91影视

The standard deviation is a statistic that estimates the distribution of information related to its mean and is estimated as the square root of the variance.

a. The standard deviation is given in the output.

$s=0.300$

Interpretation: The mercury concentration in the model cans will generally deviate by $0.300$from the mean mercury concentration.

b.

Given parameters

$$\begin{gathered} Q_1=0.071 \\ {Q}_3=0.380 \end{gathered}$$

The interquartile range is the third quartile decreased by the first quartile:

$IQR=Q_3-Q_1=0.380-0.071=0.309$

Therefore,

$$\begin{gathered} Q_3+1.5IQR=0.380+1.5(0.309)=0.8435 \\ {Q}_1-1.5IQR=0.071-1.5(0.309)=-0.3925 \end{gathered}$$

Since the maximum of 1.500 is higher than 0.8435, it can be known that the maximum is an outlier.

c.

Shape: Right-skewed, because the highest bar in the histogram is to the left and the tail of smaller bars is to the right.

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

Spread: The data take on values between $0.012\mathrm{ppm}$and $1.500\mathrm{ppm}$.