Cans of soda vary slightly in weight. Given below are the measured weights of seven cans, in pounds. Find the mean and median of these weights. Which, if any, of these weights would be considered an outlien? What are the mean and median weights if the outlier is excluded? $$ \begin{array}{lllllll} 0.8161 & 0.8194 & 0.8165 & 0.8173 & 0.7901 & 0.8141 & 0.8121 \end{array} $$ Find the mean and median of the weights. The mean is 0.81223 '. (Round to five decimal places as needed.) The median is 0.8161 . (Type an integer or a decimal.) Which, if any, of these weights would be considered an outlier? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The outlier is $$ \square $$ 0.7901 (Type an integer or a decimal.) B. None of the weights would be considered an outlier. Find the mean with the outlier excluded. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The mean without the outlier is $$ \square $$ (Round to five decimal places as needed.) B. None of the weights would be considered an outlier.

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UpStudy AI · Accepted Answer

$$ \textbf{Step 1. Order the data:} $$

The given weights are:  
$$
0.8161,\; 0.8194,\; 0.8165,\; 0.8173,\; 0.7901,\; 0.8141,\; 0.8121.
$$

First, we sort them in increasing order:  
$$
0.7901,\; 0.8121,\; 0.8141,\; 0.8161,\; 0.8165,\; 0.8173,\; 0.8194.
$$

$$ \textbf{Step 2. Calculate the Mean:} $$

The mean is given by  
$$
\text{Mean} = \frac{\text{Sum of all weights}}{\text{Number of weights}}.
$$

Sum of the weights:  
$$
0.8161 + 0.8194 + 0.8165 + 0.8173 + 0.7901 + 0.8141 + 0.8121 = 5.6856.
$$

There are 7 weights, so the mean is:  
$$
\text{Mean} = \frac{5.6856}{7} \approx 0.81223.
$$

$$ \textbf{Step 3. Determine the Median:} $$

Since there are 7 observations, the median is the 4th value in the sorted list:  
$$
\text{Median} = 0.8161.
$$

$$ \textbf{Step 4. Identify any Outlier:} $$

We can look for an outlier using the 1.5 $$\times$$ Interquartile Range (IQR) rule.

$$\textbf{Find Q1 and Q3:}$$

- The lower half of the data (not including the median) is:  
  $$
  0.7901,\; 0.8121,\; 0.8141.
  $$
  The median of this set (i.e., $$Q1$$) is $$ 0.8121 $$.

- The upper half of the data (not including the median) is:  
  $$
  0.8165,\; 0.8173,\; 0.8194.
  $$
  The median of this set (i.e., $$Q3$$) is $$ 0.8173 $$.

Now,  
$$
\text{IQR} = Q3 - Q1 = 0.8173 - 0.8121 = 0.0052.
$$

$$\textbf{Compute lower and upper fences:}$$

- Lower fence:  
  $$
  Q1 - 1.5 \times \text{IQR} = 0.8121 - 1.5(0.0052) = 0.8121 - 0.0078 = 0.8043.
  $$

- Upper fence:  
  $$
  Q3 + 1.5 \times \text{IQR} = 0.8173 + 1.5(0.0052) = 0.8173 + 0.0078 = 0.8251.
  $$

Any value less than $$ 0.8043 $$ or greater than $$ 0.8251 $$ is considered an outlier.

Since $$ 0.7901 < 0.8043 $$, the weight $$ 0.7901 $$ is an outlier.

$$ \textbf{Choice: A. The outlier is } 0.7901. $$

$$ \textbf{Step 5. Calculate the Mean with the Outlier Excluded:} $$

Remove the outlier $$ 0.7901 $$. The remaining weights are:  
$$
0.8161,\; 0.8194,\; 0.8165,\; 0.8173,\; 0.8141,\; 0.8121.
$$

Sum of the remaining weights:  
$$
0.8161 + 0.8194 + 0.8165 + 0.8173 + 0.8141 + 0.8121 = 4.8955.
$$

There are 6 weights now, so the mean is:  
$$
\text{Mean without the outlier} = \frac{4.8955}{6} \approx 0.81592.
$$

$$ \textbf{Final Answers:} $$

- Mean of the weights: $$ 0.81223 $$
- Median of the weights: $$ 0.8161 $$
- Outlier: $$ 0.7901 $$
- Mean without the outlier: $$ 0.81592 $$
Mean of the weights: 0.81223 Median of the weights: 0.8161 Outlier: 0.7901 Mean without the outlier: 0.81592

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