It seems that you "know" or otherwise assume that you have two quantiles; say you have that 42 and 666 are the 10% and 90% points for a lognormal.
The key is that almost everything is easier to do and understand on the logged (normal) scale; exponentiate as little and as late as possible.
I take as examples quantiles that are symmetrically placed on the cumulative probability scale. Then the mean on the log scale is halfway between them and the standard deviation (sd) on the log scale can be estimated using the normal quantile function.
I used Mata from Stata for these sample calculations. The backslash \ joins elements column-wise.
mean = mean(ln((42 \ 666)))
(ln(666) - mean) / invnormal(0.9)
1.078232092
SD = (ln(666) - mean) / invnormal(0.9)
The mean on the exponentiated scale is then
exp(mean + SD^2/2)
299.0981759
and the variance is left as an exercise.
(Aside: It should be as easy or easier in any other decent software. invnormal() is just qnorm() in R if I recall correctly.)
For positive data $x_1, x_2, \ldots, x_n$ let $y_i = \log(x_i)$ be their natural logarithms. Set
$$\bar{y} = \frac{1}{n}(y_1+y_2+\cdots + y_n)$$
and
$$s^2 = \frac{1}{n-1}\left((y_1 - \bar{y})^2 + \cdots + (y_n - \bar{y})^2\right);$$
these are the mean log and variance of the logs, respectively. The UMVUE for the arithmetic mean when the $x_i$ are assumed to be independent and identically distributed with a common lognormal distribution is given by
$$m(x) = \exp(\bar{y}) g_n\left(\frac{s^2}{2}\right)$$
where $g_n$ is Finney's function
$$g_n(t) = 1 + \frac{(n-1)t}{n} + \frac{(n-1)^3t^2}{2!n^2(n+1)} + \frac{(n-1)^5t^3}{3!n^3(n+1)(n+3)}+\frac{(n-1)^7t^4}{4!n^4(n+1)(n+3)(n+5)} + \cdots.$$
For the data in the question, $s^2 = 1.23594$, $g_4(s^2/2) = 1.532355$, and the UMVUE is $m(x) = 0.084519.$
Because this might take a while to converge when $s^2/2 \gg 1$, it is best implemented as an Excel macro. Such power series are straightforward to program efficiently: just maintain a version of the current term and at each step update it to the next term and add that to a cumulative sum. The term values will typically rise and then fall again; stop when they have fallen below a small positive threshold. (For less floating point error, first compute all such terms and then sum them from smallest to largest in absolute value.)
My version of this macro (in very plain vanilla VBA) follows.
'
' Finney's G (Psi) function as in Millard & Neerchal, formula 5.57
' or equivalently in Gilbert, formula 13.4 (m here = n-1 there).
'
' Typically, m is a positive integer. Z can be positive or negative.
'
' Programmed by WAH @ QD 5 March 2001
'
' This algorithm will be less accurate for large m*z. It could be replaced by
' one that separately computes the descending half of the terms,
' iterating backward over i.
'
' It can be badly inaccurate for very negative m*z.
'
' This function returns 0 (an impossible value) upon encountering
' an input error.
'
Public Function Finney(m As Integer, z As Double) As Double
Dim i As Integer ' Index variable
Dim g As Double ' Result
Dim x As Double ' z * m * m / (m+1)
Dim a As Double ' Power series coefficient
Dim iMax As Integer ' Maximum iteration count
Const aTol As Double = 0.0000000001 ' Convergence threshold
Const iterMax As Integer = 1000 ' Limits execution time
If (m <= -1) Then
' issue an error
Finney = 0#
End If
x = z * m * m / (m + 1)
If (Abs(x) < aTol) Then
Finney = 1# ' This is the correct answer.
Exit Function
End If
iMax = Abs(Int(z) + 1) + 20
If (iMax > iterMax) Then
' issue an error
Finney = 0#
Exit Function
End If
'
' Initialize
'
a = 1#
g = a ' Lead terms
For i = 1 To iMax
'
' Test for convergence
'
If (Abs(a) <= aTol * Abs(g)) Then
Exit For
End If
'
' Compute the next term
'
a = a * x / (m + 2 * (i - 1)) / i
'
' Accumulate terms
'
g = g + a
Next
Finney = g
End Function
References
Gilbert, Richard O. Statistical Methods for Environmental Pollution Monitoring. Van Nostrand Reinhold Company, 1987.
Millard, Steven P. and Nagaraj K. Neerchal, Environmental Statistics with S-Plus. CRC Press, 2001.
Appendix
For those using a vectorized implementation it pays to precompute the coefficients of $g_n$ in advance for a given value of $n$. This can also be exploited to determine in advance how many coefficients will be needed, thereby avoiding almost all the comparison operations. Here, as an example, is an R implementation. (It uses the equivalent Gamma-function formula of http://www.unc.edu/~haipeng/publication/lnmean.pdf after correcting a typographical error there: the power series argument should be $(n-1)^2t/(2n)$ rather than $(n-1)t/(2n)$ as written.)
finney <- function(t, n, eps=1.0e-20) {
u <- t * (n-1)^2 / (2*n)
tau <- max(u)
i.max <- ceiling(max(1, -log(eps), 1 + log(tau)/2))
a=lgamma((n-1)/2) - (lgamma(1:i.max+1) + lgamma((n-1)/2 + 1:i.max))
b <- exp(a[a + log(tau) * 1:i.max > log(eps)]) # Retain only terms larger than eps
x <- outer(u, 1:length(b), function(z,i) z^i) # Compute powers of u
return(x %*% b + 1) # Sum the power series
}
For example, finney(1.2359357/2, 4) produces the value $1.532355$. This implementation can compute a million values per second for $n=3$ and about $400,000$ values per second for $n=300$. As another example of its use, here is a plot of $g_4, g_8, g_{16}, g_{32}$. (The higher graphs correspond to larger values of $n$.)
par(mfrow=c(1,1))
curve(finney(x/2, 32), 0, 2, lwd=2, main="Finney g(t/2)", xlab="t", ylab="")
curve(finney(x/2, 16), add=TRUE, lwd=2, col="#2040c0")
curve(finney(x/2, 8), add=TRUE, lwd=2, col="#c02040")
curve(finney(x/2, 4), add=TRUE, lwd=2, col="#40c020")
Best Answer
Chebyshev and similar +/- one sigma intervals refer to arithmetic mean and std, not to geometric ones. If you think your $X$ is lognormal, then construct an "arithmetic" -/+ one $\sigma$ interval for $log(X)$, then exponentiate it to get the CI for $X$.
In other words, you have zero chance of constructing an interval with the expected coverage the way you did because for lognormal such interval should not be symmetric around the mean. After you exponentiate, you'll see that the CI became asymmetric wrt the arithmetic mean of $X$.