Question

Mercury in tuna Here are a dotplot and numerical summaries of the data on mercury concentration in the sampled cans (in parts per million, ppm):

Part (a). Interpret the standard deviation.

Part (b). Determine whether there are any outliers.

Part (c). Explain why the mean is so much larger than the median of the distribution.

Short answer

Part (a) The mercury concentration in the 164 sampled tuna cans was then expected to vary by about 0.300 ppm on average from the mean of 0.285 ppm.

Part (b) Outliers are present.

Part (c) The mean is not resistant, whereas the median is.

Step-by-step solution

Part (a) Step 1. Given information

A dotplot and numerical summaries of the data on mercury concentration in the sampled cans (in parts per million, ppm) are provided below:

Part (a) Step 2. The standard deviation.

In the given numerical summary, the mean is mentioned directly beneath "Mean":

Mean =$\overline{x}$

$\overline{x}=0.285$

In the given numerical summary, the standard deviation is mentioned directly beneath "SD":

Standard deviation =$s$

$s=0.300$

The average deviation of the data values from the mean is defined as the stand deviation.

The mercury concentration in the 164 sampled tuna cans was then expected to vary by about 0.300 ppm on average from the mean of 0.285 ppm.

Part (b) Step 1. Whether there are any outliers.

Given in the numerical summary:

Min = minimum

Minimum = 0.012

$Q_1$= First quartile

First quartile = 0.071

$Q_3$= Third quartile

Third quartile = 0.380

Max= maximum

Maximum = 1.500

The IQR is the difference between the third and first quartiles:

IQR = $Q_3$- $Q_1$

IQR = 0.380 -0.071=0.309

Outliers are observations that are more than 1.5 times the IQR above $Q_3$or below $Q_1$

$Q_3$+1.51QR = 0.380+ 1.5(0.309)

0.380+ 1.5(0.309) = 0.8435

$Q_1$- 1.5IQR = 0.071-1.5(0.309)

0.071-1.5(0.309) = - 0.3925

We then note that there are outliers in the data set, because the maximum 1.500 is higher than 0.8435 and thus the maximum 1.500 is an outlier (while there could be other additional outtiers).

As a result:

Outliers are present

Part (c) Step 1 The mean is significantly greater than the distribution's median.

The distribution is strongly skewed to the right in the dotplot, because the majority of the dots are to the left, with a tail of more unusual dots to the right.

The mean is not resistant, whereas the median is. This means that the mean is more influenced by the distribution's unusually large values than the median, and thus the mean is greater than the median.

Result

The mean is not resistant, whereas the median is.