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,

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


1 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 $\Pr[20X \ge 2000] = \Pr[X \ge 100]$

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 $\sum\limits_{i=1}^{20} X_i \sim N(2020,80)$

  • 1
    $\begingroup$ I see, I think I was confused on the diffirence between multiple iid samples and a single sample considered multiple times, for example, Var(x1+x2)=Var(x1)+Var(x2) not 4Var(X) $\endgroup$
    – Quality
    Feb 13, 2019 at 0:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.