Calculate Standard Deviation and Mean from 95th/5th Percentile and z-scores for other percentiles

Question

I need to determine the standard deviation and mean of a set of data (normal distribution), given the values of the 5th and 95th percentiles, and the z-scores for the 50th, 10th/90th, 5th/95th, 2.5th/97.5th, and 1st/99th percentiles:

5th percentile - 1640

95th percentile - 1870

z-scores:

50th - 0

10th or 90th - 1.28

5th or 95th - 1.64

2.5th or 97.5th - 1.96

1st or 99th - 2.32

How can I find the standard deviation and mean from this?

Answer 1

Hi Robert S!

This is Robert B! :) I tried to provide you with a video answer. However they are still working out some kinks. As I'm sure you know the z-scores are just data points that are shifted by the mean and normalized by the standard deviation. So, if your data value is X, then the corresponding z-score is (X-µ)/σ, where µ stands for mean and σ stands for standard deviation.

In this case, you are given data values for the 5^th and 95^th percentiles. X₅=1640 and X₉₅=1870. You are also given the z-scores associated with these two percentiles. Z₉₅=1.64. However, since the 5^th percentile is below the mean, the corresponding z-score must be negative. So, Z₅=-1.64. Substituting the data values in the formula for z-scores, we have

Z₅=-1.64=(1640-µ)/σ and

Z₉₅=1.64=(1870-µ)/σ

So, we have two equations with 2 unknowns, which (hopefully) we can solve. If we eliminate σ, we have that

-1640+µ=1870-µ or

µ=(1/2)(1640+1870) = 1755

Now that we have µ, we can use it to find σ. Using the equation for Z₉₅, we have that

1.64=(1870-µ)/σ = (1870-1755)/σ = 115/σ

Accordingly, σ=115/1.64. So σ=70.12.

So, there you have it! The mean is 1755, and the standard deviation is 70.12.

Hope this makes sense to you .

Let me know if you need some help with something else! :)

Thanks

Bob