Q67. - Algebra 2

Question

Find the median of each set of numbers.

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

Step-by-Step Solution

Answer

The median of the given set of numbers is 5.85.

Step 1 – State the formula

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.

Step 2 – List the given data

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

Step 3 – Calculate n

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

Step 4 – Arrange in ascending order

Arranging the given set of numbers in ascending order, 

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

Step 5 – Find the median

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,

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

 

$=\frac{11.7}{2}=5.85$ 

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