The Truly Amazing Dudes are a group of comic acrobats. The weights (in

jamessinatraaa 2021-12-11 Answered
The Truly Amazing Dudes are a group of comic acrobats. The weights (in pounds) of the ten acrobats are as follows.
146 192 174 177 159 87 178 152 144 148
(a) Find the mean and sample standard deviation of the weights. (Round your answers to one decimal place.)
mean 155.7 lbs
standard deviation ? lbs
(b) What percent of the data lies within one standard deviation of the mean? %
(c) What percent of the data lies within two standard deviations of the mean? %

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Expert Answer

enhebrevz
Answered 2021-12-12 Author has 5005 answers
Given data:
146 192 174 177 159 87 178 152 144 148
Sol (a)
\[\begin{array}{|c|c|} \hline X&X-\overline{X}&(X-\overline{X})^{2}\\ \hline 146&-9.7000&94.0900\\ \hline 192&36.3000&1317.6900\\ \hline 174&18.3000&334.8900\\ \hline 177&21.3000&453.6900\\ \hline 159&3.3000&10.8900\\ \hline 87&-68.7000&4719.6900\\ \hline 178&22.3000&497.2900\\ \hline 152&-3.7000&13.6900\\ \hline 144&-11.7000&136.8900\\ \hline 148&-7.7000&59.2900\\ \hline \sum x=1557&&\sum (x-\overline{x})^{2}=638.1000\\ \hline \end{array}\]
sample Size \(\displaystyle={n}={10}\)
Mean \(\displaystyle=\overline{{{X}}}={\frac{{\sum{x}}}{{{n}}}}={155.7}\)
Std dev. \(\displaystyle={s}=\sqrt{{{\frac{{\sum{\left({x}-\overline{{{x}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}={29.1}\)
\[\begin{array}{|c|c|} \hline \text{mean}&155.7\ lbs\\ \hline \text{standard deviation}&29.1\ lbs\\ \hline \end{array}\]
Sol (b)
one standard deviation range: \(\displaystyle{\left({155.7}-{29.1},{155.7}+{29.1}\right)}={\left({126.6},{184.8}\right)}\)
Data values in this range: [146 174 177 159 178 152 144 148]: 8 values in this range NK Percentage \(\displaystyle={\frac{{{8}}}{{{10}}}}={0.8}\)
80% of the data lies within one standard deviation of the mean
Sol (C)
two standard deviation range: \(\displaystyle{\left({155.7}-{2}\cdot{29.1},{155.7}+{2}\cdot{29.1}\right)}={\left({97.4},{214}\right)}\)
Data values in this range: [146 174 177 159 178 152 144 148 192]: 9 values in this range
Percentage \(\displaystyle={\frac{{{9}}}{{{10}}}}={0.9}\)
90 % of the data lies within two standard deviation of the mean
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