Basic question about Normal distribution

Question

I am wondering about the following,

Suppose $X \sim N(101,4) $ and we have iid copies.

What is the $\Pr[20X \ge 2000]$

So the correct answer can be found in the basic way by noticing that

$$20X \sim N(2020,80)$$

So where is the mistake here,

$$\operatorname{Var}(20X)=20^2\operatorname{Var}(X)=20^2(4)=1600 \neq 80$$

Answer 1

It rather depends on whether the $4$ in $X \sim N(101,4)$ is the variance or standard deviation

If it is the variance (as is the usual convention) then $20X\sim N(2020,1600)$ with $1600$ as the variance of $20 X$

If $4$ is the standard deviation then $80$ is the standard deviation of $20 X$

You can check your result with $\Pr[20X \ge 2000] = \Pr[X \ge 100]$

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But if you have $20$ i.i.d. samples (rather than multiplying a single sample by $20$), and the variance of each is $4$, the the variance of the sum is $80$, so you might write $\sum\limits_{i=1}^{20} X_i \sim N(2020,80)$