DeterminingtheMeanandStandardDeviationofaDistributionUsingZ-ScoresandPercentiles.

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Question: Calculate Standard Deviation and Mean from 95th/5th Percentile and z-scores for other percentiles
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 - 164095th percentile - 1870z-scores: 50th - 0 10th or 90th - 1.28 5th or 95th - 1.64 2.5th or 97.5th - 1.96 1st or 99th - 2.32How can I find the standard deviation and mean from this?

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Answered By:

Bob B.
Tutor for Algebra, Calculus, Physics, and Electrical Engineering
More information: https://www.wyzant.com/Tutors/SC/Spar...

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Written Explanation:

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 5th and 95th percentiles. X5=1640 and X95=1870. You are also given the z-scores associated with these two percentiles. Z95=1.64. However, since the 5th 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 haveZ5=-1.64=(1640-µ)/σ andZ95=1.64=(1870-µ)/σSo, we have two equations with 2 unknowns, which (hopefully) we can solve. If we eliminate σ, we have that -1640+µ=1870-µ orµ=(1/2)(1640+1870) = 1755Now 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/σAccordingly, 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! :)ThanksBob
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