R1.9. Mercury in tuna A histogram and some computer output provide information about the mercury concentration in the sampled cans (in parts per million, ppm).

(a) Interpret the standard deviation in context.

(b) Determine whether there are any outliers.

(c) Describe the shape, center, and spread of the distribution.

A statistical graph or chart is a visual representation of statistical data in graphical form.

According to the given table, the standard deviation of the data is $0.300$, implying that the mercury per can values will vary by roughly$0.300$ from the mean. As a result, according to the provided table, the standard deviation of the data is $0.300$, implying that the mercury per can values will vary by $0.300$ from the mean.

When looking at the problem's graph, it's evident that a lot of the values are on the right side of the graph, and also very far away. As a result, it is evident that the graph has outliers on the right side. The lower outlier can be calculated as below. $=Q_1-1.5(Q_3-Q_1)$

Now, put the values provided into the table. $$\begin{gathered} =0.071-1.5\times (0.380-0.071) \\ =0.071-1.5\times 0.309 \\ =-0.3925 \end{gathered}$$

Since, there is no points lies below $-0.3925$

As a result, there aren't any lower outliers. Now, as shown below, calculate the higher outlier. $=Q3+1.5(Q3-Q1)$

Now, fill in the blanks in the table with the values you've been given. $$\begin{gathered} =0.380+1.5\times (0.380-0.071) \\ =0.380+1.5\times 0.309 \\ =0.8435 \end{gathered}$$

Because there are so many points above $0.8435$, there are a lot of upper outliers. Therefore, there are many upper outliers.

The shape of the graph is correctly skewed, as evidenced by the supplied graph. Because there are outliers on the graph, the median will be the best centre of the data, and the median value is $0.180$, according to the table. Similarly, in the case of outliers, the graph's spread can be calculated using the inter quartile range (IQR). The spread varies between $0.012ppm$ and $1.500ppm$