# Median estimated from grouped data with a single class

Given the formula for grouped median:

$Median = L_m + \left [ \frac { \frac{n}{2} - F_{m-1} }{f_m} \right ] \times c$

Where:

• $L_m$: lower boundary of median class
• $c$ : size of the median class
• $F_{m-1}$ : cumulative frequency of the class before median class
• $f_m$ : frequency of the median class
• $n$ : size data

Example: What should the median be for the following:
- 100, 100, 100, 100, 100, 100, 100, 100, 100, 100 (a repeat of 100 ten times)?

Calculation:

Using a bin/class size of 0.5:

$L_m$ = 100

$c$ = 0.5

$F_{m-1}$ = 0*

$f_m$ = 10

$n$ = 10

100 + [(5-0)/10]*0.5

= 100.25

• If a sample has ten $100$'s the median is not informative. Many books would say the median is $100.$ (Certainly the mean and the mode are 100.) // Your 'grouping' table does not make practical sense. But if you want to apply the formula to it, you could do that. // This is not a duplicate of any page I know of, but you might benefit from reading about getting the median from sensibly grouped data in a few of the pages under 'Related' in the right margin. – BruceET Jan 27 '18 at 3:07

The assumption of this formula for estimating the median from grouped data is that the data are spread roughly uniformly throughout the interval. Clearly, this assumption is not met in your situation because all ten of the $100$'s lie at the lower endpoint of the interval. The idea of the formula is to estimate the median by interpolation, putting the estimate somewhere within the interval. In your case the estimated value $100.25$ is in the middle of the 'median interval' (the interval known to contain the median).
If you were trying to contrive a situation in which the estimate is even farther from the truth, you could put your ten $100$'s at the left end of an interval $[100, 120).$ With no other data, your estimate of the median would then be $110.$
Note: By contrast, the assumption usually made when trying to estimate the sample mean from grouped data is that each observation lies precisely at the midpoint of the interval that contains it. This idea gives rise to the formula $\bar X \approx \frac 1 n \sum_{i=1}^k f_jm_j,$ where there are $k$ intervals (usually of equal width), with midpoints $m_j$ and frequencies $f_j.$