# Thread: percentile for grouped data

1. ## percentile for grouped data

I am asked to find the 60th percentile of this question , so i found that 30 x 0.6 = 18 ,

So , the 60th percentile is the 18th term ? Or 19th term ? I am confused . I know whether it's 18th or 19th , it's still the same .

I just want to get my concept correct

2. ## Re: percentile for grouped data

Your sample data are discrete, so percentiles are not unique. However, we can smooth the data and create unique percentile values called smoothed empirical percentiles.

If you have $\displaystyle n$ observations, then the $\displaystyle p$th smoothed empirical percentile is found by interpolating between the two data points that are around the $\displaystyle (n\cdot p)\text{th}$ observation.

$\displaystyle \hat{\pi}_p=[(n+1)p]\text{th observation}$

If $\displaystyle (n+1)p$ is not an integer, interpolate between the order statistics before and after the $\displaystyle [(n+1)p]\text{th observation.}$

Your data have $\displaystyle n=137$ observations.
$\displaystyle \hat{\pi}_{0.60}=[(137+1)0.6]=82.8\text{th observation}$

So the 60th smoothed empirical percentile is $\displaystyle 0.2X_{(82)} + 0.8X_{(83)}$ where $\displaystyle X_{(k)}$ is the $\displaystyle k$th order statistic.

3. ## Re: percentile for grouped data

Originally Posted by abender
Your sample data are discrete, so percentiles are not unique. However, we can smooth the data and create unique percentile values called smoothed empirical percentiles.

If you have $\displaystyle n$ observations, then the $\displaystyle p$th smoothed empirical percentile is found by interpolating between the two data points that are around the $\displaystyle (n\cdot p)\text{th}$ observation.

$\displaystyle \hat{\pi}_p=[(n+1)p]\text{th observation}$

If $\displaystyle (n+1)p$ is not an integer, interpolate between the order statistics before and after the $\displaystyle [(n+1)p]\text{th observation.}$

Your data have $\displaystyle n=137$ observations.
$\displaystyle \hat{\pi}_{0.60}=[(137+1)0.6]=82.8\text{th observation}$

So the 60th smoothed empirical percentile is $\displaystyle 0.2X_{(82)} + 0.8X_{(83)}$ where $\displaystyle X_{(k)}$ is the $\displaystyle k$th order statistic.
Sorry , i still didnt get you , can you explain further ? I have only 30 observation ,not 137

4. ## Re: percentile for grouped data

As you say, 60% of 30 is 18. In your data, 18 lies between 15 and 20, that is, between marks 4 and 5. For a more "precise" answer, you can interpolate as abender suggested: 18 is (18- 15)/(20- 15)= 3/5= 0.60 of the way between 15 and 20 so the marks are .60 of the way between 4 and 5, or 4.6.

5. ## Re: percentile for grouped data

I apologize -- I overlooked "cumulative" and assumed frequency. HallsofIvy has the idea.

6. ## Re: percentile for grouped data

Originally Posted by abender
I apologize -- I overlooked "cumulative" and assumed frequency. HallsofIvy has the idea.
No , i dont want very precise answer .

This is discrete data , how possible to get 4.6 ?

7. ## Re: percentile for grouped data

Originally Posted by HallsofIvy
As you say, 60% of 30 is 18. In your data, 18 lies between 15 and 20, that is, between marks 4 and 5. For a more "precise" answer, you can interpolate as abender suggested: 18 is (18- 15)/(20- 15)= 3/5= 0.60 of the way between 15 and 20 so the marks are .60 of the way between 4 and 5, or 4.6.
No , i dont want very precise answer .

This is discrete data , how possible to get 4.6 ?