Not precisely.
About histograms, KDEs and ECDFs.
(1) Roughly speaking, a histogram (on a density scale so that the sum of areas of bars is unity) can be viewed as a estimate of the density function. A KDE is a more sophisticated method of density estimation. Generally speaking one cannot reconstruct the exact values of the data for either a histogram or a KDE.
(2) By contrast an empirical CDF (ECDF) retains exact information about all of the data. An ECDF is made as
follows: (a) sort the data from smallest to largest, (b) make a stair-step function that begins at 0 below the
minimum and increases by $1/n$ at each data value, where $n$ is the sample size. If $k$ values are tied then the increase is $k/n$ at the tied value.
Thus the ECDF approximates the CDF of the distribution,
with increasingly accurate approximations for samples of increasing size. Generally speaking an ECDF gives a better approximation to the population CDF than a histogram gives for the density function. (Information
is lost in binning data to make a histogram.)
[By suitable manipulation (a kind of numerical integration), information in a KDE could be used to make a function that imitates
the population CDF, but it does not use the actual data values. In my experience, this is rarely done.]
Graphical illustrations.
(1) A sample of size $n = 100$ from $$\mathsf{Gamma}(\text{shape} = \alpha = 5,\,\text{rate} = \lambda = 1/6)$$ is simulated. The figure shows a density histogram (blue bars), the default KDE from R statistical software (red curve), and the population density function (black).
set.seed(930)
x = rgamma(100, 5, 1/6)
summary(x)
hist(x, prob=T, ylim=c(0,.035),
col="skyblue2", main="n = 100")
rug(x) # tick marks below x-axis
lines(density(x), lwd=2, lty="dotted", col="red")
curve(dgamma(x, 5, 1/6), add=T)
(2) Sampling from the same distribution, we show the ECDF for a sample of size $n = 20,$ so that the
steps are easy to see.
set.seed(2019)
x = rgamma(20, 5, 1/6)
plot(ecdf(x), main="n = 20", col="blue"); rug(x)
curve(pgamma(x, 5, 1/6), add=T, lwd=2)
Best Answer
Let us denote distribution functions by $F$, and density functions by $f$. Then, \begin{align*} F_X(x) &= P(X \leqslant x)\\ &= P\left(Y \geqslant \frac{1-x}{2}\right)\\ &= 1 - F_Y\left(\frac{1-x}{2}\right)\\ &= 1 - \int_{-\infty}^{\frac{1-x}{2}} f_Y(t) dt\\ &= 1 - \int_{0}^{\frac{1-x}{2}} f_Y(t) dt - \int_{-\infty}^{0} f_Y(t) dt\\ &= \frac{1}{2} - \int_{0}^{\frac{1-x}{2}} f_Y(t) dt \end{align*} Therefore, by Fundamental Theorem of Calculus, we have : $$ f_X(x) = \frac{1}{2} \cdot f_Y\left(\frac{1-x}{2}\right) = \frac{1}{6\sqrt{2\pi}} \exp\left[-\frac{(x+7)^2}{72}\right]$$ So, $~X \sim N(-7,36)$ . Hope it helps.