In order to find the median, we must first put the data in numerical order from least to greatest:

 $\displaystyle \small 9mi,10mi,11mi,12mi,12mi,12mi,12mi,13mi,14mi,14mi,14mi,14mi,15mi,16mi$

Next, we must find the number that is in the middle, but since there are fourteen numbers, there is no number that falls exactly in the center.  When this happens, we add the two middle numbers together and divide by two.  In this case, our middle numbers are $\displaystyle \small 12$ and $\displaystyle \small 13$.

$\displaystyle \small \frac{12+13}{2}=12.5$

Brandon's median distance is $\displaystyle 12.5\: miles$.

Each data set has five elements; the median is the element in the middle after the elements are arranged in ascending sequence, which both are. In each case, the median will be the third-highest element. Since in both data sets, this element is 100, the medians are equal.

The median of a data set with an even number of elements is the mean of the two elements in the middle when they are arranged in ascending order. The data set has six elements, so the median is the mean of the third-highest and third-lowest elements:

$\displaystyle M = \frac{8+16}{2} = \frac{24}{2} = 12$

It is impossible to tell which is greater.

For example, if all nine students weight 150 pounds, their median weight is 150, and their total weight is equal $\displaystyle 9 \times 150 = 1,350$ pounds. 

If, however, their weights are 

$\displaystyle \left \{ 125, 125, 125, 125, 150, 155, 155, 155, 155\right \}$

then the median of their weights - the middle value - is still 150 pounds, but the sum of their weights is 

$\displaystyle 125 \times 4 + 150 + 155 \times 4 = 500 + 150 + 620 = 1,270 < 1,350$

The median of eleven elements is the element in the middle when arranged in ascending order - that is, the sixth-lowest element. If the box is replaced with a 40, the data set becomes

$\displaystyle \left \{ 20, 30, 30, 40, 40, 40,40, 40, 50, 50, 60 \right \}$

making 40 the sixth element.

If the box is replaced with a value less than 40, then the lowest five values are 20, 30, 30, 40, and the new element, and the sixth element is 40.

If the box is replaced with a value greater than 40, then the lowest five values are 20, 30, 30, 40, 40, and the sixth element is 40.

No matter what, the median is 40.

In ascending order, their weights are:

$\displaystyle \left \{ 145, 153, 159,166, 172, 201\right \}$

The median is the average of the two numbers in the middle of the set, which are 159 and 166:

$\displaystyle \frac{159 + 166}{2} = \frac{325}{2} = 162.5$

The median weight is 162.5 pounds.

The median of eleven elements is the element in the middle, assuming the numbers are arranged in order.

If the box is replaced with a 40, the data set becomes

$\displaystyle \left \{ 20, 30, 30, 40, 40, 40,40, 40, 50, 50, 60 \right \}$,

making 40 the sixth element.

If the box is replaced with a value less than 40, then the lowest five values are 20, 30, 30, 40, and the median is 40.

If the box is replaced with a value greater than 40, then the lowest five values are 20, 30, 30, 40, 40, and the median is 40.

No matter what, the median is 40.

In ascending order, their weights are:

$\displaystyle \left \{ 145, 153, 159,166, 172, 201\right \}$

The median is the average of the two numbers in the middle of the set, which are $\displaystyle 159$ and $\displaystyle 166$:

$\displaystyle \frac{159 + 166}{2} = \frac{325}{2} = 162.5$

The median weight is $\displaystyle 162.5$ pounds.

Arrange the scores from least to greatest.

$\displaystyle \left \{ 61, 67, 68, 72, 73, 76, 80, 90 \right \}$

There are an even number (eight) of scores, so the median is the arithmetic mean of the middle two scores, 72 and 73. This makes the median

$\displaystyle \frac{72 + 73}{2} = 72.5$

First, order the numbers from least to greatest.

$\displaystyle 159, 456, 654, 741, 753, 852, 963$

Then, identify the middle number: $\displaystyle 741.$