The 16th percentile of a normal distributed variable has a value of 25 and the 97.5th percentile has a value of 40.

The 16th percentile of a normal distributed variable has a value of 25 and the 97.5th percentile has a value of 40.

Tutors, please sign in to answer this question.

The easiest way to answer this question is to set up a system of equations using the definition of the z-score. The z-score is defined as:

*z = (x - μ)/σ*

Where *x *is the value, *μ* is the mean, and *σ* is the standard diviation. For a normal distribution, the precentile can be calculated from the z-score. These value of the percentile for each z-score can be found in various tables (see:http://www.regentsprep.org/Regents/math/algtrig/ATS7/ZChart.htm). For the 15.9th percentile, the z-score is -1.0 and for the 97.7th precentile, the z-score is +2.0. Thus, we have the following system of equations:

-1.0 = (25 - μ)/σ => μ - σ = 25

+2.0 = (40 - μ)/σ => μ + 2σ = 40

Solving this system of equations, one finds the mean to be 30 and the standard diviation to be 5.

## Comments

Statistics can seem complex to many, and this (above) is a good answer. To teach yourself more about Statistic and related math areas, her's a great self-study video: http://www.khanacademy.org/math/statistics/v/statistics--standard-deviation.

Also, watch videos both before and after this one (same website) - then you will have a full understanding.