Problem 1.2| Introduction to Statistics and Data Analysis| Probability & Statistics for Engineers & Scientists 8th Ed| Walpole, Myers, Myers, Ye

 

According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber is taken and the absorbency on each piece was measured. The following are the absorbency values:

18.71 21.41 20.72 21.81 19.29 22.43 20.17
23.71 19.44 20.50 18.92 20.33 23.00 22.85
19.25 21.77 22.11 19.77 18.04 21.12  

a. Calculate the sample mean and median for the above sample values.

b. Compute the 10% trimmed mean.

c. Do a dot plot of the absorbency data. 

SOLUTION:

Part a

We are given

n=20

\sum f=18.71+21.41+20.72+21.81+19.29+22.43+20.17+23.71+19.44+20.50+18.92+20.33+23.00+22.85+19.25+19.25+21.77+22.11+19.77+18.04+21.12

\sum f=415.35

The mean is 

\overline{x}=\frac{415.35}{20}

\overline{x}=20.7675

If we arrange the data from least to greatest, the 10th and 11th data are 20.5 and 20.72. The median therefore is

\widetilde{x}=\frac{20.5+20.72}{2}

\widetilde{x}=20.61

Part b

If we removed the 10% from the bottom and 10% from the top of the data arranged from least to greatest, the 4 values will be eliminated. The values eliminated are 18.04, 18.71, 23, and 23.71. Therefore, the sum of the remaining data is 

\sum f=331.89

n=16

The trimmed mean is 

\overline{x}_{tr10}=\frac{331.89}{16}

\overline{x}_{tr10}=20.743

Part c

A dotplot is shown below

dotplot 1.2c

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