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The standartization of $X$ is given by $$ Z=\frac{X-\mu}{\sigma}, $$ where $\mu=\operatorname EX$ and $\sigma^2=\operatorname{Var}X$. We have that $$ P(Z\le z)=P\biggl(\frac{X-\mu}{\sigma}\le z\biggr)=P(X\le \sigma z+\mu) $$ and we know that $X\sim\mathcal N(\mu,\sigma^2)$. Hence, $$ P(X\le \sigma z+\mu)=\frac1{\sigma\sqrt{2\pi}}\int_{-\infty}^{\sigma ...


Suppose there are $n$ flowers. The number of flowers still blossoming at time $t$ is a random variable $X(t)$, equal to $$X(t)=\sum_{f=1}^n I(\text{flower $f$ is alive at time $t$})\tag1$$ where the indicator function $I(A)$ equals one if event $A$ is true, zero otherwise. The expectation of $X(t)$ is then $$E(X(t))=n P(\text{flower $1$ is alive at time ...


Hint:$$\Phi^{-1}(U)\leq x\iff U\leq \Phi(x)$$ Here $\Phi$ denotes the CDF corresponding with standard normal distribution.


It is a one-dimensional normal distribution. This follows from an alternative definition of multi-dimensional Gaussian distribution. EDIT: see the Wikipedia article (for future reference/anyone new to the question)

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