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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 can`t 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 I`m 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.