Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 1E from Chapter 6 from Devore's Probability and Statistics for Engineering and Science.
Problem 1E
Chapter:
Problem:
The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2. ... a. Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used. [Hint: ∑xi = 219.8.] b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50%, and state which estimator you used...
Step-by-Step Solution
Step 1
We are given with flexural strengths of concrete beams.
Step 2: (a)
Point estimate of the mean value
\[\begin{array}{l}\bar X = \dfrac{{\sum\limits_{i = 1}^n {{X_i}} }}{n}\\ = \dfrac{{\left[ \begin{array}{l}5.9 + 7.2 + 7.3 + 6.3 + 8.1 + 6.8 + 7.0 + \\7.6 + 6.8 + 6.5 + 7.0 + 6.3 + 7.9 + 9.0 + \\8.2 + 8.7 + 7.8 + 9.7 + 7.4 + 7.7 + 9.7 + \\7.8 + 7.7 + 11.6 + 11.3 + 11.8 + 10.7\end{array} \right]}}{{27}}\\ = \dfrac{{219.8}}{{27}}\\ = 8.14\end{array}\]Therefore, \[\bar X = 8.14\]
Step 3: (b)
Point estimate of the strength that separates the weakest 50\% of the values if given by the medium of the data. Arrange the data and find the value that liest at center location (i.e. 14)
https://imgur.com/2CMjrn8
Therefore, \[m = 7.70\]
Step 4: (c)
Population Standard Deviation. Consider the data and find the sum of x values and their squares
https://imgur.com/xaax22J
\[\begin{array}{l}s = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {x_i^2 - \dfrac{{{{\left( {\Sigma {x_i}} \right)}^2}}}{n}} }}{{n - 1}}} \\ = \sqrt {\dfrac{{1860.94 - \dfrac{{{{\left( {219.80} \right)}^2}}}{{27}}}}{{27 - 1}}} \\ = \sqrt {\dfrac{{1860.94 - 1789.35}}{{26}}} \\ = \sqrt {2.75} \\ = 1.66\end{array}\]Therefore, \[s = 1.66\]
Step 5: (d)
The point estimate of the proportion of beams whose flexural strength exceed 10 is given by the ratio of number of beams with flexural strength more than 10 and total number of beams. Consider the arranged data, we can see that there are only four such beams
https://imgur.com/xaax22J
\[\begin{array}{l}p = \dfrac{{{n_{x > 10}}}}{{27}}\\ = \dfrac{4}{{27}}\\ = 0.148\end{array}\]Therefore, \[p = 0.148\]
Step 6: (e)
The point estimate of the population coefficient of variation.
\[\begin{array}{l}\dfrac{\sigma }{\mu } = \dfrac{S}{{\bar x}}\\ = \dfrac{{1.66}}{{8.14}}\\ = 0.204\end{array}\]Therefore, \[\dfrac{\sigma }{\mu } = 0.204\]
We are given with flexural strengths of concrete beams.
Step 2: (a)
Point estimate of the mean value
\[\begin{array}{l}\bar X = \dfrac{{\sum\limits_{i = 1}^n {{X_i}} }}{n}\\ = \dfrac{{\left[ \begin{array}{l}5.9 + 7.2 + 7.3 + 6.3 + 8.1 + 6.8 + 7.0 + \\7.6 + 6.8 + 6.5 + 7.0 + 6.3 + 7.9 + 9.0 + \\8.2 + 8.7 + 7.8 + 9.7 + 7.4 + 7.7 + 9.7 + \\7.8 + 7.7 + 11.6 + 11.3 + 11.8 + 10.7\end{array} \right]}}{{27}}\\ = \dfrac{{219.8}}{{27}}\\ = 8.14\end{array}\]Therefore, \[\bar X = 8.14\]
Step 3: (b)
Point estimate of the strength that separates the weakest 50\% of the values if given by the medium of the data. Arrange the data and find the value that liest at center location (i.e. 14)
https://imgur.com/2CMjrn8
Therefore, \[m = 7.70\]
Step 4: (c)
Population Standard Deviation. Consider the data and find the sum of x values and their squares
https://imgur.com/xaax22J
\[\begin{array}{l}s = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {x_i^2 - \dfrac{{{{\left( {\Sigma {x_i}} \right)}^2}}}{n}} }}{{n - 1}}} \\ = \sqrt {\dfrac{{1860.94 - \dfrac{{{{\left( {219.80} \right)}^2}}}{{27}}}}{{27 - 1}}} \\ = \sqrt {\dfrac{{1860.94 - 1789.35}}{{26}}} \\ = \sqrt {2.75} \\ = 1.66\end{array}\]Therefore, \[s = 1.66\]
Step 5: (d)
The point estimate of the proportion of beams whose flexural strength exceed 10 is given by the ratio of number of beams with flexural strength more than 10 and total number of beams. Consider the arranged data, we can see that there are only four such beams
https://imgur.com/xaax22J
\[\begin{array}{l}p = \dfrac{{{n_{x > 10}}}}{{27}}\\ = \dfrac{4}{{27}}\\ = 0.148\end{array}\]Therefore, \[p = 0.148\]
Step 6: (e)
The point estimate of the population coefficient of variation.
\[\begin{array}{l}\dfrac{\sigma }{\mu } = \dfrac{S}{{\bar x}}\\ = \dfrac{{1.66}}{{8.14}}\\ = 0.204\end{array}\]Therefore, \[\dfrac{\sigma }{\mu } = 0.204\]