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We have random data, which is exponentially distributed.

Data = exp($\lambda$), where $\lambda$ = 0.5.

If it is possible to change exponential distribution into the normal distribution. Then what would be our mean and std of the normal distribution, with prove, please?

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    $\begingroup$ You can transform any fully specified univariate continuous distribution into any other fully specified univariate continuous distribution using the approach outlined at the start of both answers here; that is, to convert to a uniform using the probability integral transform (the transformation is the cdf of the distribution you start with) and then convert the uniform to the desired distribution by transforming by the inverse cdf of the distribution you want. $\endgroup$ – Glen_b -Reinstate Monica Aug 10 '19 at 11:11
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    $\begingroup$ If $F$ is the distribution you start with and $G$ is the distribution you want, then $t = G^{-1}(F)$ is the desired transformation. See Wikipedia: Probability integral transform and Inverse transform sampling. You can prove it by relating simple probability statements in the $P(X\leq x)$ form $\endgroup$ – Glen_b -Reinstate Monica Aug 10 '19 at 11:13
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    $\begingroup$ @Glen's comments have established the possibility of such a transformation. Concerning the second part of the question, the mean and SD of the resulting normal distribution would be whatever you want them to be. After all, if $f:\mathbb{R}\to\mathbb{R}$ is a function for which $f(X)$ has a Normal$(\mu,\sigma)$ distribution whenever $X$ has an Exponential$(\lambda)$ distribution, then the function $g(x)=\tau(f(x)-\mu)/\sigma + \nu$ has a Normal$(\nu,|\tau|)$ distribution. $\endgroup$ – whuber Aug 10 '19 at 14:59
  • $\begingroup$ ... the reason I didn't simply post an answer myself is because I believe a question of the same sort has already been answered before in essentially the same generality as my comment (though some searching didn't locate what I was looking for). I was hoping to either locate such a duplicate and perhaps make sure it was sufficiently canonical for any future questions of this kind, or to edit this into a canonical question and give a canonical answer. $\endgroup$ – Glen_b -Reinstate Monica Aug 11 '19 at 11:21
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I will show how to use R to follow the plan outlined in the Comments, leaving the mathematical proofs to you.

Because a random variable $Y \sim \mathsf{Exp}(\text{rate} = \lambda = 3)$ has CDF $F_X(t) = 1 - e^{-3t}$ for $t > 0,$ the relationship $F_X(t) = U$ leads to $U = 1 - e^{-3X}.$ In R, we generate a sample of size $n = 50,000$ from $\mathsf{Exp}(3)$ and transform it to $U \sim \mathsf{Unif}(0,1).$

set.seed(1234)  # for reproducibility
n = 50000;  lam = 3
x = rexp(n, lam)
mean(x);  sd(x)
[1] 0.3311246    # aprx E(X) = 1/3
[1] 0.3297832    # aprx SD(X) = 1/3
u = 1 - exp(-3*x)
mean(u);  var(u);  1/12
[1] 0.4988082    # aprx E(U) = 1/2
[1] 0.08298005   # aprx Var(U) = 1/12
[1] 0.08333333   # 1/12

Then we use the quantile function qnorm in R, to transform the uniform vector u to a vector of $n = 50,000$ observations randomly taken from standard normal distribution. The CDF $\Phi(\cdot)$ and its inverse, the quantile function, of a standard normal distribution cannot be expressed in closed form. However qnorm uses Michael Wichura's rational approximation to $\Phi^{-1},$ which is accurate up to the double-precision arithmetic used by R.

z = qnorm(u)
mean(z);  sd(z)
[1] -0.004416826  # aprx E(Z) = 0
[1] 0.995895      # aprx SD(Z) = 1

We have shown means, SDs, and variances along the way to show that the simulation and transformations are going well. With 50,000 observations you can expect sample means to agree with population means to about 2 decimal places.

You can use a linear transformation to transform $Z$ to a normal distribution with mean and SD of your choice.

As further verification of this concept, here are histograms of the three samples, along with the density functions of their respective distributions.

enter image description here

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  • $\begingroup$ So in all cases, the calculated mean and std of normal distribution will be always $0$ and $1$ respectively. $\endgroup$ – dtc348 Aug 11 '19 at 20:26
  • $\begingroup$ Don't know what you mean by 'all cases': If u is standard uniform, then qnorm(u) is standard normal. which has population mean 0 and SD 1. That is, $U \sim UNIF(0,1)$ implies $\Phi^{-1}(U) \sim NORM(0,1).$ $\endgroup$ – BruceET Aug 11 '19 at 20:33

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