Chapter 1, Problem 9CRE

(a)

To determine

To interpret: the standard deviation

Answer to Problem 9CRE

The standard deviation of the data is 0.300

Explanation of Solution

Given:

  

Calculation:

According to the provided table, it is clear that the standard deviation of the data is 0.300 which insures that the values of the mercury per can will vary about 0.300 from the mean

Conclusion:

Therefore, According to the provided table, it is clear that the standard deviation of the data is 0.300 which insures that the values of the mercury per can will vary about 0.300 from the mean

(b)

To determine

To identify: whether there are any outliers

Answer to Problem 9CRE

There are many upper outliers.

Explanation of Solution

Calculation:

On seeing the provided graph of the problem, it is clear that a lot of the values is lying on the right of the graph and also very far away. So, it is clear that the graph is having the outliers on the right side. The lower outlier can be calculated as below.$=Q_1-1.5\left(Q_3-Q_1\right)$

Now, put the values provided into the table.

  $\begin{array}{l}=0.071-1.5\times (0.380-0.071)\hfill \\ =0.071-1.5\times 0.309\hfill \\ =-0.3925\hfill \end{array}$

Since, there is no points lies below $-0.3925,$ hence there is no lower outliers. Now, calculate the upper outlier as below.$=Q_3+1.5\left(Q_3-Q_1\right)$

Now, put the values provided into the table.

  $=0.380+1.5\times (0.380-0.071)$

  $=0.380+1.5\times 0.309$

  $=0.8435$

Since, there is many points lies above 0.8435, hence there are many upper outliers.

Conclusion:

Therefore, there are many upper outliers.

(c)

To determine

To describe: the shape, center, and spread of the distribution.

Answer to Problem 9CRE

The shape of the graph is rightly skewed.

Center: the mean is 0.285ppm and the median is 0.180ppm

Spread: between 0.012ppm and 1.500ppm

Explanation of Solution

Calculation:

On seeing the provided graph it is clear that the shape of the graph is rightly skewed. Since, in the graph outliers exist hence the median will be the best center of the data and according to the provided table, the value of the median is 0.180. Similarly, in the case of outliers, the spread of the graph can be obtained by inter quartile range (IQR). The value of spread is between 0.012ppm and 1.500ppm