The quantile or probit function, as you can see from the link (see "Computatuon"), is computed with inverse gaussian error function 
which I hope is downloadable for calculators like TI-89. Look here for instance.
The quantile or probit function, as you can see from the link (see "Computatuon"), is computed with inverse gaussian error function which I hope is downloadable for calculators like TI-89. Look here for instance.
2nd Vars (Distr)>"InvNorm" next you subtract 1-% and enter this into your Inverse Norm along with your Mean and standard deviation.
Ex: Find the third quartile Q3 which is the IQ score separating the top 25% from the others. With a Mean of 100 and a Standard Deviation of 15.
1-.25=.75 in Inv Norm (.75,100,15)=110 My answer is 110
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 :
$$ S ≈ \frac {q_{3} - q_{1}}{2 \Phi^{-1}\left(\frac{0.75n-0.125}{n+0.25}\right) }$$
where 
is the first quartile, 
the median, 
is the 3rd quartile and 
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.
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.
Possibly really dumb question, but say the average on a test was 86% with a SD of 2.3%, does that mean everyone who scored lower than 1 SD away is in 3rd quartile and everyone who scored 2 SD or lower is in 4th quartile?
Or do SD and quartiles not really correlate?
It's mathematically impossible to deduce mean or standard deviation from median/quartiles, because medians and quartiles discard most of the data on which the mean and standard deviation are based.
Example:
data frequency
0 50
1.4 4
2 50
That has a mean of 1.0 and standard deviation of 0.9. (I'm using 2 significant figures so I don't have to go into population versus sample standard deviation.)
data frequency
0 30
1.4 44
2 30
That data also has the median and quartiles the same as in your example, but now the mean is 1.2 and the standard deviation is 0.8.
data frequency
0 30
1.4 3
2 70
10000000 1
Now I've changed my maximum without changing the median or quartiles, you can see even more clearly how the median and quartiles exclude extreme data, because the mean is now 96000 and the standard deviation is 98000 (still 2 sig.fig.).
can i assume a normal distribution given that data is based on quartiles?
no you cannot
can I assume that the 25th and 75th quartile are equivalent to the limits of 50% confidence interval (CI)?
no you cannot
once I get the equivalent CIs, I could obtain the number of standard errors in a 50% CI based on a z-score for normal distribution:
se = 0.674 on a one tail and 1.348 on a two tail
so, replacing values on the formulae: sd = sqrt(104) x (0.0 - 2.0)/ 1.348 sd= -15.13
this has to be wrong since st.dev cannot be negative
is all the above correct?...
how could I now obtain the mean?
In short - you can't. The best you can say is that this distribution is asymetric. Likely it's skewed, with a tail to the left. (since 0 is farther from 1.4 than 2 is). Thus the mean is very likely smaller than 1.4, since means are sensitive to to tails of distributions and medians are not.