# Can You Have 2 Numerical, Identical Values To Represent the nth Percentile?

For an assignment, I am required to find the value of a piece of data in the 70th percentile using mean and standard deviation.

Here is my organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.

The image link attached below called 'Percentiles With Normal Distribution' is the work I have done to find the value. (For some reason I cant seem to upload the actual image here...)

Percentiles with Normal Distribution

Just in case you cannot access the image, I will explain what I did.

1. I created the normal distribution graph and shaded approximately 70% of it, leaving a little part on the right side empty. I am finding b, where P(X > b) = p, where p = 0.7. I reversed this and it is now P(X > b) = 1 - p (where p = 0.7). This means to find the (1-p)Th percentile for X.

2. Next, I found the corresponding percentile for Z by looking in the body of the Z-Score Table, and finding the probability that is closest to p = 0.7, to which I did this: (1-0.7) = 0.3 so the closest value to this is 0.3015, which falls under row = -0.5 and column = 0.02. This means the 70th percentile for Z is equal to -0.52.

3. Lastly, I just changed the Z-Score value back into an x-value (original units). x = Mean + Z(Standard Deviation). I substiuted my mean value of -0.55, Z-Score value of -0.52, and standard deviation of 2.4 and solved for x. x = -2.

Now, if you look at the organized data set I provided above, youd see that there are two -2s. This is confusing me because Im not sure whether there can be two numerical values to represent the value of a certain percentile.

Three points:

1. If you look at your picture, you want $$+0.52$$ for the $$70$$th percentile of a standard normal distribution, but you used $$-0.52$$ which in fact corresponds to the $$30$$th percentile. Correcting this would have given you about $$0.7$$ rather than $$-1.8$$

2. There is no problem with a percentile corresponding to a value from the sample which happens multiple times. For example with the data 3, 5, 5, 5, 7 it is obvious that the $$50$$th percentile or median is $$5$$

3. I am not sure why you need to fit a normal distribution unless it is to fit the question's "using mean and standard deviation". As an alternative, sorting your $$20$$ data points to -4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, you have $$25\%$$ of the values below $$-2$$ and $$65\%$$ of the values above $$-2$$ so you could say $$-2$$ is the $$30$$th percentile. Similarly you have $$40\%$$ of the values below $$0$$ and $$25\%$$ above $$0$$ so you could say $$0$$ is the $$70$$th percentile.

• This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2. – Yashvi Shah Jan 4 at 3:40
• @YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations. – Henry Jan 4 at 8:03
• To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer – Henry Jan 4 at 8:31
• Im not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something? – Yashvi Shah Jan 4 at 17:47
• Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/… – Yashvi Shah Jan 4 at 20:31