Question: Hull failures of oil tankers.Owing to several major ocean oil spills by tank vessels, Congress passed the Oil Pollution Act, which requires all tankers to be designed with thicker hulls. Further improvements in the structural design of a tank vessel have been proposed since then, each with the objective of reducing the likelihood of an oil spill and decreasing the amount of outflow in the event of a hull puncture. To aid in this development, Marine Technology (Jan. 1995) reported on the spillage amount (in thousands of metric tons) and cause of puncture for 50 major oil spills from tankers and carriers. [Note: Cause of puncture is classified as either collision (C), fire/explosion (FE), hull failure (HF), or grounding (G).] The data are saved in the accompanying file.

a.Use a graphical method to describe the cause of oil spillage for the 50 tankers. Does the graph suggest that any one cause is more likely to occur than any other? How is this information of value to the design engineers?

b.Find and interpret descriptive statistics for the 50 spillage amounts. Use this information to form an interval that can be used to predict the spillage amount of the next major oil spill.

The Pareto chart is useful in this scenario to describe the reason for oil spills for the 50 ships. If we merely examine the number of reasons (frequent), grounding is the most significant and has a higher incline to happen than the others.

However, in this scenario, the amount of leakage from each source is just as crucial as the variety of causes. If we add the similar in the chart, one may assume that the puncture caused by the fire is the most serious.

Designers and engineers may use this knowledge to determine the likelihood of reasons and avoid the most probable causes from occurring through corrective actions and protective measures. This assists the design engineers in meeting the goal of decreasing the possibility of an oil leak as well as minimizing the number of outflows in the case of a hull puncture.

Mean, $\overline{X} = \frac{\sum X}{n} = \frac{2991}{50}$=59820 metric tons

The arithmetic mean spillage quantity is the total of all quantities divided by the number of quantities participating. The mean is the most often used metric of central tendency because it indicates the distribution's balancing point as well as a center of mass.

Median = ${\left(\frac{n+1}{2}\right)}^{th}$item when the elements are ordered in ascending and descending size sequence.

= ${\left(\frac{50+1}{2}\right)}^{th}$item = 50.5^th item

= 39500Metric tons

The median is the data point in the centre, computed when all elements are grouped in ascending or descending of size. In other words, the median is the numerical level above and below which 50% of all spilled quantities fall.

$$\begin{gathered} \text{Mode = Most frequently occurring item} \\ \text{= 31000 Metric tons} \end{gathered}$$

The Mode is essentially the most common item in a distribution.

$$\begin{gathered} \text{Range = Highest observed value -- Lowest observed value} \\ \text{= 257}-\text{21} \\ \text{= 236 metric tones} \end{gathered}$$

Range, rather than an average, reflects the distribution of the quantity of spillage utilized among individuals.

$$\begin{gathered} Samplevariance=\sum {\left(X_i-\overline{X}\right)}^2 = \frac{1}{50-1}\times 139525.38 \\ =2847457Metrictons \end{gathered}$$

The variance number is described as the average squared deviation of the spilled quantities as well as may be used to contrast various distribution sizes. Variance indicates how much the quantities in the data deviate from the average spilled quantity.

$$\begin{gathered} \text{Sample standard deviation}=\sqrt{\frac{1}{n-1}\sum {\left(X_i-\overline{X}\right)}^2} \\ = \sqrt{2847457} \\ =53362\text{Metric tons} \end{gathered}$$

The standard deviation is the square root of the variance and measures how much the data deviates from the average value. A high standard deviation implies that the measured values are distant from the mean, whereas a low standard deviation shows that they are closely packed around the mean.

The variance and standard deviations are both measurements of the distribution's dispersion around the mean. The 95 percent confidence interval can be used to determine the quantity of spillage. One may use the normally distributed to generate CI because the sample is pretty large.

The confidence interval is given by $\left(\overline{X}-Z_{\alpha /2}\frac{S}{\sqrt{n}},\overline{X}+Z_{\alpha /2}\frac{S}{\sqrt{n}}\right)$

Thus, we may declare with 95% certainty that the population mean quantity of spillage is within (45029, 74611) metric tonnes.