In order to find the median, the data must first be ordered. So we should write:

\(\displaystyle \left \{ 1,2,2,3,3,3,4,4,4,7 \right \}\)

When the number of values is even, the median is the mean of the two middle values. In this problem we have \(\displaystyle 10\) values, so the median would be the mean of the \(\displaystyle 5th\) and \(\displaystyle 6th\) values:

\(\displaystyle Median=\frac{3+3}{2}=3\)

The data should first be ordered:

\(\displaystyle \left \{ t, t+1, t+2, t+4, t+5, t+8 \right \}\)

When the number of values is even, the median is the mean of the two middle values. So in this problem we need to find the mean of the \(\displaystyle 3th\) and \(\displaystyle 4th\) values:

\(\displaystyle Median=\frac{(t+2)+(t+4)}{2}=\frac{2t+6}{2}=t+3\)

The mean is the sum of the data values divided by the number of values or as a formula we have:

\(\displaystyle \bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i}\)

Where:

\(\displaystyle \bar{x}\) is the mean of a data set, \(\displaystyle \sum\) indicates the sum of the data values \(\displaystyle x_{i}\) and \(\displaystyle n\) is the number of data values. So we can write:

\(\displaystyle \bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i}=\frac{(64+78+76+80+82+83+75+x)}{8}\)

\(\displaystyle \Rightarrow \bar{x}=\frac{538+x}{8}=78\Rightarrow 538+x=78\times 8\)

\(\displaystyle \Rightarrow 538+x=624\Rightarrow x=624-538=86\ in\)

In order to find the median, the data must first be ordered:

\(\displaystyle \left \{ 64,75,76,78,80,82,83,86 \right \}\)

Since the number of values is even, the median is the mean of the two middle values. So we get:

\(\displaystyle Median=\frac{78+80}{2}=79\ in\)

Arrange the eight known values from least to greatest.

\(\displaystyle \left \{ 24, 27, 34, 37, 50, 57,61, 63\right \}\)

For \(\displaystyle 50\) to be the median of the nine elements, it muct be the fifth-greatest, This happens if \(\displaystyle N \geq 50\).

The median of this nine-element set is its fifth-highest element. Of the eight known elements, the fourth-highest and fifth-highest elements are both 20. Regardless of the value of \(\displaystyle N\), 20 is the fifth-highest element of the nine.

The "stem" of this data set represents the tens digits of the data values; the "leaves" represent the units digits. 

There are 22 elements, so the median is the arithmetic mean of the eleventh- and twelfth-highest elements, which are 64 and 65, the middle two "leaves". Their mean is \(\displaystyle \left (64 + 65 \right )\div 2 = 64.5\).

To determine the median of a set of numbers, you first need to order them from least to greatest:

\(\displaystyle 79, 80, 81, 88, 93, 93, 95\)

Since there is an odd number of scores, the median is the score that falls exactly in the middle of the new list. Thus, the median is 88.

To determine the median of a set of numbers, you first need to order them from least to greatest:

\(\displaystyle 27, 34, 36, 44, 51, 55\)

Since there is an even amount of numbers, the median is determined by finding the average of the two numbers in the middle - 36 and 44.

\(\displaystyle 36+44 = 80 \div 2 = 40\)

Thus, the median is 40.

The median is the center number when the data points are listed in ascending or descending order. To find the median, reorder the values in numerical order:

\(\displaystyle 2,3,4,7,14\)

In this problem, the middle number, or median, is the third number, which is \(\displaystyle 4.\)