Chapter 1, Problem R1.9RE

a.

To determine

To interpret: The value of the standard deviation

Explanation of Solution

Given:

The dot plot is:

  

Numerical summaries are:

  

From the provided output, the standard deviation is 0.300. It shows that concentration of the mercury in the provided sample cans would differ by 0.300 from the mean concentration of 0.285.

b.

To determine

To find: Whether there any outlier exists.

Answer to Problem R1.9RE

Yes, outliers are present in the data set.

Explanation of Solution

Here,

  $Q_3=0.071,Q_1=0.380$

IQR can be calculated as:

  $\begin{array}{c}IQR=Q_3-Q_1\\ =0.380-0.071\\ =0.309\end{array}$

Now,

  $\begin{array}{c}{Q}_3+1.5\left(IQR\right)=0.380+1.5\left(0.309\right)\\ =0.8435\\ Q_1-1.5\left(IQR\right)=0.071-1.5\left(0.309\right)\\ =-0.3925\end{array}$

From the provided output, the maximum value is greater than the value 0.8435. Thus, it could be said that outlier is present in the data set.

c.

To determine

To explain: The reason of mean being greater than the median.

Explanation of Solution

From the provided dot plot, it is clear that the distribution is skewed to the right. For the right skewed the value of mean is always greater than the median. Thus, it could be said that mean is greater than median because the distribution is skewed to the right.