Three independent random variables having Normal distribution

[DottZakapa]

Homework Statement

Let $X_1 X_2 X_3$ be three independent random variables having Normal(Gaussian ) distribution, all with mean $\mu$=20 and variance $\sigma^2$=9. Also let $S=X_1+ X_2 +X_3$ and let $N$ be the number of the $X_i$ assuming values greater than 25.

Relevant Equations

Probability

Let $X_1 X_2 X_3$ be three independent random variables having Normal(Gaussian ) distribution, all with mean $\mu$=20 and variance $\sigma^2$=9. Also let $S=X_1+ X_2 +X_3$ and let $N$ be the number of the $X_i$ assuming values greater than 25.

$E\left[N\right]$=?

I did not understand how to proceed also in the evaluation of the Probability P[N>25]

 

[DottZakapa]

Should I consider that the total number of $X_i$ is 3, so, among these three random variables i have to check how many of them will satisfy the condition of being larger than $25$?
in such way $N$ is a binomial distribution. with first parameter$= 3$ (because the set of random variables is finite and is equal to $3$) the second parameter can be found solving:

$p= P\left[X_i>25\right]=P\left[Z>\frac {25-\mu}{\sigma}\right]=P\left[Z>1,66\right]=1-P\left[Z<1,66\right]= 1-0,9515= 0,05$

now i have all the parameters
$N\sim Bin(n=3, p=0,05)$

So

$E\left[N \right]$= $n*p$=3*0,05

 

[vela]

Looks good except $1-0.9515 = 0.0485$ not 0.05.

 

[DottZakapa]

Looks good except $1-0.9515 = 0.0485$ not 0.05.

rounding at two digits

 

[vela]

I figured, but there's no good reason to. Also, if you're going to round, it's generally best to do that as a last step, not in the middle of a calculation where the error introduced by rounding will propagate forward.

 

[DottZakapa]

I figured, but there's no good reason to. Also, if you're going to round, it's generally best to do that as a last step, not in the middle of a calculation where the error introduced by rounding will propagate forward.

I agree with you but it was explicitly requested in the text, wasn't my choice. other way there would be no match with the possible answers.

 

[haruspex]

I agree with you but it was explicitly requested in the text, wasn't my choice. other way there would be no match with the possible answers.

That is illogical.
@vela is saying you should have written 3*0.0485=0.1455 and then rounded to 0.15. As it happens, it would have made no difference, but suppose p had had the value 0.0475 instead. Your way you would still have got 0.15, but found no match in the offered answers because you should have got 0.14.

 

[DottZakapa]

That is illogical.
@vela is saying you should have written 3*0.0485=0.1455 and then rounded to 0.15. As it happens, it would have made no difference, but suppose p had had the value 0.0475 instead. Your way you would still have got 0.15, but found no match in the offered answers because you should have got 0.14.

In the following, the screenshot of the exam paper, logical or not logica, I did not write the exam.