The Student Room Group

Stats - Normal Distribution-Modulus

X~N(15,4)

Find w to 2 d.p.

P(mod(X-15) < w ) = 0.9

Im confused. Any help pleeeaase
Reply 1
Expand the modulus bracket inside the probability.
Reply 2
I have, i get stuck @:

P(-(x-15)< w < (x-15) )

Next I modelled probability onto Z~N(0,1)

P(((-x-15)-15)/2) < Z < (((x-15)-15)/2) =0.9

P((x-30)/2)-P(-x/2) = 0.9
P((x-30)/2) + P( x/2) - 1 = 0.9

P( 1.2815 > w ) = 0.9

(x-30)/2 + x/2 - 1 = 1.2815

rearrage etc

x = 11.521

Plugged ths into P(mod(x-15) < w ) = 0.9

w = 3.479

The answer is 3.29

And im not entirely sure you can do half the steps i just used
Reply 3
Look at it this way, Let Y be a normal random variable such that, Y = X - 15, now Y~N(0,4)

And you want P(|y|<w); This means that the only area you are not looking for is 1-0.9 = 0.1; but since the normal is symmetric you have 0.05 of that on each side.

So now look up the Z value for a probability of 0.05 and plug it in to the standardization equation: Z = (w - 0)/2. This obtains the answer.
I still do not understand.

Can anyone please go through this question:

If the random variable X is distributed as N(5,4), calculate:

P(modulus(X -5) > 3 ).


Thanks
Original post by TheNightmare
I still do not understand.

Can anyone please go through this question:

If the random variable X is distributed as N(5,4), calculate:

P(modulus(X -5) > 3 ).


Thanks


l,

ok so this is the same as p (x-5>3)+p(x-5<-3)

so its basically p(x>8)+p(x<2)

Still want help with finding w?
Original post by TheNightmare
I still do not understand.

Can anyone please go through this question:

If the random variable X is distributed as N(5,4), calculate:

P(modulus(X -5) > 3 ).


Thanks


This is basically:
P((x - 5) > 3) + P((x - 5) < -3)

= P(x > 8) + P(x < 2)
Original post by falcon pluse
l,

ok so this is the same as p (x-5>3)+p(x-5<-3)

so its basically p(x>8)+p(x<2)

Still want help with finding w?


Thanks a lot. I think I got the hang of it now!

So the general rule is:

P(modulus(X-y) > q) = P(X-y > q ) + P(X-y < -q) right?

What if it was like this:

P(modulus(X-y) < q) would this be = P(X-y < q ) + P(X-y > -q) ?
Original post by MathematicsKiller
This is basically:
P((x - 5) > 3) + P((x - 5) < -3)

= P(x > 8) + P(x < 2)


What is the answer?

According to my text book it is: 0.0668

but i'm somehow getting : 0.1336

What am I doing WRONG???
Original post by TheNightmare
What is the answer?

According to my text book it is: 0.0668

but i'm somehow getting : 0.1336

What am I doing WRONG???


Are you sure you posted the question correctly?
I get the same answer as you.

I believe the answer in the text book is the answer you would get if there were no modulus.
(edited 12 years ago)
Original post by MathematicsKiller
Are you sure you posted the question correctly?
I get the same answer as you.

I believe the answer in the text book is the answer you would get if there were no modulus.


(Modulus is Bold and Underlined )

Yes the question is P( X-5 > 3).

So is the text book answer wrong?
(edited 12 years ago)
Original post by falcon pluse
l,

ok so this is the same as p (x-5>3)+p(x-5<-3)

so its basically p(x>8)+p(x<2)

Still want help with finding w?


What did you get as an answer?

According to my text book the answer is: 0.0668

but I am getting : 0.1336

Is the text book answer wrong?
I miss statistics :frown:
Original post by littleone271
I miss statistics :frown:


Will you be able this question for me:

If the random variable X is distributed as N(5,4), calculate:

P((X -5) > 3 ).

Modulus is in BOLD and is Underlined

Thanks :smile:
Original post by TheNightmare
Will you be able this question for me:

If the random variable X is distributed as N(5,4), calculate:

P((X -5) > 3 ).

Modulus is in BOLD and is Underlined

Thanks :smile:


I used to be good at it and I did an AS in pure statistics but that was a couple of years ago and I havn't really done it properly since so I can't remember how to do it :frown:... Sorry...

I remember this book being pretty amazing though because it's got worked examples and everything in it. I had this one for s1b and the s2 and s3 ones and they were all really good.

http://www.amazon.co.uk/Advancing-Maths-AQA-Statistics-2nd/dp/0435513389/ref=sr_1_1?s=books&ie=UTF8&qid=1331519123&sr=1-1
Original post by littleone271
I used to be good at it and I did an AS in pure statistics but that was a couple of years ago and I havn't really done it properly since so I can't remember how to do it :frown:... Sorry...

I remember this book being pretty amazing though because it's got worked examples and everything in it. I had this one for s1b and the s2 and s3 ones and they were all really good.

http://www.amazon.co.uk/Advancing-Maths-AQA-Statistics-2nd/dp/0435513389/ref=sr_1_1?s=books&ie=UTF8&qid=1331519123&sr=1-1


Ahh, I'm doing Statistics 1 with Edexcel not AQA. But never mind, thanks anyway. :smile:
Can anyone answer this?

X~N(5,4) -------------------------------- Calculate: P( |X-5| > 3 ).

Quick Reply

Latest