The Student Room Group

Quick S2 question

If we have X B(12,0.25) X ~ B(12, 0.25) and asked to find P(uo<X<u+o) P(u-o < X < u+o)
(u = mu, o = sigma)

I got
P(1.5<X<4.5) P(1.5 < X < 4.5)
I'm confused as to how that would then be P(1X4) P(1 \leq X \leq 4) , if X is greater than 1.5 why can it also be greater than or equal to 1? Wouldn't it be greater than or equal to 2?
Thanks!
Reply 1
Original post by IgorYakov
If we have X B(12,0.25) X ~ B(12, 0.25) and asked to find P(uo<X<u+o) P(u-o < X < u+o)
(u = mu, o = sigma)

I got
P(1.5<X<4.5) P(1.5 < X < 4.5)
I'm confused as to how that would then be P(1X4) P(1 \leq X \leq 4) , if X is greater than 1.5 why can it also be greater than or equal to 1? Wouldn't it be greater than or equal to 2?
Thanks!


Yes P(1.5<X<4.5)=P(2X4)P(1.5<X<4.5)=P(2 \le X \le 4)

By the way for sigma and mu just put \sigma and \mu in your LaTeX.
Reply 2
Original post by IgorYakov
If we have X B(12,0.25) X ~ B(12, 0.25) and asked to find P(uo<X<u+o) P(u-o < X < u+o)
(u = mu, o = sigma)

I got
P(1.5<X<4.5) P(1.5 < X < 4.5)
I'm confused as to how that would then be P(1X4) P(1 \leq X \leq 4) , if X is greater than 1.5 why can it also be greater than or equal to 1? Wouldn't it be greater than or equal to 2?
Thanks!


I agree with you, I think it is
Unparseable latex formula:

P(1.5&lt;X&lt;4.5)=P(2 \le X \le 4)

.
Reply 3
Original post by BabyMaths
Yes P(1.5<X<4.5)=P(2X4)P(1.5<X<4.5)=P(2 \le X \le 4)

By the way for sigma and mu just put \sigma and \mu in your LaTeX.

PRSOM

Thanks!

Original post by fGDu
I agree with you, I think it is
Unparseable latex formula:

P(1.5&lt;X&lt;4.5)=P(2 \le X \le 4)

.


Thanks!

Quick Reply

Latest