# Find the median from this data?

$150 < t \le 200$

#### Explanation:

The median is the "middle answer" in a list of answers. For instance, if we have the numbers $1 , 2 , 6 , 7 , 9$, the median is the middle value, or 6.

For the data above, we can take the frequencies as the number of values we'll be lining up - the middle value will sit in the interval that holds the middle value. Like this:

$10 , 8 , 24 , 29 , 32 , 10$

We can add the values and divide by 2:

$\frac{10 + 8 + 24 + 29 + 32 + 10}{2} = 56.5$

The 56th/57th values sit:

$10 + 8 + 24 = 42$
$10 + 8 + 24 + 29 = 71 \leftarrow \text{in here}$

And so the interval with the median sits in the $150 < t \le 200$

Feb 11, 2018

The median is the ${57}^{t h}$ value which lies in the interval

$150 < t \le 200$

#### Explanation:

If you have an odd number of values, then the median is slightly easier to find than if you have an even number of values.

The median is the middle value in a set of data.

• Add the frequencies to find the total number of lightbulbs,

$10 + 8 + 24 + 29 + 32 + 10 = 113$

Think of $113$ as having the same number on each side, with one value in the middle: (that will be the median)

$113 \div 2 = 56 \frac{1}{2}$

$113 = \left(56 + \textcolor{b l u e}{\frac{1}{2}}\right) + \left(\textcolor{b l u e}{\frac{1}{2}} + 56\right)$

$113 = 56 \text{ "+" "color(blue)(1)" "+" } 56$

• Arrange the given frequencies to look like this.

$10 + 8 + 24 = 42$, if you now add $29$ you will have more than $56$

$10 + 8 + 24 + 29 = 71$

• Split up the $29 \rightarrow 56 - 42 = 14$ more needed

$\text{ } 10 + 8 + 24 + \textcolor{b l u e}{29} + 32 + 10$

$10 + 8 + 24 + \textcolor{b l u e}{14 + 1 + 14} + 32 + 10$

This gives us what we need:

$\textcolor{red}{\left(10 + 8 + 24 + 14\right)} \text{ " color(blue)( +1)" } \textcolor{red}{\left(+ 14 + 32 + 10\right)}$
$\textcolor{w h i t e}{w w w w w} \downarrow \textcolor{w h i t e}{w w w w w w} \downarrow \textcolor{w h i t e}{w w w w w w w w} \downarrow$
$\textcolor{w h i t e}{w w w w w} \textcolor{red}{56} \textcolor{w h i t e}{w w w w} \textcolor{b l u e}{\text{the median}} \textcolor{w h i t e}{w w w w w} \textcolor{red}{56}$

The median therefore lies in the interval which has the frequency of $29$ because we had to split up $29$ to get the equal numbers on each side.

This is $150 < t \le 200$

Note that the median is the ${57}^{t h}$ value.