I have 1st quartile, median and 3rd quartile. I want to find mean and SD. I want to use Bland’s method1 but i do not have max, min values of the data. How can i solve this problem? Is there any R package?

1 Bland, Martin. 2014. “Estimating Mean and Standard Deviation from the Sample Size, Three Quartiles, Minimum, and Maximum.” International Journal of Statistics in Medical Research 4 (1): 57–64.

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    $\begingroup$ Without some knowledge of the population distribution you cannot determine the mean. If the mean exists it doesn't even have to fall between the first and third quantile. $\endgroup$ – Michael R. Chernick Jan 16 '17 at 6:44
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    $\begingroup$ When you say "quantile" do you mean "quartile"? $\endgroup$ – Nick Cox Jan 16 '17 at 7:44
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    $\begingroup$ If you have literally nothing else, then why do you think it's either necessary or useful to estimate the mean and SD? What would do with them? $\endgroup$ – Nick Cox Jan 16 '17 at 13:09
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    $\begingroup$ I recently posted data sets where the mean didn't even lay between the 10th and 90th percentiles... without additional assumptions that rule out such possibilities, this is not going to give sensible answers. $\endgroup$ – Glen_b Jan 18 '17 at 11:06
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    $\begingroup$ I'd flag the simple points that one should always (1) flag that this is at best an approximation to be treated very circumspectly (2) cite median and quartiles explicitly or make them accessible otherwise (3) use the spacing of the quartiles around the median as a check on symmetry (4) use what else is known about the measurement scale: even when minimum and maximum are not known there should be some idea of theoretical or likely limits. $\endgroup$ – Nick Cox Aug 29 '17 at 15:14

You can check Wan et al. (2014)*. They build on Bland (2014) to estimate these parameters according to the data summaries available. See scenario C3 in their paper :

$$ \bar{X} ≈ \frac {q_{1} + m + q_{3}}{3}$$

$$ S ≈ \frac {q_{3} - q_{1}}{1.35}$$

or, if you have the sample size :

$$ S ≈ \frac {q_{3} - q_{1}}{2 \Phi^{-1}(\frac{0.75n-0.125}{n+0.25}) }$$

where $q_{1}$ is the first quartile, $m$ the median, $q_{3}$ is the 3rd quartile and $\Phi^{-1}(z)$ the upper zth percentile of the standard normal distribution.

So, in R :

q1 <- 0.02
q3 <- 0.04
n <- 100

(s <- (q3 - q1) / (2 * (qnorm((0.75 * n - 0.125) / (n + 0.25)))))
#[1] 0.0150441

* Wan, Xiang, Wenqian Wang, Jiming Liu, and Tiejun Tong. 2014. “Estimating the Sample Mean and Standard Deviation from the Sample Size, Median, Range And/or Interquartile Range.” BMC Medical Research Methodology 14 (135). doi:10.1186/1471-2288-14-135.

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    $\begingroup$ No matter how you slice it these papers assume the sample comes fro a normal distribution. Then the population mean=median and of course it is between the 1st and 3rd quartiles! But nowhere did the OP assume normality. $\endgroup$ – Michael R. Chernick Jan 16 '17 at 14:08
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    $\begingroup$ They test their formulas on normal and skewed data ; but I'm no specialist... $\endgroup$ – mdag02 Jan 16 '17 at 14:35

Adding to Michael Chernick's comment, here's an example.

x <- runif(1000,0,1)
summary(x)  #1st Q = 0.27  3rd = 0.77  mean = .51

x1 <- c(x,100)
summary(x1) #1Q = 0.27  3rd = 0.77  mean = .61

x2 <- c(rnorm(100,0,1), rnorm(10,10,.1))
summary(x2)  # 1st = -.85  3rd = 0.69, mean = 0.71

With the first pair, note that a single outlier affects the mean but not the quartiles. The last example is one where the mean is larger than the 3rd quartile.

One real world case where the mean could be greater than the third quartile is income.


There is a detailed publication on this topic from Greco et al, How to impute study-specific standard deviations in meta-analyses of skewed continuous endpoints? World Journal of Meta-Analysis 2015;3(5):215-224.

The main findings of this work are that it is acceptable to approximate "missing values of mean and SD with the correspondent values for median and interquartile range".


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