The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is, \(\mu\)

\(\hat{\mu}=\bar{x}\)

\(=\frac{\sum x_{i}}{n}\)

\(=\frac{219.5}{27}\)

\(=8.12963\)

The estimator is \(\bar{X}\)

Part a

The point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion is 8.12963. The estimator is \(\bar{X}\)

Step 2

First arrange the data into ascending order.

\(\begin{array}{|c|c|}\hline A & B & C \\ \hline 5.3 & & 8.1 \\ \hline 6.3 & & 8.1 \\ \hline 6.3 & & 8.1 \\ \hline 6.5 & & 8.5 \\ \hline 6.8 & & 8.7 \\ \hline 6.8 & & 9 \\ \hline 7 & & 9.7 \\ \hline 7 & & 9.7 \\ \hline 7.2 & & 10.7 \\ \hline 7.3 & & 11.3 \\ \hline 7.3 & & 11.6 \\ \hline 7.4 & & 11.8 \\ \hline 7.7 & & \\ \hline 7.7 & & \\ \hline 7.8 & & \\ \hline \end{array}\)

Here, the number of observations is odd. So, the median is the middle value of the of the given data.

Hence, the middle value is 7.7.

The estimator is \(x\)

Part b

The point estimate of the strength value that separates the weakest \(50\%\) of all such beams from the strongest \(50\%\) MPa is 7.7. The estimator is \(x\)

Step 3

The point estimate of the population standard deviation is,

\(\hat{\sigma}=s=\sqrt{\frac{\sum x_{i}^{2}-(\sum x_{i})^{2}/n}{n-1}}\)

\(=\sqrt{\frac{1859.53-(219.5)^{2}/27}{27-1}}\)

\(=\sqrt{\frac{1859.53-1784.454}{26}}\)

\(s=1.699\)

The estimator is \(s\)

Part c

The point estimate of the population standard deviation is 1.699. The estimator is \(s\)

Step 4

Let \(x\) denote the number all such beams whose flexural strength exceeds 10 MPA.

From the given information number of beams whose flexural strength exceeds 10 MPA is 4.

The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is,

\(\hat{p}=\frac{x}{n}\)

\(=\frac{4}{27}\)

\(=0.148\)

Part d

The point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa is 0.148.

Step 5

The point estimate of the proportion coefficient of variation is,

\(\frac{\sigma}{\mu}=\frac{S}{\bar{x}}\times100\)

\(=\frac{1.699}{8.12963}\times100\)

\(=20.9\)

The estimator is,

\(\frac{\sigma}{\mu}=\frac{S}{\bar{x}}\)

Part e

The point estimate of the proportion coefficient of variation is 20.9. The estimator is \(\frac{s}{\bar{x}}\)

Coefficient variation is the ratio of the standard deviation and mean and multiply with 100.