be given as x*(11)= 1

trimmed percentage x = 9.09%

After trimming, the mean is reduced and the number of data is reduced from 11 to 9 . hence trimmed median would be 36.6

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