Find the median of each set of numbers.

$\left\{\mathbf{2}.\mathbf{5},\mathbf{7}.\mathbf{8},\mathbf{5}.\mathbf{5},\mathbf{2}.\mathbf{3},\mathbf{6}.\mathbf{2},\mathbf{7}.\mathbf{8}\right\}$

The median of a set of $n$numbers arranged in ascending order, is ${\left(\frac{n+1}{2}\right)}^{\text{th}}$number in the set if $n$is odd and $\frac{{\left(\frac{n}{2}\right)}^{\text{th}}\text{number}+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{number}}{2}$

if $n$is even.

The given set of numbers is $\left\{2.5,7.8,5.5,2.3,6.2,7.8\right\}$.

Counting, it is found that there are $6$ numbers in the given set. Then, $n=6$.

Arranging the given set of numbers in ascending order,

$\left\{2.3,2.5,5.5,6.2,7.8,7.8\right\}$

Put $n=6$in $\frac{{\left(\frac{n}{2}\right)}^{\text{th}}\text{number}+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{number}}{2}$to get,

$\frac{{\left(\frac{6}{2}\right)}^{\text{th}}\text{number}+{\left(\frac{6}{2}+1\right)}^{\text{th}}\text{number}}{2}$

$\frac{{\left(3\right)}^{\text{rd}}\text{number}+{\left(3+1\right)}^{\text{th}}\text{number}}{2}$

$\frac{{\text{3}}^{\text{rd}}\text{number}+4^{\text{th}}\text{number}}{2}$

The ${\text{3}}^{\text{rd}}\text{number}$and the ${\text{4}}^{\text{th}}\text{number}$in the given set after it is arranged in ascending order are 5.5 and 6.2 respectively.

Put ${\text{3}}^{\text{rd}}\text{number}=5.5$and ${\text{4}}^{\text{th}}\text{number}=6.2$in $\frac{{\text{3}}^{\text{rd}}\text{number}+4^{\text{th}}\text{number}}{2}$to get,

$\frac{5.5+6.2}{2}$

localid="1648124140350" $=\frac{11.7}{2}=5.85$

So, the median of the given set of numbers is 5.85.