Without using chi-square, if $M(t)= e^{500t +5000t^2}$
find $P(27060\leq (X-500)^2\leq 50240)$
Please, help

Question

Without using chi-square, if $M(t)= e^{500t +5000t^2}$
find $P(27060\leq (X-500)^2\leq 50240)$
Please, help

loser66 · Answer

@kirbykirby

loser66 · Answer

From moment generating function, we have $\mu_x =500$ and $\delta ^2=10000$
then??

perl · Answer

you cant integrate the function ?

loser66 · Answer

page113 , problem 3.3-12

zarkon · Answer

solve the inequality for X


the mgf is from a normal distribution

loser66 · Answer

but X is not just X, it is (X-500)^2, if applying $Z=\dfrac{X-\mu}{\delta}$,  is it not that using chi-sqruare??

zarkon · Answer

solve the inequality for X...then standardize it

zarkon · Answer

and it is $\sigma$ not $\delta$

loser66 · Answer

:)

zarkon · Answer

you are using the chi square if you work with $Z^2$

you are just going to work with Z

loser66 · Answer

$(P({2.7060}\leq Z^2\leq 5.0240\(P(\sqrt{2.7060}\leq Z\leq \sqrt{5.0240}$ for N (0,1)
Am I right?

zarkon · Answer

Z could be negative

for example if $$1<z^2<4$$

then $1<z<2$ and $-2<z<-1$

zarkon · Answer

not 'and' ... it's 'or'

loser66 · Answer

But they have the same probability, right?

loser66 · Answer

the same mean and variance also.

zarkon · Answer

yes...same prob...so you have to multiply by two

loser66 · Answer

I got it. Thank you. :)