Mean Deviation Percent Number of of Individual Sample K+ Observations Results from Mean 1 5.12 5 0.13, 0.09, 0.08, 0.06, 0.08 7.09 3 0.09, 0.08, 0.12 3 3.98 4 0.02, 0.17, 0.05, 0.12 4 4.73 0.12, 0.06, 0.05, 0.11 5.96 0.08, 0.06, 0.14, 0.10, 0.08 (a) Evaluate the standard deviation s for each sample. (b) Obtain a pooled estimate for s.

A First Course in Probability (10th Edition)
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Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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Mean
Deviation
Percent
Number of
of Individual
Sample K+
Observations Results from Mean
1
5.12
5
0.13, 0.09, 0.08, 0.06,
0.08
2
7.09
3
0.09, 0.08, 0.12
3
3.98
4
0.02, 0.17, 0.05, 0.12
4
4.73
4
0.12, 0.06, 0.05, 0.11
5
5.96
5
0.08, 0.06, 0.14, 0.10,
0.08
(a) Evaluate the standard deviation s for each
sample.
(b) Obtain a p0oled estimate for s.
Transcribed Image Text:Mean Deviation Percent Number of of Individual Sample K+ Observations Results from Mean 1 5.12 5 0.13, 0.09, 0.08, 0.06, 0.08 2 7.09 3 0.09, 0.08, 0.12 3 3.98 4 0.02, 0.17, 0.05, 0.12 4 4.73 4 0.12, 0.06, 0.05, 0.11 5 5.96 5 0.08, 0.06, 0.14, 0.10, 0.08 (a) Evaluate the standard deviation s for each sample. (b) Obtain a p0oled estimate for s.
Analysis of several plant-food preparations for
potassium ion yielded the following data:
Transcribed Image Text:Analysis of several plant-food preparations for potassium ion yielded the following data:
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