Comment (but probably not solution) too long for the comment field: The series of the exponential gives an alternating series for the given function
$$e^{-x^2}=1-x^2+\tfrac12 x^2-\tfrac16x^4+\tfrac1{24}x^8+\dots+\tfrac{1}{k!}(-x^2)^k+\dots$$
By the quantitative statements of the Leibniz criterion,
$$e_{2n-1}(-x^2)\le e_{2n+1}(-x^2)\le e^{-x^2}\le e_{2n}(-x^2),$$
where $e_n$ is the $n$-th partial sum of the exponential series and the estimate is true for $x^2<2n$. This gives an absolute error of
$$\frac1{(2n)!}x^{4n}\left(1-\tfrac{x^2}{2n+1}\right)\le\left|e_{2n-1}(-x^2)- e^{-x^2}\right|\le\frac1{(2n)!}x^{4n}$$
which shows that in terms of polynomial approximation, one can not do much better than the partials of the exponential series. To get a small relative error, one now needs to chose $n$ large enough to get
$$\frac{e^{15^2}(15)^{4n}}{(2n)!}<ϵ$$
or about
$$2n(\ln(2n)-\ln(225e))>225+|\ln(ϵ)|$$
Which means that you need $n=406,...,419$ to get $ϵ=10^{-4},...,10^{-18}$
Using arithmetic rules of the exponential, one notices that $e^{-x^2}=\left(e^{-(\tfrac{x}{m})^2}\right)^{m^2}$, for instance with $m=15$, and the relative error $ϵ$ is guaranteed if the partial sum $e_n$ is taken such that the relative error of $e_n(-x^2)$ to $e^{-x^2}$ on $[-\tfrac{15}m,\tfrac{15}m]$ is smaller than $\tfrac{ϵ}{2m^2}$. With $m=15$ this requires the easier to control inequality
$$\frac{e^1}{(2n)!}<\tfrac{ϵ}{450}$$
which gives the usual precisions for $n=6,...,12$.
$e_{811}(-15^2)$ on Magma has a gross error, instead of ...e-98 it shows 8e73.
$e_{21}(-\tfrac{x^2}{225})^{225}$ has relative error 5e-19 at $x=15$.
If only basic operations are allowed, then we go back to basic divide and conquer or half-and-square in this case. Take in the above calculus $m$ a nice power of $2$, $m=16$ could work, but let's take $m=32$. Then the error estimate is still largest at the interval end, and the condition at the even larger $x=16$ is
$$\frac{e^{\frac14} \left(\frac14\right)^{2n}}{(2n)!}<\tfrac{ϵ}{2048}
\iff
\frac{\sqrt[4]e}{2^{4n-11}(2n)!}<ϵ$$
Trying out some small values leads to $n=5$ as a likely candidate, and indeed, $e_9(-\tfrac{15^2}{1024})\cdot\exp(15^2)-1\approx -1e-10$
$n=4$ with $e_7$ works as well with an actual error of about $2e-7$, with error bound of actually $1e-6$.
With less than 30 operations, replace 7 by 9 for higher accuracy:
xx=-x*x/1024;
a=xx;
e=1+a;
for k=2 to 7 do a*=xx/k; e+=a; end for;
for k=1 to 10 do e=e*e; end for;
return e
or with a division
xx=x*x/1024;
a=xx;
e=1+a;
for k=2 to 7 do a*=xx/k; e+=a; end for;
e=1/e;
for k=1 to 10 do e=e*e; end for;
return e
(a) Yes. If $X \sim \operatorname{Normal}(\mu, \sigma^2)$, then the PDF of $X$ is given by $$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$ We can also readily observe that $X$ is a location-scale transformation of a standard normal random variable $Z \sim \operatorname{Normal}(0,1)$, namely $$X = \sigma Z + \mu,$$ or equivalently, $$Z = \frac{X - \mu}{\sigma},$$ and the density of $Z$ is simply $$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty.$$ Therefore, if $m$ is the median of $X$, then the median of $Z$ is $m^* = (m - \mu)/\sigma$. But we also know that $m^*$ satisfies $$F_Z(m^*) = \int_{z=-\infty}^{m^*} f_Z(z) \, dz = \Phi(m^*) = \frac{1}{2}.$$ But since $f_Z(z) = f_Z(-z)$ for all $z$, the substitution $$u = -z, \quad du = -dz$$ readily yields $$F_Z(m^*) = -\int_{u=\infty}^{-m^*} f_Z(-u) \, du = \int_{u=-m^*}^\infty f_Z(u) \, du = 1 - F_Z(-m^*),$$ and since both of these must equal $1/2$, we conclude $F_Z(m^*) = F_Z(-m^*)$, or $m^* = -m^*$, or $m^* = 0$. From this, we recover the median of $X$: $m = \sigma m^* + \mu = \mu$.
(b) The interquartile range is equal to $q_3 - q_1$, where $q_3$ satisfies $F_X(q_3) = \frac{3}{4}$ and $F_X(q_1) = \frac{1}{4}$. Again, using the location-scale relationship to $Z$, we first find the IQR of $Z$, then transform back to get the IQR of $X$. The conditions $$\Phi(q_1^*) = \frac{1}{4}, \quad \Phi(q_3^*) = \frac{3}{4}$$ are clearly symmetric (see part a). We can look up in a normal distribution table that $\Phi(-0.67449) \approx 0.25$, or to more precision with a computer, $$q_1^* \approx -0.67448975019608174320.$$ It follows that the IQR of $Z$ is $$q_3^* - q_1^* \approx 1.3489795003921634864,$$ hence the IQR of $X$ is $$q_3 - q_1 = (\sigma q_3^* + \mu) - (\sigma q_1^* + \mu) \approx 1.3489795 \sigma,$$ and so the desired ratio is simply approximately $$0.74130110925280093027.$$ Note this quantity does not depend on the parameters. Your error is that you performed the subtraction incorrectly.
I am given the following in a worked example (see picture).
Best Answer
$d = 57 -27 =30\ne 20$
$\frac {36.5 - m}3 = \frac {22}{30}$
$36.5 - m = 2.2$
$m = 34.3$.