## Estimating the mean and standard deviation from the median and the range

While preparing the data for a meta-analysis, I run into the problem that a few of my sources did not report the outcome of interest as means and standard deviations, but rather as medians and range of values. After looking around, I found this interesting paper which derived (and validated through simple simulations), simple formulas that can be used to convert the median/range into a mean and a variance in a distribution free fashion.  With

• a = min of the data
• b = max of the data
• m = median
• n = size of the sample

the formulas are as follows:

Mean  $\bar{m} = \frac{a+2 m+b}{4} +\frac{a-2 m+b}{4 n}$

Variance  $\frac{1}{n-1} \Big(a^2+m^2+b^2+\frac{n-3}{2} \frac{(a+m)^2+(b+m)^2}{4}-n \bar{m} \Big)$

The following R function will carry out these calculations

f<-function(a,m,b,n)
{
mn<-(a+2*m+b)/4+(a-2*m+b)/(4*n)
s=sqrt((a*a+m*m+b*b+(n-3)*((a+m)^2+(m+b)^2)/8-n*mn*mn)/(n-1))
c(mn,s)
}

Edit

Surfing around arxiv, I found another paper that handles additional scenarios and proposes alternative formulas

### 5 Responses to “Estimating the mean and standard deviation from the median and the range”

1. Carl Witthoft Says:

You should mention this was tested only for “We drew samples from five different distributions, Normal, Log-normal, Beta, Exponential and Weibull. ” My impression is that these formulas are primarily intended for situations where the full dataset is not available, as the reduction in CPU load is not gigantic.