What is the mode of $Z=\sqrt{X-Y}+4$? Given $X$ and $Y$.

Asked Jan 15 '15 at 14:34

Active Jan 15 '15 at 14:53

Viewed 33 times

Given that $$\begin{array}{c|lcr} \text{X} & 2 & 6 \\ \hline \text{p} & 0.3 & 0.7 \\ \end{array}$$ and $$\begin{array}{c|lcr} \text{Y} & 1 & 2 \\ \hline \text{p} & 0.1 & 0.9 \\ \end{array}$$

What is the mode of $Z=\sqrt{X-Y}+4$?

So far I have got that:

$$\begin{array}{c|lcr} \sqrt{X-Y}+4 & 4 & 5 & 6 & \sqrt{5}+4 \\ \hline \text{p} & 0.03 & 0.27&0.63&0.07 \\ \end{array}$$ (Am I right this far?)

If one would ask for mode of this random variable $Z$ would it be correct to say that mode is ${4,5,6,\sqrt{5}+4}$?

EDIT: Discard the last question, I forgot, that I have to look for value of $Z$ that has the highest probability.

I guess the mode would be $\sqrt{5}+4$, because of it's $0.63$ probability, right?

edited Jan 15 '15 at 14:53

asked Jan 15 '15 at 14:34

Kristians Kuhta

1

Except that you've mixed up which value of $Z$ has what probability, yes, you're correct (it's $Z = 6$ that has probability $0.63$, since it corresponds to $X = 6$ and $Y = 2$). – Arthur Jan 15 '15 at 14:51
Thank you, Now I got it right! – Kristians Kuhta Jan 15 '15 at 14:53

What is the mode of $Z=\sqrt{X-Y}+4$? Given $X$ and $Y$.

0 Answers0