Is there a distribution or can I work from another distribution to create a distribution like that in the image below (apologies for the bad drawings)?

where I give a number (0.2, 0.5 and 0.9 in the examples) for where the peak should be and a standard deviation (sigma) that makes the function wider or less wide.P.S.: When the given number is 0.5 the distribution is a normal distribution.

**Answer**

One possible choice is the beta distribution, but re-parametrized in terms of mean μ and precision ϕ, that is, “for fixed μ, the larger the value of ϕ, the smaller the variance of y” (see Ferrari, and Cribari-Neto, 2004). The probability density function is constructed by replacing the standard parameters of beta distribution with α=ϕμ and β=ϕ(1−μ)

f(y)=1B(ϕμ,ϕ(1−μ))yϕμ−1(1−y)ϕ(1−μ)−1

where E(Y)=μ and Var(Y)=μ(1−μ)1+ϕ.

Alternatively, you can calculate appropriate α and β parameters that would lead to beta distribution with pre-defined mean and variance. However, notice that there are restrictions on possible values of variance that are valid for beta distribution. For me personally, the parametrization using precision is more intuitive (think of x/ϕ proportions in binomially distributed X, with sample size ϕ and the probability of success μ).

Kumaraswamy distribution is another bounded continuous distribution, but it would be harder to re-parametrize like above.

As others have noticed, it is *not* normal since normal distribution has the (−∞,∞) support, so at best you could use the truncated normal as an approximation.

_{
Ferrari, S., & Cribari-Neto, F. (2004). Beta regression for modelling rates and proportions. Journal of Applied Statistics, 31(7), 799-815.
}

**Attribution***Source : Link , Question Author : Stan Callewaert , Answer Author : Tim*