Question 1b522

Aug 11, 2016

Median= $34.1$

Explanation:

This is simply a method which needs to be followed.
Make sure you have a frequency table drawn up from the values given.

There are 180 students. (the sum of the frequencies)

The Median is the value in the middle of a set of data arranged in order.

$\frac{180}{2} = 90$

There are 2 groups of students: 90 lie below the median, 90 lie above the median.

The median is the value between the 90th and the 91st student.

In which interval does the median lie?

Count down the frequency column until you get to 90.

$10 + 60 = 70$
$70 + 20 = 90$ The median lies in the $\textcolor{g r e e n}{30 \to 40}$ interval. This is a class width of $\textcolor{m a \ge n t a}{10}$.

There are $\textcolor{b l u e}{50}$ students in that interval, 20 will fall below the median and 30 will lie above the median.

The median lies between the 20th and 21st students in the 30 to 40 interval. (color(red)(20.5))#

We assume that the marks are evenly distributed throughout the interval, so we now know how far along the interval the median will lie.

$\frac{\textcolor{red}{20.5}}{\textcolor{b l u e}{50}} \times \textcolor{m a \ge n t a}{10} = 4.1$

The median= 30 + 4.1 = 34.1